START WITH and CONNECT BY PRIOR

Hello

Database: Oracle 11g

SELECT ename, empno, mgr

WCP

START WITH empno = 7839

CONNECT BY PRIOR MGR = EMPNO

Result set:

EMPNO, ENAME MGR

-------- ---------- ----------

KING 7839

SELECT empno.ename, Bishop

WCP

START WITH mgr = 7839

CONNECT BY PRIOR MGR = EMPNO

Result set:

EMPNO, ENAME MGR

-------- ---------- ----------

7566 JONES 7839

7698 BLAKE 7839

7782 CLARK 7839

KING 7839

My questions are:

Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839

Q2. What is the difference between

CONNECTION BY MGR PRIOR = EMPNO and

CONNECT BY PRIOR EMPNO = MGR?

can someone please help me here?

Thank you

Hello

A CONNECT BY query looks like an operation UNION ALL of the data of different levels, numbered 1, 2, 3 and more.

Level 1 is filled with the START WITH clause.

If there are data on level N, then N + 1 level is filled, using the CONNECT BY clause, which generally refers to something on the N level via the PRIOR operator. Another way to put it is that level N + 1 is filled by a self-join with lines that have already chosen the level N.

If there is no data on the level of N, the query stops.

Let's see how this applies to your queries.

Level being such an important concept in CONNECT BY queries, you might want to see in all your CONNECT BY queries all test and debug the.

1 query, including the level are:

SELECT ename, empno, mgr

LEVEL

FROM scott.emp

START WITH empno = 7839

Empno = mgr PRIOR CONNECTION

;

You will notice that I have re-arranged the CONNECT BY clause. I find it a little more clear medium. Of course, it never changes the results just if you say "x = y" or "y = x.

The results, including the level, are:

LEVEL OF ARCHBISHOP EMPNO, ENAME

---------- ---------- ---------- ----------

7839 KING 1

What happened to produce these results?

First level 1 has been met using the START WITH clause. Level 1 is identical to the results of

SELECT ename, empno, mgr

AS LEVEL 1

FROM scott.emp

WHERE empno = 7839 - same as your START WITH clause

;

It happens to be only 1 row in the table scott.emp who met the empno = 7839 condition, and we show a few columns of this line.

That's all that need the level 1. Something has been on level 1, so we're trying now to complete level 2, using the CONNECT BY condition.

Any line that is included in the level 2 meets the empno = mgr PREREQUISITE condition, where the PREVIOUS operator refers to a line of level 1. In this case, there is only 1 row at level 1, this line gets to have a NULL mgr. Given that PRIOR mgr is NULL in this case, the condition to connect BY

EmpNo = mgr BEFORE equals

EmpNo = NULL and who obviously won't be true for any line, so nothing is added to level 2, and ends the query.

Now let's look at application 2. I'll add another column of debugging, called path, which I'll describe later:

SELECT ename, empno, mgr

LEVEL

, SYS_CONNECT_BY_PATH (empno, "/") AS path

FROM scott.emp

START WITH mgr = 7839

CONNECT BY PRIOR Mgr = empno

LEVEL CONTROL

path

;

Output:

EMPNO, ENAME MGR LEVEL PATH

---------- ---------- ---------- ---------- ---------------

7566 7839 1 7566 JONES

7698 7839 1 7698 BLAKE

7782 7839 1 7782 CLARK

7839 KING 2/7566/7839

7839 KING 2/7698/7839

7839 KING 2/7782/7839

Again, we'll study how people got 1 level. It happens to be 3 scott.emp lines that meet this condition START WITH, so there are 3 lines in the game at level 1.

Given that the data on the level 1, the test of the query to complete level 2, referring to some PRIOR line on level 1. Any line that meets the condition to connect BY, with a line any level 1 in the PREVIOUS line, will appear at level 2.

Let's look at the line at level 1 where ename = 'JONES '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "JONES"? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

Let's look at the line at level 1 where ename = 'BLAKE '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "BLAKE"? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

Let's look at the line at level 1 where ename = 'CLARK '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with 'CLARK '? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.

There are thus 3 rows at level 2. They happen to all be on the same line of the table emp; It is correct. Remember, CONNECT BY is like a UNION ALL (not just a UNION). It is a UNION of

lines that are at level 1, because him meets the condition to BEGIN WITH, and

lines that are at level 2 because puts it CONNECT BY condition regarding the 'JONES', and

lines that are at level 2, because they meet the condition to connect BY regarding the "BLAKE", and

lines that are at level 2, because they meet the condition to connect BY regarding the "CLARK".

SYS_CONNECT_BY_PATH can enlighten us on that. SYS_CONNECT_BY_PATH (empno, ' / ') shows the empno of each level that caused this line appears in the result set. It's a delimited list /, where the nth element (i.e. the empno after bar oblique nth) is the empno who found the N level.

Since there were data at level 2, the quert now trying to complete level 3.

Is there all the rows in the table that satisfy the CONNECT BY condition (mgr PRIOR = empno) with respect to any line level 2? No, Bishop is be NULL on all lines of level 2, so no line can satisfy this condition CONNECT BY, no lines are added at level 3, and ends the query.

I hope that answers the question:

Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839

I'll try to not be so detailed answering

Q2. What is the difference between

CONNECTION BY MGR PRIOR = EMPNO and

CONNECT BY PRIOR EMPNO = MGR?

These 2 CONNECT BY conditions are different where you put the PRIOR operator. The operator PRIOR to switching as it change the direction, upward or downward, which move you through the tree you get from level to level.

Bishop PRÉALABLE = empno means the employee on level N + 1 will be the same as the Manager of level N. This means that higher level numbers will be the most senior people in the hierarchy. This is called a query from the bottom up. (Both of the queries that you have posted this same CONNECT BY exact state; both are requests from bottom to top).

Mgr = empno PREREQUISITE or equivalent

PRIOR empno = mgr means exactly the opposite. When you move from level N to level N + 1 in the query, you will move to an older person, to a junior position in the hierarchy. This is called a query from top to bottom. The employee level N will be the Manager of wover is a level N + 1.

Tags: Database

Similar Questions

START WITH and CONNECT BY in Oracle SQL (hierarchical)

Hi, the original table as below

Customer_ID         Account_ID          Paying_Account_ID         Parent_Account_ID          Company_ID
158                    158                    158                         158                     0
159                    159                    158                         158                     0
160                    160                    158                         158                     0
181                    181                    181                         181                     0
183                    183                    183                         183                     0
24669                  24669                  24669                       24669                   0         
24671                  24671                  24671                       24669                   0
24670                  24670                  24670                       24669                   0     
3385127                3385127                3385127                     24670                   0

To identify the hierarchical relationship of the data, which are PARENT_ACCOUNT_ID & ACCOUNT_ID, here's the query I used.

 select  lpad(' ', 2*level) || A.ACCOUNT_ID AS LEVEL_LABEL, CONNECT_BY_ISCYCLE "Cycle", LEVEL, A.* from ACCOUNT A
START WITH parent_account_id = account_id
CONNECT BY NOCYCLE  PRIOR A.ACCOUNT_ID = A.PARENT_ACCOUNT_ID
AND account_id != parent_account_id
;

It is the result of the query

Level_Label              Level          Cycle        Customer_ID             Account_ID        Paying_Account_ID      Parent_Account_ID      Company_ID
158                         1             0              158                     158              158                   158                     0
   159                      2             0              159                     159              158                   158                     0
   160                      2             0              160                     160              158                   158                     0
181                         1             0              181                     181              181                   181                     0
183                         1             0              183                     183              183                   183                     0
24669                       1             0              24669                   24669            24669                 24669                   0       
    24671                   2             0              24671                   24671            24671                 24669                   0
    24670                   2             0              24670                   24670            24670                 24669                   0
        3385127             3             0              3385127                 3385127          3385127               24670                   0

My question is how can I changed the query to calculate the values for:

My_Total_PR - number of my accounts to child PR which do not include himself.
Total_PR - Total number of accounts PR in the overall structure
My_Total_NPR - number of my accounts of child NPR which do not include himself.
Total_NPR - Total number of accounts NPR in the overall structure

* PR stand for responsible for payment, for example the responsible account payment 158 158 (Paying_Account_ID), therefore the Total_PR to 158 is 3 (158, 159, 160)
* NPR stand responsible for Non-payment, for example the responsible account payment 159 is 158 (Paying_Account_ID), so the Total_NPR for 159 1

This is the expected result, any advice appreciated. Thank you

Level_Label                     Level           Cycle           My_Total_PR     Total_PR     My_Total_NPR     Total_NPR     Paying_Account
158                               1                0                  2              3          0              0              158
    159                           2                0                  0              0          0              1              158
    160                           2                0                  0              0          0              1              158
181                               1                0                  0              1          0              0              181
183                               1                0                  0              1          0              0              183
24669                             1                0                  0              1          3              3              24669                   
    24671                         2                0                  0              1          0              0              24671
    24670                         2                0                  0              1          1              1              24670
        3385127                   3                0                  0              1          0              0              3385127

Published by: user11432758 on February 14, 2012 01:00

Published by: user11432758 on February 14, 2012 07:05

Hello

user11432758 wrote:
Hi here is the statement DDL, thank you
CREATE TABLE "SYSTEM"."ACCOUNT" ...

Do not create your own objects in the diagram of the SYSTEM or any scheme that comes with the database. Create a separate schema and place your items. You'll have fewer security problems, and the migration to a new database will be easier.

Here's a way to can get the aggregates you want:

WITH     got_descendants          AS
(
     SELECT     CONNECT_BY_ROOT account_id     AS ancestor_id
     ,     paying_account_id
     ,     LEVEL                    AS lvl
     FROM     account
     CONNECT BY NOCYCLE     PRIOR account_id     = parent_account_id
          AND          account_id          != parent_account_id
)
SELECT       ancestor_id
,       COUNT (CASE WHEN lvl             > 1
                  AND  ancestor_id  = paying_account_id THEN 1 END)     AS my_total_pr
,       COUNT (CASE WHEN ancestor_id  = paying_account_id THEN 1 END)     AS total_pr
,       COUNT (CASE WHEN lvl             > 1
                  AND  ancestor_id != paying_account_id THEN 1 END)     AS my_total_npr
,       COUNT (CASE WHEN ancestor_id != paying_account_id THEN 1 END)     AS total_npr
FROM       got_descendants
GROUP BY  ancestor_id
;

Output:

`             MY_         MY_
            TOTAL TOTAL TOTAL TOTAL
ANCESTOR_ID   _PR   _PR  _NPR  _NPR
----------- ----- ----- ----- -----
        158     2     3     0     0
        159     0     0     0     1
        160     0     0     0     1
        181     0     1     0     0
        183     0     1     0     0
      24669     0     1     3     3
      24670     0     1     1     1
      24671     0     1     0     0
    3385217     0     1     0     0

This gives the correct numbers, but how can bring us in an order that reflects the hierarchy, with the columns (for example lvl) that come from the hierarchy?
A solution would be to make two CONNECT BY queries; a service without START WITH clause (like the one above) who collects the aggregates and the other with a START WITH clause (as your original query), which is in the right order and columns such as level_label and level. We could join result sets and get exactly what we want. I'll leave that as an exercise.

Here is another way, which gets good results with only one CONNECTION PER request:

WITH     got_descendants          AS
(
     SELECT     CONNECT_BY_ROOT account_id     AS ancestor_id
     ,     paying_account_id
     ,     account_id
     ,     LEVEL                    AS lvl
     ,     CONNECT_BY_ISCYCLE          AS cycle
     ,     CASE
              WHEN  CONNECT_BY_ROOT account_id
                  = CONNECT_BY_ROOT parent_account_id
              THEN  ROWNUM
          END                    AS r_num
     FROM     account
     CONNECT BY NOCYCLE     PRIOR account_id     = parent_account_id
          AND          account_id          != parent_account_id
     ORDER SIBLINGS BY     account_id     -- Optional
)
,     got_o_num     AS
(
     SELECT     got_descendants.*
     ,     MIN (r_num) OVER (PARTITION BY  account_id)     AS o_num
     ,     MAX (lvl)   OVER (PARTITION BY  account_id)      AS max_lvl
     FROM     got_descendants
)
SELECT       LPAD ( ' '
            , 2 * (MIN (max_lvl) - 1)
            )  || ancestor_id                         AS level_label
,       MIN (max_lvl)                                AS "Level"
,       MIN (cycle)                                   AS "Cycle"
,       COUNT (CASE WHEN lvl             > 1
                  AND  ancestor_id  = paying_account_id THEN 1 END)     AS my_total_pr
,       COUNT (CASE WHEN ancestor_id  = paying_account_id THEN 1 END)     AS total_pr
,       COUNT (CASE WHEN lvl             > 1
                  AND  ancestor_id != paying_account_id THEN 1 END)     AS my_total_npr
,       COUNT (CASE WHEN ancestor_id != paying_account_id THEN 1 END)     AS total_npr
,       MIN (paying_account_id)                                    AS paying_account
FROM       got_o_num
GROUP BY  ancestor_id
ORDER BY  MIN (o_num)
;

Output:

`                             MY_         MY_
                            TOTAL TOTAL TOTAL TOTAL  PAYING_
LEVEL_LABEL     Level Cycle   _PR   _PR  _NPR  _NPR  ACCOUNT
--------------- ----- ----- ----- ----- ----- ----- --------
158                 1     0     2     3     0     0      158
  159               2     0     0     0     0     1      158
  160               2     0     0     0     0     1      158
181                 1     0     0     1     0     0      181
183                 1     0     0     1     0     0      183
24669               1     0     0     1     3     3    24669
  24670             2     0     0     1     1     1    24670
    3385217         3     0     0     1     0     0  3385217
  24671             2     0     0     1     0     0    24671

That's exactly what you asked for, except that you have posted the line with level_label =' 24671' before the line with level_label = "24671 '. You may not care about who comes first, but if it's important, explains why these lines should be in descending order of account_id, while "159 and 160" are in ascending order. You will need change the ORDERBY brothers and SŒURS clause accordingly.

Join the two trees connect by prior Start With and return only common records?
Oracle 10g Release 2 (10.2)

I have two tables which have structured data. The results, when running queries individually are correct, but I need to join tree a tree two to get only the common records between them.

-Trees a
SELECT ip_entity_name, entity_code, hier_level, entity_parent
Of ip_hierarchy
WHERE hier_level > = 3
CONNECT BY PRIOR Entity_code = entity_parent
START WITH entity_code = "MEWWD";

-The two tree
SELECT ip_entity_name, entity_code, hier_level, entity_parent
Of ipt_hierarchy
WHERE hier_level > = 3
CONNECT BY PRIOR Entity_code = entity_parent
START WITH entity_code = "IPNAM";

If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?

A WITH clause is an option?

If possible, I don't want to put a selection in a database to display object and join against other queries.

Thank you.
Hello

jtp51 wrote:
Oracle 10g Release 2 (10.2)
...
If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?

Before Oracle 9 it was true. Since you're using Oracle 10, you can if you wish; but I'm guessing that you don't want in this case.

A WITH clause is an option?

If possible, I don't want to put a selection in a database to display object and join against other queries.

Yes, a WITH clause that is an option. Viewed online, as Zhxiang has shown, are another option. Either way gives you a regular display effect without creating a database object.

You did not show a sample and the results, so no one can tell if a join is really what you want. Other possibilities include INTERSECT or an IN subquery.

Hierarchical connect by and start with and joined?

I have an Employees table and a table of identifiers. The table of identifiers is hierarchical, with parents and children. Each employee has one or more identifiers, but that a unique identifier is considered to be the "primary" or root for each employee identifier. Unfortunately, the employee table can point to one of the children identifier lines and not the root. I need a quick query to reach employees with their most recent ID (root).

Here's the code to define the problem.

create table employees (employeeid varchar2(8), fakeNationalID varchar2(9), empname varchar2(30));
insert into employees (employeeid, fakeNationalID, empname) values (1,'001000001','John Smith');
insert into employees (employeeid, fakeNationalID, empname) values (2,'002000002','James Jones');

create table realids (realidkey NUMBER, fakeNationalID VARCHAR2(9) not null, 
   realNationalID VARCHAR2(9) UNIQUE, parent_realidkey number);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
   (1,'001000001','111111111',3);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
   (2,'002000002','222222222',null);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
   (3,'003000003','333333333',null);
commit;   

create or replace function get_parentid (fakeID in VARCHAR2) return varchar2 is
   tempid VARCHAR2(9);
   begin
      select realNationalID into tempid 
         from (
           select realNationalID, fakeNationalID
              from realids
              start with fakeNationalID = fakeID
              connect by nocycle prior parent_realidkey = realidkey
              order by level desc)
              where rownum = 1;
      return tempid;
      exception 
         when NO_DATA_FOUND then
            return NULL;
         when others then raise;
    end;

    
select get_parentid('001000001') from dual; -- returns 333333333 because its linked to a parent
select get_parentid('002000002') from dual; -- returns 222222222 because there is only one child
select get_parentid('003000003') from dual; -- returns 333333333 because it is the parent

What I want is to put the parent node above realids for each line of employees...

It works, but it is NOT very effective:

select employeeid, get_parentid(fakeNationalID) realid, empname from employees;
employeeid   realid       empname
----------   -----------  ------------
1            333333333     John Smith
2            222222222     James Jones

You can imagine what it would be like with 100K lines or more. It takes about 3 minutes to run.

It seemed like a good way to do it, but with a sub query.

select e.employeeid, e.fakenationalid, e.empname, sub.realnationalid
   from employees, 
      (select realidkey, fakenationalid, realnationalid, parent_realidkey
         from realids r
         start with r.fakenationalid = e.fakenationalid
         connect by prior r.parent_realidkey = r.realidkey) sub

Unfortunately, it produces an invalid identifier on e.fakenationalid (in the beginning with the clause).

Anyone has any ideas on how to get top most parent node of the realids for each row in the employees table? In real life, there are 6 or more employees tables across multiple remote instances of what children in the realids table and how much to the parents. We always want the highest parent of the page realid. Any help would be appreciated.

Hello

Thanks for posting the sample data in a convenient form!
It is always useful to post your version of Oracle, too, especially when it comes with CONNECT BY queries.

What follows is what you asked for in Oracle 10:

WITH     got_roots   AS
(
     SELECT     CONNECT_BY_ROOT     fakenationalid     AS leaf_id
     ,     realnationalid
     FROM     realids
     WHERE     CONNECT_BY_ISLEAF     = 1
     START WITH      fakenationalid IN ( SELECT  fakenationalid
                                          FROM    employees
                           )
     CONNECT BY     realidKEY     = PRIOR parent_realidkey
)
SELECT     e.employeeid
,     r.realnationalid
,     e.empname
FROM     employees     e
JOIN     got_roots     r     ON     r.leaf_id     = e.fakenationalid
;

In any query, call a function defined by the user for each line is going to be slow. Fortunately, Oracle now has the built-in functions and operators that can take the place of get_parentid. The CONNECT_BY_ROOT operator, which was introduced in Oracle 10, is the key to the problem. In Oracle 9, you can get the same results using SYS_CONNECT_BY_PATH.

It is generally faster to CONNECT BY query separately, and then join some other tables you need for results.

You had a good idea in your last query. The problem was that void and employees were equal tables in the FROM clause, and you cannot establish a correlation between equals. You can only correlate a subquery to its Super application. You could make to this general idea work by changing void in a scalar sub-requete, which can be connected to the employees, but I think it would be much less effective than what I posted above.

SatelliteU300 will not start with power connected and fully charged battery

I just bought a U300-13U. It works fine when charging the battery and AC is connected. However, if left to recharge completely it does not start with connected AC. Unpluging the power or remove the battery and running on AC make it work very well. Is a 'feature' to stop an overload or overheating of the battery, or just a problem with this machine. I want to run with AC and battery than both connected all the time like I did with all my other laptops.

Any help appreciated.

Hello Andy

Of my friends has this small U300 and everything works well. I got it a few weeks ago to test the family Windows XP Setup and it works perfectly.

Please test again very carefully and if the same thing happen again once I recommend you contact the Service partner in your country. They check your laptop. I have to be honest and say that describes the problem of strange sounds to me and please, don't wait too long. Contact the service as soon as possible.

Bye and good luck!

BlackBerry 8250 new smartphones does not start with the connected SIM card

I'm having a problem with my new Blackberry of Virgin Media (mobile). I just got it and when I have a connected SIM card I only get the loading screen timer and the blackberry does not start. If I have no connected SIM card the device starts in the operating system, but was told that there is just no SIM connected (obviously). Once I put the SIM card in again once it is stuck on the loading screen. As you probably already figured out that I did many resets hard with the battery pull and it has not solved the problem. A consultant because it would be greatly appreciated because the only advise I got support was remove the boot of sim and put it back and start again duh!

Carrier - Virgin Media / Vodafone
Model - 8520
Version of the OS - 4.6.1
Free space - 130679088 bytes

BT device - I have the details as HDW - 22736-002 Rev 1 Ver 2 4509 (is that correct?)

Apps & free space - nothing installed new device installation is not complete.
Blackberry 4.6.1 Basic application
Basque 4.6.1
Galician 4.6.1
softwqare system BlackBerry 4.6.1
1.0.1.20 Application Center
Attachment Service 4.6.1
4.6.1 cards
Messenger 4.6.0
BrickBreaker 4.6.1
browser 4.6.1
documents to go 1.003.018
configuration of e-mail application 4.6.1
help 4.6.1
Klondike 1.4.8
4.6.1 Memopad
password keeper 4.6.1
Phone 4.6.1
Sudoku 1.0.11
4.6.1 tasks
King of hold'em Texas 2.3.0
Word Mole 4.6.1

1 is the news provider's SIM card?

Old to another device?

2. do not place you a memory card in the camera?

New media card?

or from another device? What device?

Laptop does not start with the connected usb device

Hello
Voila, I have a Sp6100 Pro Satellite, when I try to start whith a usb printer is connected, the laptop does not start. I have the latest version of the bios 1.90.

THX,
Barzot @+.

Hello

What you mean with start won t at all? Is there any kind of reaction if you press the power button / stop?

7 does not start with a connected external Western Digital hardrives.

I have two external hardrives WD, connected to a pc to the bridge under windows7. When the hard drives are connected, Windows7 does not start. It hangs on the black screen. Without the disks attached, it starts very well. Anyone know what I can do to not having to pick up leads outside whenever I want to start?

Solved.

Use Windows Update to install the update from Western Digital Technologies WD SES device released by Microsoft in January 2011.

My system hangs is no longer with USB Legacy Support set to 'Auto '.

Laptop does not start with LCD connected via an HDMI cable

I have a laptop Toshiba A660 running windows 7 Home Premium. I use and external LCD connected via an HDMI cable. If I leave the cable connected to the computer when I turn it on most of the time it does not start. It hangs at the windows from logo. The only way I can boot is power off (by operating switch for a while) and remove the HDMI cable.

Hello

To update the display driver, following the steps outlined in the article below:

Updated a hardware driver that is not working properly

http://Windows.Microsoft.com/en-us/Windows7/update-a-driver-for-hardware-that-isn ' t-work correctly

Update drivers: recommended links

http://Windows.Microsoft.com/en-us/Windows7/update-drivers-recommended-links

See the articles on how to set up two monitors:

http://support.Microsoft.com/kb/976064

http://www.Microsoft.com/athome/organization/twomonitors.aspx#fBid=f5W9Hvf07IV

I hope it helps

Start - and connect SQL to the lowest level in the tree or the hierarchy
Hello

In a tree structure as shown below how can I get all the child (tree) level lowest records as a - E, H, G, and D.
A a B, C, D to level 2
B E and F at level 3, and C G at level 3
F a H level 4

A
| l | l | l
B C D
| l | l | l
E F G
| l
H

I have used start with and coonect by to give all levels of the tree as follows: -.
SELECT task_id, parent_task_id, top_task_id, level
OF pa_tasks
WHERE project =: p
START WITH parent_task_id IS NULL
CONNECT BY PRIOR task_id = parent_task_id

Thank you

Published by: user10648285 on October 17, 2011 23:42

Published by: user10648285 on October 17, 2011 23:42

Published by: user10648285 on October 17, 2011 23:42

Published by: user10648285 on October 17, 2011 23:43
Oracle a [url http://download.oracle.com/docs/cd/E11882_01/server.112/e26088/pseudocolumns001.htm#i1009313] nickname for hierarchical queries:
```
SELECT task_id,parent_task_id, top_task_id,level
FROM pa_tasks
WHERE project_id = :bind_variable
and connect_by_isleaf = 1
START WITH parent_task_id IS NULL
CONNECT BY PRIOR task_id = parent_task_id
```

Windows does not start with the second internal HDD connected

Hi all, I recently got around a few major problems with a new buuild, but I found another issue. Windows does not start with my connected storage disk.

hard drive 500 GB of important data that I can't save on it and need access. It is a hitachi drive, which works fine on other systems but usually works in this one. any suggestions are appreciated, thanks in advance

fixed the problem of swapping of sata ports and reconnect the cables, thanks for the advice but

Understanding "CONNECT BY" and "START WITH".
OK, I'm trying to update the rows with values determined by lines joined in a recursive relationship of unknown depth. I am told that "CONNECT BY" and "START WITH" can be useful in this, but I don't see how to get the value I'm looking for.

In my case, there are 3 values in my table.
ID
ID of the parent
Invoice

On some lines, the Bill is null. For records, you get the ID of the invoice by searching for the invoice of the parent folder. I'm trying to update the table so that all THE rows in the table have an ID of invoice.

Here is an example of table and the lines.
```
CREATE TABLE DISTRIBUTION (
ID            INT,
INV_NUM       INT,
PARENT_ID     INT
)

INSERT INTO DISTRIBUTION 1, 111, NULL;
INSERT INTO DISTRIBUTION 2, 112, NULL;
INSERT INTO DISTRIBUTION 3, NULL, 2;
INSERT INTO DISTRIBUTION 4, 113, NULL;
INSERT INTO DISTRIBUTION 5, NULL, 4;
INSERT INTO DISTRIBUTION 6, NULL, 5;
INSERT INTO DISTRIBUTION 7, NULL, 6;
```
What I would do is update the inv_num column in the table so that a select statement * would look like this...
```
ID        INV_NUM    PARENT_ID
-----     -------------     ---------------
1            111           null
2            112           null
3            112              2
4            113           null
5            113              4
6            113              5
7            113              6
```
You can provide any help would be greatly appreciated.
Hello

Thank you post the CREATE TABLE and INSERT instructions, but please make sure that they work.
None of the INSERT statements; I think you meant something like the statements shown after the query.

Here's a way to get the desired results:
```
UPDATE  distribution     m
SET     inv_num = ( SELECT  inv_num
              FROM    distribution
              WHERE   CONNECT_BY_ISLEAF     = 1
              START WITH     id          = m.id
              CONNECT BY     id          = PRIOR parent_id
                       AND     PRIOR inv_num     IS NULL
               )
WHERE   inv_num       IS NULL
;
```
This statement is Bottom-Up of subqueries, where we START WITH the lines that need to update, and process to the top of the tree, until you get to an ancestor who was an inv_num.
In your sample data, only the roots (the lines that have no parents) have inv_num. In this case, it might be a little easier (but only a little) to make a request from top to bottom , where we START WITH the roots and low process in the tree to find their subordinates.
If we add some data examples where a nonroot has inv_num:
```
INSERT INTO DISTRIBUTION (id, inv_num, parent_id) VALUES ( 91, 910,   1);
INSERT INTO DISTRIBUTION (id, inv_num, parent_id) VALUES ( 92, NULL, 91);
```
What results would you like?
Using the UPDATE statement above, id = 92 would get his inv_nuym of the closest ancestor (in the case of thios, parent) who had an inv_num:
```
.       ID    INV_NUM  PARENT_ID
---------- ---------- ----------
         1        111
        91        910          1
        92        910         91
         2        112
         3        112          2
         4        113
         5        113          4
         6        113          5
         7        113          6
```
Either the row with id = 92 gets inv_num = 910, no 111.

Rebuild the parent/child (connect by prior and where clause)

Hi guys, you have a quick question for you - code as follows:

with t1 as)
Select 1 id, 'Name1' some_name, null, 'Y' parent_id of double some_flag
Union select 2 id, 'Name2', 1 parent_id, 'Y' some_flag of the double
Union select 3 id, "Name3", 2 parent_id, "n" of the double some_flag
Union select id 4, 'Name4.1', 3 parent_id, 'Y' some_flag of the double
Union select id 5, 'Name4.2', 3 parent_id, 'Y' some_flag of the double
Union select id 6, 'Name4.3', 3 parent_id, 'Y' some_flag of the double
Union select id 7, "Name5", 6 parent_id, 'Y' some_flag of the double
)
SELECT id, the_name, parent_id, null new_parent_id
de)
SELECT id, lpad (' ', (level 1) * 3) | some_name the_name, parent_id
from t1
where some_flag = 'Y '.
Start with id = 1
connect by parent_id = prior id
brothers and sisters of order by some_name
)
Output:
ID, THE_NAME, PARENT_ID, NEW_PARENT_ID
1, Name1,
2, name2, 1,.
4, Name4.1,3,
5, Name4.2,3.
6, Name4.3, 3,.
7, name5, 6,.
As you can see, id = 3 is not displayed (because of where some_flag = 'Y'), but parent_id for id in (4,5,6) shows 3.
-Only by using SQL - display 'current parent in valid results' as new_parent_id?
In this example that would be showing new_parent_id = 2 for the id in (4,5,6)
Currently on Oracle 11 g 1 material.

Similar theme (manipulate path), but seems to work with the brothers and SŒURS of ORDER BY.

SELECT id, the_name, parent_id, some_flag, new_parent_id

FROM (SELECT id, lpad (' ', (LEVEL - 1) * 3): some_name the_name,)

some_flag, parent_id,

REGEXP_SUBSTR)

REGEXP_SUBSTR)

SYS_CONNECT_BY_PATH)

DECODE (some_flag, 'Y', id), ' / ').

'[0-9]+[/]+[0-9]+$'),

(0-9] +') new_parent_id

FROM t1

WHERE some_flag = 'Y '.

START WITH id = 1

CONNECT BY PRIOR ID = parent_id

ORDER OF brothers and SŒURS some_name);

Application of the terms of registration parent when you use START WITH / CONNECT BY FRONT
Hello

I'm trying to understand how to apply when only apparent conditions when using records to begin with... connect by prior logic.

Here is an example...

Table:
```
CREATE TABLE TEMP_BTL
(
  TRANS_ID               NUMBER(22,20),
  PARENT_TRANS_ID  NUMBER(22,20),
  TYPE_ID                 NUMBER,
  STATUS_ID             NUMBER
);
```
Records:
```
SET DEFINE OFF;
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (1, NULL, 1, 2);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (2, 1, 2, 1);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (3, 2, 3, 4);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (4, 3, 4, 3);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (5, NULL, 2, 3);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (6, 5, 4, 1);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (7, 6, 5, 3);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (8, 7, 6, 2);
Insert into TEMP_BTL
   (TRANS_ID, PARENT_TRANS_ID, TYPE_ID, STATUS_ID)
 Values
   (9, NULL, 1, 3);
```
If my query rules are:

PARENT_TRANS_ID IS NULL = parent_record

I have to limit my results so that only considered hierarchical groups are groups that record the type_id of the parent = 1 (single parent record limit to this condition, no children)

Then I need to find the max (trans_id) of each subset for groups where the parent record type_id = 1

Thus, according to the data from the example above, would be results that I would look:

TRANS_ID = 1 group parent, I don't want to return trans_id = 4

I would not return anything for the trans_id = 5 parent group because the type_id of that parent is not 1

For trans_id 9, is the parent and the only record for this group so it is type_id = 1 I'm not going back 9 as the max (trans_id) for this game.

And then, eventually, I'll limit my results to the place where the max id batch trans = 3.

Any help is appreciated...

Thank you
Christine
Hi, Christine.

cad0227 wrote:
Hello

I'm trying to understand how to apply when only apparent conditions when using records to begin with... connect by prior logic.

In your example of data 2 is the parent of 3, and 3 is the parent of 4.
Use 'root' to describe nodes (for example, 1 and 5) that have no parents.
Here is an example...

Table:
```
CREATE TABLE TEMP_BTL ...
```
Thank you for including CREATE TABLE and INSERT statements. It's very useful!

If my query rules are:

PARENT_TRANS_ID IS NULL = parent_record

I have to limit my results so that only considered hierarchical groups are groups that record the type_id of the parent = 1 (single parent record limit to this condition, no children)

The START WITH clause is where to put the conditions that apply only to the roots.

Then I need to find the max (trans_id) of each subset for groups where the parent record type_id = 1

Thus, according to the data from the example above, would be results that I would look:

TRANS_ID = 1 group parent, I don't want to return trans_id = 4

I would not return anything for the trans_id = 5 parent group because the type_id of that parent is not 1

For trans_id 9, is the parent and the only record for this group so it is type_id = 1 I'm not going back 9 as the max (trans_id) for this game.

And then, eventually, I'll limit my results to the place where the max id batch trans = 3.

Any help is appreciated...

It would be useful that you reported the exact output desired. Describing the output is great, especially when it is as clear as your description, but describe the results, in addition to, not instead not to display.
Here are the results you want?
```
`  ROOT_ID MAX_TRANS_ID
---------- ------------
         1            4
         9            9
```
Here's a way to get them:
```
WITH     got_root_id     AS
(
     SELECT     trans_id
     ,     CONNECT_BY_ROOT trans_id          AS root_id
     FROM     temp_btl
     WHERE     status_id     = 3
     START WITH     type_id          = 1
                 AND     parent_trans_id     IS NULL
     CONNECT BY     parent_trans_id     = PRIOR trans_id
)
SELECT       root_id
,       MAX (trans_id)     AS max_trans_id
FROM       got_root_id
GROUP BY  root_id
;
```
Published by: Frank Kulash on 13 August 2012 13:58

SQL SELECT to hierarchical tables: START WITH... CONNECT BY...
This seems to be a simple problem, but I think that I have enough intelligence SQL to solve.

I have a table called DE_DOMAIN. The columns of interest are:
DOMAIN_ID - PK
NAME
PARENT_ID - FK (can be NULL), poining to CF of the parent

What I want is: Returns a hierarchical list, containing: column 3 above as well as the NAME of the parent.

Regardless of the name of the parent, it works fine:
SELECT DOMAIN_ID, PARENT_ID, level
OF DE_DOMAIN
WHERE SUPERDOMAINE = 2673
Start by PARENT_ID IS NULL
Connect DOMAIN_ID PARENT_ID = prior
BROTHERS AND SŒURS ORDER BY NAME ASC

and I get:
11 rec_11 1 Null
15 rec_15 1 Null
16 1 Null rec_16
17 1 Null rec_17
22 17 2 rec_22
1 2 17 rec_1
rec_25 25 17 2
2 2 17 rec_2

i.e. records with PK = 1, 22, 25, 2 have all like parent record with PK = 17, then the new column name, they must bear the name of the parent (i.e. rec_17).

A simple idea?
Thank you very much.
Hello

You can use the FIRST operator in the SELECT clause.
I don't have a version of your table, so I'll use scott.emp to illustrate:
```
SELECT     empno
,     ename
,     mgr
,     PRIOR ename    AS mgr_name
FROM     scott.emp
START WITH     mgr     IS NULL
CONNECT BY     mgr     = PRIOR empno
ORDER SIBLINGS BY     ename
;
```
Output:
```
.    EMPNO ENAME             MGR MGR_NAME
---------- ---------- ---------- ----------
      7839 KING
      7698 BLAKE            7839 KING
      7499 ALLEN            7698 BLAKE
      7900 JAMES            7698 BLAKE
      7654 MARTIN           7698 BLAKE
      7844 TURNER           7698 BLAKE
      7521 WARD             7698 BLAKE
      7782 CLARK            7839 KING
      7934 MILLER           7782 CLARK
      7566 JONES            7839 KING
      7902 FORD             7566 JONES
      7369 SMITH            7902 FORD
      7788 SCOTT            7566 JONES
      7876 ADAMS            7788 SCOTT
```

START WITH and CONNECT BY PRIOR

Similar Questions

Maybe you are looking for