connection of a level query

date sample:

EMPLOYEE_NUMBER, CODE

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

100001 A, B

100006 C, D, E

Power required:

EMPLOYEE_NUMBER CODE

100001 A

100001 B

C 100006

D 100006

100006 E

more question I tried but still stuck with logic

WITH sample_data AS

(SELECT 100001 employee_number, "A, B" code FROM dual)

UNION ALL

SELECT 100006, 'C, D' THE double code

)

SELECT DISTINCT employee_number,

CASE

WHEN level = 1

THEN SUBSTR (code, 0, instr (code,', ', 1)-1)

Of OTHER SUBSTR (code, instr(code,',',1,level-1) + 1)

END,

LEVEL

OF sample_data

CONNECTION level < = 2

Hello

2742751 wrote:

Divide the CTE expression into anchor and recursive members.

Run the members of anchor creating the first invocation or base of (T,₀) result set.

Run the recursive member with T_I as an input and T_{i + 1} as an output.

Repeat step 3 until an empty set is returned.

Returns the set of results. It's a UNION ALL of T₀ to T_n operation

I use above the algorithm for the recursive comprehension sub query. Please see exhibit attached, if the interpretation is correct or not. I'm new on sub recursive query

...

Right! It's a good explanation of how works a recursive query, and you seem to understand.

Tags: Database

Similar Questions

Performance issue with connection of a level query

Hi I have a problem with connection by level in oracle.

My table is:

J_USER_CALENDAR
--------------------------

USER_NAME FROM_DATE TO TO_DATE COMMENTS
-------------------------------------------------------------------------------------------------
Comment1 5 November 09 uma Shankar 2 November 09
Veera 11 November 09 13 November 09 Comment2
Commnet3 17 December 09 uma Shankar 15 December 09
Vinod 20 October 09 21 October 09 Comments4

The table above is the calendar of user authorization.

Now I need to view the users who are on leave between November 1, 2009 to November 30, 2009

The output should look like:

USER_NAME FROM_DATE COMMENTS
---------------------------------------------------------------------------------
Comment1 2 November 09 uma Shankar
Comment1 3 November 09 uma Shankar
Comment1 4 November 09 uma Shankar
Comment1 5 November 09 uma Shankar
Comment2 Veera 11 November 09
Comment2 Veera 12 November 09
Comment2 Veera 13 November 09

For this, I tried with the following query, but it takes too long to run.
Select FROM_DATE, user_name, comments from (SELECT distinct FROM_DATE, user_name,
Comments FROM (SELECT (LEVEL) + FROM_DATE-1 FROM_DATE, TO_DATE, FIRST_NAME |) » '|| LAST_NAME
user_name, COMMENTS OF J_USER_CALENDAR
where
and J_USER_CALENDAR.IS_DELETED = 0
CONNECT BY LEVEL < = TO_DATE - FROM_DATE + 1) has) where ("FROM_DATE = 1st November 2009 ' or FROM_DATE = November 30, 2009")
or FROM_DATE between November 1, 2009 "-November 30, 2009") order of from_Date, lower (user_name)

Please help me.

Thanks in advance.

Kind regards
Phanikanth

I have not tried to analyze the SQL statement.

Here is a test set in place:

CREATE TABLE T1(
  USERNAME VARCHAR2(30),
  FROM_DATE DATE,
  TO_DATE DATE,
  COMMENTS VARCHAR2(100));

INSERT INTO T1 VALUES ('Uma Shankar', '02-Nov-09','05-Nov-09','Comment1');
INSERT INTO T1 VALUES ('Veera','11-Nov-09','13-Nov-09','Comment2');
INSERT INTO T1 VALUES ('Uma Shankar','15-Dec-09','17-Dec-09','Commnet3');
INSERT INTO T1 VALUES ('Vinod','20-Oct-09','21-Oct-09','Comments4');
INSERT INTO T1 VALUES ('Mo','20-Oct-09','05-NOV-09','Comments4');

COMMIT;

Note that I have included an additional line, when the person starts their vacation in the previous month and ends in November.

You could approach the problem like this:
Assume that you want to appear every day in a given month:

SELECT
  TO_DATE('01-NOV-2009','DD-MON-YYYY')+(ROWNUM-1) MONTH_DAY
FROM
  DUAL
CONNECT BY
  LEVEL<=ADD_MONTHS(TO_DATE('01-NOV-2009','DD-MON-YYYY'),1)-TO_DATE('01-NOV-2009','DD-MON-YYYY');

Note that the above attempts to calculate the number of days in the month of November - if we know that the month has a certain number of days, 30 for example, you can rewrite the CONNECT BY clause like this:

CONNECT BY
  LEVEL<=30

Now, he must pick up lines of interest to the table:

SELECT
  *
FROM
  T1 T
WHERE
  (T.FROM_DATE BETWEEN TO_DATE('01-NOV-2009','DD-MON-YYYY') AND TO_DATE('30-NOV-2009','DD-MON-YYYY')
    OR T.TO_DATE BETWEEN TO_DATE('01-NOV-2009','DD-MON-YYYY') AND TO_DATE('30-NOV-2009','DD-MON-YYYY'));

USERNAME        FROM_DATE TO_DATE   COMMENTS
--------------- --------- --------- ----------
Uma Shankar     02-NOV-09 05-NOV-09 Comment1
Veera           11-NOV-09 13-NOV-09 Comment2
Mo              20-OCT-09 05-NOV-09 Comments4

If join us then both sets of results, we have the following query:

SELECT
  *
FROM
  T1 T,
  (SELECT
    TO_DATE('01-NOV-2009','DD-MON-YYYY')+(ROWNUM-1) MONTH_DAY
  FROM
    DUAL
  CONNECT BY
    LEVEL<=ADD_MONTHS(TO_DATE('01-NOV-2009','DD-MON-YYYY'),1)-TO_DATE('01-NOV-2009','DD-MON-YYYY')) V
WHERE
  (T.FROM_DATE BETWEEN TO_DATE('01-NOV-2009','DD-MON-YYYY') AND TO_DATE('30-NOV-2009','DD-MON-YYYY')
    OR T.TO_DATE BETWEEN TO_DATE('01-NOV-2009','DD-MON-YYYY') AND TO_DATE('30-NOV-2009','DD-MON-YYYY'))
  AND V.MONTH_DAY BETWEEN T.FROM_DATE AND T.TO_DATE
ORDER BY
  USERNAME,
  MONTH_DAY;

USERNAME        FROM_DATE TO_DATE   COMMENTS   MONTH_DAY
--------------- --------- --------- ---------- ---------
Mo              20-OCT-09 05-NOV-09 Comments4  01-NOV-09
Mo              20-OCT-09 05-NOV-09 Comments4  02-NOV-09
Mo              20-OCT-09 05-NOV-09 Comments4  03-NOV-09
Mo              20-OCT-09 05-NOV-09 Comments4  04-NOV-09
Mo              20-OCT-09 05-NOV-09 Comments4  05-NOV-09
Uma Shankar     02-NOV-09 05-NOV-09 Comment1   02-NOV-09
Uma Shankar     02-NOV-09 05-NOV-09 Comment1   03-NOV-09
Uma Shankar     02-NOV-09 05-NOV-09 Comment1   04-NOV-09
Uma Shankar     02-NOV-09 05-NOV-09 Comment1   05-NOV-09
Veera           11-NOV-09 13-NOV-09 Comment2   11-NOV-09
Veera           11-NOV-09 13-NOV-09 Comment2   12-NOV-09
Veera           11-NOV-09 13-NOV-09 Comment2   13-NOV-09

Charles Hooper
IT Manager/Oracle DBA
K & M-making Machine, Inc.

SQL query help (we connect by clause level)

Hi all
I have this application developed with data with the clause.

 With dat As
(
  select '@AAA @SSS @DDD' col1 from dual union all
  select '@ZZZ @XXX @TTT @RRR @ZZA' col1 from dual 
)
Select regexp_substr( col1 , '[^@][A-Z]+',1,level) Show from dat
connect by level  <= regexp_count(col1, '@');

Output current: -.

SHOW
-----------------------
AAA
SSS
DDD
RRR
ZZA
TTT
RRR
ZZA
XXX
DDD
RRR

SHOW
-----------------------
ZZA
TTT
RRR
ZZA
. . .
. . .

1st row comes very well, but the next line data copy. And the number of total records = 30. I tried with some, but not worked.
Expected results: -.

SHOW
-----------------------
AAA
SSS
DDD
ZZZ 
XXX 
TTT 
RRR 
ZZA

I need some changes on my request and I am not able to see that. So anyone can add to that or can also provide a different solution also.

Thank you!
Ashutosh

Thanks for providing the loan to the use of query. :)

Here's a solution :-(tested on 10 g, do not have 11 g at hand)

For 11g, just use regexp_count instead of functions of the length.

With dat As
(
  select '@AAA @SSS @DDD' col from dual union all
  select '@ZZZ @XXX @TTT @RRR @ZZA' col1 from dual
)
Select regexp_substr( col, '[^@][A-Z]+',1,level) Show from dat
connect by nocycle level  <= length(col) - length(translate(col, 'A@', 'A'))
       and col = prior col
       and prior sys_guid() is not null;

SHOW
------------------------
AAA
SSS
DDD
ZZZ
XXX
TTT
RRR
ZZA                      

 8 rows selected

Can the vcenter server that an ESX host is connected to determine through query of the esx host cmd line?

Can the vcenter server that an ESX host is connected to determine through query of the esx host cmd line?
All points will be awarded. Thxs.

Look in
```
/etc/opt/vmware/vpxa/vpxa.cfg
```
you will see that the IP VCMS listed something like
```
1.1.1.1
```

connection pool and inefficient query plan

There is a single query that covers almost 90% of cpu DB.
Select * from employee e, Department d where e.departement_id = d.department_id
Us will gather statistics for this table in two, and the issues is resolved.
This occurs every 2 weeks. That said, the query runs fine for 2 weeks and then we have questions... we will bring together the statistics of these two tables... and things are good for another 2 weeks.
This query is a proc to store oracle pl - sql, which is called by the JDBC code. I introduced 2 months behind connection pooling, and today, we are facing this problem.
The query has been accounting for 5 years with no problems.
Before that I presented the connection pool, the jdbc code created a new connection before calling the store proc.
Do you think that my connection to the connection pool has introduced this problem.
The DBA tell me that the query runs a bad plan. Oracle recovers not the more effective plan (and that leads to this high CPU utilization).
I guess that after 2 weeks Oracle begins to pick up plans that are effective in.
Do you think all that this question never has nothing to do with my connection pooling code.
I use Oracle 10 G.

Hi Mike and ground beach thanks for your response.

It was just a doubt... that you guys allowed.

Thank you

m

Connect by - no start Query
Is it possible to get the hierarchy correctly in "connect by" without using 'Start with' request?
CP says:
I use "Connect By" query in a Join Condition.
```
SELECT employee_id
FROM employee e, dept d
WHERE e.dept_id = d.dept_id
AND e.employee_id = 100
START WITH manager_id IS NULL
CONNECT BY PRIOR employee_id = manager_id;
```
Director ID may or may not be there. If the Manager id is NOT NULL, then this query fails. So need to connect by query without a Clause to start.
You seem a little confused on the hierarchical queries.
Let me put it this way: hierarchical queries are return of rows in a table, climbing from root to branches (or sheets if you prefer) or descendant of a branch (or sheets) to a root.

When you start, you have to say when you start from. This can be any value to your table.
I'll use the employees of HR schema provided with Oracle standard table here below.

Suppose you want to select the top-down hierarchy (or roots-sheets if you prefer) from employees_id = 108.
In this case, you can do like this:
```
 SELECT employee_id, manager_id
   FROM employees e
  START WITH employee_id=108
CONNECT BY PRIOR employee_id = manager_id;

EMPLOYEE_ID MANAGER_ID
----------- ----------
        108        101
        109        108
        110        108
        111        108
        112        108
        113        108
```
Now you see that number 108 is also a Manager. If I change my request and I say I want to start with manager_id = 108, I'll start with the records having manager_id = 108. This is why employee_id = 108 will be excluded:
```
 SELECT employee_id, manager_id
   FROM employees e
  START WITH manager_id=108
CONNECT BY PRIOR employee_id = manager_id;

EMPLOYEE_ID MANAGER_ID
----------- ----------
        109        108
        110        108
        111        108
        112        108
        113        108
```
If you do not specify a START WITH clause, for each record in the table the top-down hierarchical query will be executed.
```
 SELECT employee_id, manager_id
   FROM employees e
--  START WITH employee_id=100
CONNECT BY PRIOR employee_id = manager_id
order by employee_id;
EMPLOYEE_ID MANAGER_ID
----------- ----------
        100
        101        100
        101        100
        102        100
        102        100
....
        206        205
        206        205
        206        205
        206        205

315 rows selected.
```
I hope it's clear. If this isn't the case, please post more details about your question.

Kind regards.
Al

connect by prior (hierarchical query)
Hi all

Some1 asked me a question that goes like this:

Source

emp_no dep_no
10 110
20 110
30 110
40 110
10 120
10 130
130 20

write a query to get the following output from the source above

emp_no dept_no
110 10203040
10 120
130 1020

Now I stumbled upon solutions with the terms "connect by front" but I'm nt able to understand how oracle produces this result, could someone point me to a good article or a text that can explain this concept very well (I searched on google and have seen many articles, but I could not able to understand since these articles were not all explain)

Concerning
Rahul
Hi, Rahul,

Welcome to the forum!

What you want to do is called chain aggregation . This page shows the different ways to do it:
http://www.Oracle-base.com/articles/10G/StringAggregationTechniques.php

SYS_CONNECT_BY_PATH (which requires a CONNECT BY query) is just a medium. (From Oracle 11.2, LISTAGG is clearly the best way to string aggregation. Before that, SYS_CONNECT_BY_PATH seems to be the most popular, especially if the results must be in order.)

The following pages are introductions to CONNECT BY queries:
http://www.adp-GmbH.ch/ora/SQL/connect_by.html
http://www.oradev.com/connect_by.jsp

PhotoShop CS 3 / CS 4 level query

Hello
I have a few copies of legitimate of After Effects and Photoshop CS4 - however, I just found out that they are only upgraded. What is the most easy/less expensive method to acquire licenses of CS 3?
Here is the background: I work for a television company in London. The software was never recorded as publishers who used it were not on an online network (to protect the broadcast servers and multimedia). We stayed at Adobe CC and I got the software for sale. I don't have the proof of purchase, although I have all the serial numbers. We bought it forms a reseller as well as other software and hardware. All the IT Department who worked when we bought it is no longer with the company. It's a perfect storm... I have had a few conversations with Adobe customer service online, and while it is useful, they won't do anything without proof of purchase...
Summary: I need to purchase legitimate licenses for ch & AE CS3 allow my CS4 updates work - what is the easiest way to do this?

Thanks for you response JJ.

I don't think I was clear in my original post. I don't have the serial number of the previous version, and soap operas I have for the CS4 update are certainly not registered according to Adobe... Following several discussions with Adobe, I think I really need to get the PoP of the dealer.

Thanks again,

Connect By hierarchical query clause-
Hello

I wrote a query using the Connect By clause to generate the table data structure tree with appropriate levels.

But given could not correct as there may be data that did the trick. As Table1 is parent from table 2 and table 2 is parent from table 3 and table 3 is again Parent from table 1. If these incorrect data then I get error Oracle "ORA-01436: CONNECT loop in the user data" which is correct.

What I need is, whenever this error message is triggered I give my own error instead of this Oracle error message in the SQL statement. Please note that I do not use any PLSQL block, I have just one SQL statement. Appreciate any help on this topic. Thank you.
Hello

Outside of PL/SQL, I don't know anyway to substitute your own error messages for those that oracle provides.

In Oracle 10 (and more), you can write a query using the CONNECT_BY_ISCYCLE pseudo-column. Using CONNECTION BY NOCYCLE, the query will never actually throw the error ORA-01436, but you can get it to display your message in the case there, you not used CONNECT BY NOCYCLE.

Here's how:
Equip your actual query, CONNECT_BY_ISCYCLE, the WITH clause.
Get this subquery (CONNECT_BY_ISCYCLE) MAX. We'll call this max_cbi.
Do a UNION ALL of two pins:
(1) selects everything (except, pehaps, CONNECT_BY_ISCYCLE), subquery, WHERE max_cbi = 0
(2) allows you to select your 'error message' double, WHERE max_cbi = 1.

Published by: Frank Kulash, November 26, 2008 14:09

Dreamweaver offers a secure connection multi level feature?

Hello:
I am trying to create a connection so multi level when guest log in, they see the content specific to their level of play.

This http://www.kingluddite.com/tools/crud-in-dreamweaver will show you how to do what you want to do using deprecated technology.

SQL query to retrieve the same record because no. boxes

Hi all
I have a table of Bill XXINV.
the columns are
INVH_NO INVH_ITEM_CODE INVH_QTY INVH_BOXES
1122                XXITEM1                    1                    5
1123                XXITEM2                     2                   6
I can do a sql query so that I can retrieve the same records repeated as number INVH_BOXES.
Select * from XXINV
where invh_no = 1122
output:
INVH_NO INVH_ITEM_CODE INVH_QTY INVH_BOXES
1122                XXITEM1                    1                    5
Desired output:
INVH_NO INVH_ITEM_CODE INVH_QTY INVH_BOXES
1122                XXITEM1                    1                    5
1122                XXITEM1                    1                    5
1122                XXITEM1                    1                    5
1122                XXITEM1                    1                    5
1122                XXITEM1                    1                    5
or
INVH_NO INVH_ITEM_CODE INVH_QTY INVH_BOXES
1122 XXITEM1 1 1 of 5
1122 XXITEM1 1 2 of 5
1122 1 3 5 XXITEM1
1122 XXITEM1 1 4 of 5
1122 XXITEM1 1 5 of 5
Please suggest. Like I need to print each line in a single page to stick them on boxes so that packaging.
Kind regards
Afzal.

Just add INVH_ITEM_CODE to the connection by the clause.

Select invh_no

invh_item_code

invh_qty

to_char (, 'fm99') | 'from ' | TO_CHAR (invh_boxes, 'fm99') invh_boxes

of xxinv

where invh_no = 1122

connect

by level<=>

and prior invh_item_code = invh_item_code

and prior sys_guid() is not null

Problem with pseudo-HIERARCHICAL query

Hi all

I'm writing a piece of SQL to validate the check digit in wines

(http://en.wikipedia.org/wiki/Vehicle_identification_number#Check_digit_calculation)

using
Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - 64bi
PL/SQL Release 10.2.0.5.0 - Production

The logic in the wikipedia article can be encapsulated as follows:

with val as (select '0' ch, 0 val from dual
             union all
             select '1', 1 from dual
             union all 
             select 'A',1 from dual
             union all 
             select 'J',1 from dual
             union all 
             select '2',2 from dual
             union all 
             select 'B',2 from dual
             union all 
             select 'K',2 from dual
             union all 
             select 'S',2 from dual
             union all 
             select '3',3 from dual
             union all 
             select 'C',3 from dual
             union all 
             select 'L',3 from dual
             union all 
             select 'T',3 from dual
             union all 
             select '4',4 from dual
             union all 
             select 'D',4 from dual
             union all 
             select 'M',4 from dual
             union all 
             select 'U',4 from dual
             union all 
             select '5',5 from dual
             union all 
             select 'E',5 from dual
             union all 
             select 'N',5 from dual
             union all 
             select 'V',5 from dual
             union all 
             select '6',6 from dual
             union all 
             select 'F',6 from dual
             union all 
             select 'W',6 from dual
             union all 
             select '7',7 from dual
             union all 
             select 'G',7 from dual
             union all 
             select 'P',7 from dual
             union all 
             select 'X',7 from dual
             union all 
             select '8',8 from dual
             union all 
             select 'H',8 from dual
             union all 
             select 'Y',8 from dual
             union all 
             select '9',9 from dual
             union all 
             select 'R',9 from dual
             union all 
             select 'Z',9 from dual)
   ,weight as (select 1 pos, 8 weight from dual
                union all 
                select 2, 7 from dual
                union all 
                select 3, 6 from dual
                union all 
                select 4, 5 from dual
                union all 
                select 5, 4 from dual
                union all 
                select 6, 3 from dual
                union all 
                select 7, 2 from dual
                union all 
                select 8, 10 from dual
                union all 
                select 9, 0 from dual
                union all 
                select 10, 9 from dual
                union all 
                select 11, 8 from dual
                union all 
                select 12, 7 from dual
                union all 
                select 13, 6 from dual
                union all 
                select 14, 5 from dual
                union all 
                select 15, 4 from dual
                union all 
                select 16, 3 from dual
                union all 
                select 17, 2 from dual)

I can easily get the check digit for a WINE of the following logic:

    ,vm as (select row_number() over(partition by vin order by level) rn
                   , substr(vin, level,1) v
                   , vin
                from (select 'ZAR91600006043427' vin
                        from dual
                     )
             connect by level <= length(vin))
select vin
     , mod(sum(checksum),11) calculated_check
     , substr(vin,9,1) existing_check
     , regexp_instr(substr(vin,9,1),mod(sum(checksum),11)) correct
  from (select val.val * weight.weight checksum
              ,vm.vin 
          from vm
          join val on (val.ch = vm.v)
          join weight on (weight.pos = vm.rn)
  order by vm.rn)
group by vin;

which produces:

VIN                              CALCULATED_CHECK                       EXISTING_CHECK                   CORRECT                                
ZAR91600006043427                4                                      0                                0

However, I am having a little problem when I try to combine several wines of a table which I want to do rather than calculate each wine by programming:

This, for example, produces a line for each letter of the wine ready to match with the other two CTES:

with vehicle as (select 1000 as vehicle_id, 'ZAR91600006043427' as vin from dual
                   union all 
                 select 1001, 'XXX91600006043427' from dual)
select level
       , substr(vin, level,1) v
       , vin
    from (select *
            from vehicle
           where vehicle_id = 1000
         )
 connect by level <= length(vin);

I want ideally is a line for each character of each WINE. but this:

with vehicle as (select 1000 as vehicle_id, 'ZAR91600006043427' as vin from dual
                   union all 
                 select 1001, 'XXX91600006043427' from dual)
select level
       , substr(vin, level,1) v
       , vin
    from (select *
            from vehicle
           where vehicle_id in (1000,1001)
         )
 connect by level <= length(vin);

product 262142 lines!

and adding

where vin = prior vin

just reduced to 131070 lines...

place a different one on the SQL produced the result that I want, but it is not scalable product lines by adding a third vehicle are exponentially more than two.

so, in short: what is happening with my connect by statement that I'm missing?

Hello

Samir says:
... I want ideally is a line for each character of each WINE. but this:
with vehicle as (select 1000 as vehicle_id, 'ZAR91600006043427' as vin from dual
union all
select 1001, 'XXX91600006043427' from dual)
select level
, substr(vin, level,1) v
, vin
from (select *
from vehicle
where vehicle_id in (1000,1001)
)
connect by level <= length(vin);
product 262142 lines!

and adding
where vin = prior vin
just reduced to 131070 lines...

place a different one on the SQL produced the result that I want, but it is not scalable product lines by adding a third vehicle are exponentially more than two.

so, in short: what is happening with my connect by statement that I'm missing?

Whenever you want to debug a CONNECT BY query, put SYS_CONNECT_BY_PATH in the SELECT clause, to see how the lines are related.
We will also cut the sting down to only 3 characters, in order to make things faster and easier to read and add ROWNUM just to make it easier to talk about the results:

with vehicle as (select 1000 as vehicle_id, 'ABC' as vin from dual
                   union all
                 select 1001, 'XYZ' from dual)
select level
       , substr(vin, level,1) v
       , vin
       , sys_connect_by_path ( vin
                    , ','
                    )     AS path_3
       , sys_connect_by_path ( substr (vin, level, 1)
                    , ','
                    )     AS path_1
       , ROWNUM
    from (select *
            from vehicle
           where vehicle_id in (1000,1001)
         )
-- where     vin     = PRIOR vin
 connect by level <= length(vin);

Output:

`    LEVEL V    VIN PATH_3               PATH_1         ROWNUM
---------- ---- --- -------------------- ---------- ----------
         1 A    ABC ,ABC                 ,A                  1
         2 B    ABC ,ABC,ABC             ,A,B                2
         3 C    ABC ,ABC,ABC,ABC         ,A,B,C              3
         3 Z    XYZ ,ABC,ABC,XYZ         ,A,B,Z              4
         2 Y    XYZ ,ABC,XYZ             ,A,Y                5
         3 C    ABC ,ABC,XYZ,ABC         ,A,Y,C              6
         3 Z    XYZ ,ABC,XYZ,XYZ         ,A,Y,Z              7
         1 X    XYZ ,XYZ                 ,X                  8
         2 B    ABC ,XYZ,ABC             ,X,B                9
         3 C    ABC ,XYZ,ABC,ABC         ,X,B,C             10
         3 Z    XYZ ,XYZ,ABC,XYZ         ,X,B,Z             11
         2 Y    XYZ ,XYZ,XYZ             ,X,Y               12
         3 C    ABC ,XYZ,XYZ,ABC         ,X,Y,C             13
         3 Z    XYZ ,XYZ,XYZ,XYZ         ,X,Y,Z             14

14 rows selected.

There is no START WITH clause, so it will start by each of the 2 rows. It is very good; Take the first case of 'ABC'.
Level = 1, we have row "ABC", so the substring added to this level is 'A'
The lines are connected to this line to LEVEL = 2? All rows that satisfy the condition of LEVEL<= length="" (vin),="" which,="" in="" this="" sample="" set,="" means="" both="" rows.="" let's="" take="" 'abc'="" first,="" which="" is="" what="" happened="" on="" the="" line="" where="" rownum="">
The lines are connected to this line (the line at the end of the path_3 ", ABC, ABC' ') = 3 level? All rows that satisfy the condition of LEVEL<= length="" (vin),="" which,="" in="" this="" sample="" set,="" means="" both="" rows.="" let's="" take="" 'abc'="" first,="" which="" is="" what="" happened="" on="" the="" line="" where="" rownum="">
So far, so good; the results were exactly what we wanted. But here's where it won't.
The lines are connected to the line at the end of the path ", ABC, ABC, ABC' level = 4?" All rows that satisfy the condition of LEVEL<= length="" (vin),="" which,="" in="" this="" sample="" set,="" means="" no="" rows.="" so="" let's="" back="" up="" the="" tree="" and="" see="" if="" any="" of="" the="" nodes="" we've="" already="" vistited="" has="" any="" other="" children.="" we="" back="" up="" to="" the="" row="" where="" path_3=',ABC,ABC' does="" 'abc'="" have="" any="" other="" children="" (that="" is,="" rows="" that="" meet="" the="" condition="" 3="">< length="" (vin)?="" yes,="" 'xyz'="" meets="" that="" condition,="" so="" 'xyz'="" is="" a="" child="" of="" 'abc',="" and="" we="" get="" the="" output="" row="" where="" rownum="">

A CONNECT BY query WHERE clause is applied after the CONNECT BY is done. Adding a WHERE clause only deletes the rows from the result set once they have been generated. If we have a comment the WHERE clause above, the results are:

`    LEVEL V    VIN PATH_3               PATH_1         ROWNUM
---------- ---- --- -------------------- ---------- ----------
         2 B    ABC ,ABC,ABC             ,A,B                1
         3 C    ABC ,ABC,ABC,ABC         ,A,B,C              2
         3 Z    XYZ ,ABC,XYZ,XYZ         ,A,Y,Z              3
         3 C    ABC ,XYZ,ABC,ABC         ,X,B,C              4
         2 Y    XYZ ,XYZ,XYZ             ,X,Y                5
         3 Z    XYZ ,XYZ,XYZ,XYZ         ,X,Y,Z              6

Better, but it is still bad. All the 14 ranks were always generated, and then some of them have been excluded by the WHERE clause.
Why the lines like the one with ROWNUM = included 3? Because the WHERE 'wine = wine PRIOR' condition is true: on this line, the wine is "XYZ", and wine PRIOR is also "XYZ". The status of research only in the last 2 levels. (PREREQUISITE may only be 1 level.) It does not matter that, earlier in the path, there is a knot where wine! = wine BEFOREHAND; This condition applies only to the finished result set. So the output row with path_3 =', ABC, XYZ "was excluded the result defined by the WHERE clause, but the descendants of this line can remain.

Moral: If you do a query CONNECT BY and provided to connect BY is 'LEVEL '.<= x",="" then="" the="" table="" had="" better="" have="" only="" one="">
Odie showed how to use CONNECT BY in this problem, by making the relative to the double CONNECT BY and joining your table of multiple row results.

condition "after" versus "after CONNECT".
What is the difference in sematics/sense/result if condition "T.COL1 is null" is written after paragraph "connect by", or, after "where" - would adopt?
See the query below.
Thank you.

Select *.
FROM table1 T
where T.COL1 is null
Start with T.PARENT_ID is null
connect by prior g.PARENT_ID = T.ID and T.COL1 is null;

Published by: CharlesRoos on January 11, 2010 02:20
Hello
The CONNECT BY of a query part is made before the clause WHERE is evaluated. (Except for the join conditions. See message from Rob, 2nd after this one).
Conditions in the CONNECT BY clause may remove a node and its descendants (i.e. any node that is linked to the root through the removed node), but conditions in a clause WHERE only a single line.

There are so few NULL values in the table scott.emp "col1 IS NULL" condition will not be a good example.
There is nothing special about this condition, so I'll use the condition "jpob!" = 'MANAGER' ' instead. If that confuses you, create a copy of the table scott.emp, but change the work with null for all managers.

This is a CONNECT BY querying base, which includes the entire hierarchy starting with the KING:
```
PROMPT     Query 1: No condition concerning job

SELECT     LPAD ( ' '
          , 3 * (LEVEL - 1)
          ) || ename          AS i_ename
,     LEVEL
,     empno
,     mgr
,     job
FROM     scott.emp
START WITH     ename     = 'KING'
CONNECT BY     mgr     = PRIOR empno
;

Query 1: No condition concerning job

I_ENAME                   LEVEL      EMPNO        MGR JOB
-------------------- ---------- ---------- ---------- ---------
KING                          1       7839            PRESIDENT
   JONES                      2       7566       7839 MANAGER
      SCOTT                   3       7788       7566 ANALYST
         ADAMS                4       7876       7788 CLERK
      FORD                    3       7902       7566 ANALYST
         SMITH                4       7369       7902 CLERK
   BLAKE                      2       7698       7839 MANAGER
      ALLEN                   3       7499       7698 SALESMAN
      WARD                    3       7521       7698 SALESMAN
      MARTIN                  3       7654       7698 SALESMAN
      TURNER                  3       7844       7698 SALESMAN
      JAMES                   3       7900       7698 CLERK
   CLARK                      2       7782       7839 MANAGER
      MILLER                  3       7934       7782 CLERK

14 rows selected.
```
The condition "job! = 'MANAGER' in the WHERE clause eliminates only the 3 rows where job = "MANAGER":
{code}
PROMPT query 2: Condition in the WHERE clause

SELECT LPAD (' ' ')
3 * (LEVEL - 1)
) || Ename AS i_ename
LEVEL
empno
Bishop
work
FROM scott.emp
WHERE job! = 'MANAGER '.
START WITH ename = 'KING '.
CONNECT BY PRIOR empno = mgr
;

Query 2: Condition in the WHERE clause

I_ENAME LEVEL EMPNO, MGR JOB
-------------------- ---------- ---------- ---------- ---------
KING 1 7839 PRESIDENT
ANALYST SCOTT 3-7788-7566
4 7876 7788 CLERK ADAMS
3-7902-7566 FORD ANALYST
CLERK 4 7369 7902 SMITH
SELLER OF ALLEN 3-7499-7698
SELLER OF WARD 3-7521-7698
3 7654 7698 SELLER MARTIN
SELLER OF 7698 7844 3 TURNER
3 7900 7698 CLERK JAMES
CLERK 3 7934 7782 MILLER

11 selected lines.
{code}
Put the same condition in the CONNECT BY clause means that the State is part of what determines a descendant.
X is not a child of is x job is "MANAGER"; so, X will not succeed in the result set and none of the descendants of x will either.
{code}
PROMPT query 3: status of CONNECT BY clause

SELECT LPAD (' ' ')
3 * (LEVEL - 1)
) || Ename AS i_ename
LEVEL
empno
Bishop
work
FROM scott.emp
START WITH ename = 'KING '.
CONNECT BY PRIOR empno = mgr
AND job! = 'MANAGER '.
;

Query 3: Condition CONNECT BY clause

I_ENAME LEVEL EMPNO, MGR JOB
-------------------- ---------- ---------- ---------- ---------
KING 1 7839 PRESIDENT
{code}
Note that this does not affect the roots. In query 4, below is the same 3 query above, except that the START WITH JONES, makes no mention of KING.
{code}
PROMPT query 4: CONNECT BY clause Condition does not affect the root

SELECT LPAD (' ' ')
3 * (LEVEL - 1)
) || Ename AS i_ename
LEVEL
empno
Bishop
work
FROM scott.emp
START WITH ename = 'JONES '.
CONNECT BY PRIOR empno = mgr
AND job! = 'MANAGER '.
;

Request 4: CONNECT BY clause Condition does not affect the root

I_ENAME LEVEL EMPNO, MGR JOB
-------------------- ---------- ---------- ---------- ---------
1 7566 7839 JONES MANAGER
2 7788 7566 ANALYST SCOTT
3 7876 7788 CLERK ADAMS
2 7902 7566 FORD ANALYST
CLERK 3 7369 7902 SMITH
{code}
JONES is in the result set, despite being a MANAGER.
It is not that JONES is not considered a child of someone else. JONES has not raised through the CONNECT BY clause, only the START WITH clause.

Published by: Frank Kulash, 11 January 2010 09:46

As Rob said, join in the WHERE clause conditions are performed before START WITH and CONNECT BY. Sorry, I forgot to mention that.
When you join a CONNECT BY query, it is usually more effective to do CONNECT BY in a subquery, then join the results for the other tables, if possible.

Conversion of a microphone to a level signal level signal line.

Friday, June 24, 2016

Re: My computer: Mac mini (mid-2010)

The model number is A1347

Very well. I wonder if the community of Apple Support can help me with this problem.

I bought a version a SADES 810 Gaming Headset which is supposed to be compatible with Mac in 2016. I tried to plug the headphones on my Mac mini in the hope that I would be able to have conversations using a magic Jack. While I hear people talking to me perfectly clear with the magic Jack, they did not intend to talk into the microphone of the headset.

I tested the helmet with a chip "phone and the microphone and the headphones worked well." Thus, the microphone is not defective. I discovered that my Mac mini to model requires a microphone level line. The microphone on the "2016 new Version for PS4 SADES SA - 810 Gaming PC Headset Headset" is not level line, it seems, but the lower level of the microphone. Thus, the microphone of the headset is not turn off force, that he needs to work in making the Mac mini line level microphone.

On a site I saw this quote:

"Mic Level is the amount of signal you get a microphone. It is much lower than the level of the line and therefore need to be stimulated by a preamp to be usable in a recording situation. »

What I need is the gadget that will allow me to connect the microphone level 3.5 mm headphones microphone, on my Mac mini, which will convert then the microphone level low in a line level signal, when plugged (at the level of the line) microphone jack/plug of the Mac mini. Is there such a gadget?

T he one or two 3.5 mm Sockets?

The headphone jack on the Mac should include a contact for a microphone. To check see it:

How to use the Apple earphones with remote and mic with your Mac - Apple Support

Execution of Mac OSX 10.7.3 and impossible to switch to Firefox 11.0. Continuous attempts to connect to the upgrade server. Suggestions?

I got the message "connection to server level" as long as a few hours. I tried to launch the upgrade directly from mozilla.org and get the same thing. Running Firefox 8.0.1

Sorry to hear that, but if you have a problem with the firefox update the best way is to download and install the new version.

1. download a new copy of the Firefox program

2 Firefox 11.0 of: http://www.mozilla.org/en-US/firefox/all.html

3. Trash , install the current application of Firefox to do a cleaning.

4. install the new version you downloaded.

Do not select to delete your personal data, your profile data is stored in the Firefox profile folder, so you will not lose your bookmarks or other personal data.

See also: Installation of Firefox on Mac

Thank you

Please check 'Resolved' the answer really solve the problem, to help others with a similar problem.

connection of a level query

Similar Questions

Maybe you are looking for