How to order a tree balanced with SQL hierarchical queries

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Re: Generate tree balanced with SQL hierarchical queries

is there a way I can order this tree in alphabetical order so that the result looks like

LEVEL INDENTED_ENAME
---------- --------------------
1 KING BED
2 BLAKE
3 ALLEN
3 JAMES
MARTIN 3
3 TURNER
WARD 3
2 CLARK
3 MILLER
2 JONES
3 FORD
4 SMITH
3 SCOTT
4 ADAMS

-the original query-

SELECT THE LEVEL
, LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
FROM scott.emp
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
;

LEVEL INDENTED_ENAME
---------- --------------------
1 KING BED
2 JONES
3 SCOTT
4 ADAMS
3 FORD
4 SMITH
2 BLAKE
3 ALLEN
WARD 3
MARTIN 3
3 TURNER
3 JAMES
2 CLARK
3 MILLER

Hello

Bodin wrote:
Hi Frank, I can order it selectively depending on the level, which means that only siblings stopped at third level, but rest of the brothers and sisters remain Nations United ordered

It's actually quite difficult. You can "ORDER of brothers and SŒURS BY CASE... ', like this:

SELECT  LEVEL
,      LPAD (' ', 3 * LEVEL) || ename     AS indented_ename
FROM      scott.emp
START WITH        mgr     IS NULL
CONNECT BY         mgr      = PRIOR empno
ORDER SIBLINGS BY  CASE
               WHEN  job = 'MANAGER'  THEN  ename
                                          ELSE  NULL
             END
;

In this case to get desired results in table scott.emp, as the lines are LEVEL = 2 if and only if use = "MANAGER".
But if you reference LEVEL in the CASE expression (for example, if you replace ' job = 'MANAGER' ' with "2 LEVEL =" above "), then you will get the error" ORA-00976: LEVEL, PRIOR or ROWNUM not allowed here. "
The best way I can think to do exactly what you asked is to do 2 CONNECT BY queries; one just to get the LEVEL and the other for the brothers and SŒURS ORDER BY:
{code}
WITH got_lvl AS
(
SELECT LEVEL AS lvl
Bishop
empno
ename
FROM scott.emp
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
)
SELECT lvl
, LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
OF got_lvl
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
BROTHERS AND SŒURS OF ORDER OF CASES
ONCE lvl = 2 THEN ename
ANOTHER NULL
END
;
{code}
Why you can't simply "Brothers and SŒURS of ORDER BY ename" at all levels? If all you care is the order of the items of LEVEL = 2, then this is probably the most effective and simplest way. It really hurt anything if nodes on levels 3, 4, 5,... are in order, too?

Here's something you can do if you want to order by different unique things to different levels:
{code}
WITH got_sort_key AS
(
SELECT LEVEL AS lvl
, LPAD (' ', 3 * LEVEL) | Ename AS indented_ename
empno
SYS_CONNECT_BY_PATH (LPAD (CASE
WHEN LEVEL = 2
THEN ename
Of OTHER TO_CHAR (empno)
END
10
)
, ','
) AS sort_key
FROM scott.emp
START WITH mgr IS NULL
CONNECT BY PRIOR empno = mgr
)
SELECT lvl
indented_ename
empno
OF got_sort_key
ORDER BY sort_key
;
{code}
However, all possible values of the CASE expression must uniquely identify the node; otherwise, the output won't necessarily hierarchical order. You can assign arbitrary unique IDS to the lines (using the ROW_NUMBER analytic function, for example), but that requires another subquery and is also complex and perhaps as ineffective as the solution above with 2 garages CONNECT.

Tags: Database

Similar Questions

Building the tree balanced with SQL hierarchical queries

Hi all

I have the following hierarchical data with different levels of subtree:

A0
-A001
-A00101
A1
-A101
A2
-A201
-A20101
-A201010001

A0 subtree has 3 levels, A1 subtree has 2 levels and subtree of the A3 is level 4. I want to generate a tree balanced on the data with all levels of the subtree equal to the maximum number of levels available in the whole tree which, in this particular case, is 4.

I don't know that it is possible with SQL. Script to generate the above mentioned are as below:

CREATE TABLE codes_tree
(node_id VARCHAR2 (10))
parent_node_id VARCHAR2 (10)
);
INSERT INTO codes_tree VALUES ('A0', NULL);
INSERT INTO codes_tree VALUES ('A001', 'A0');
INSERT INTO codes_tree VALUES ('A00101', 'A001');
---
INSERT INTO codes_tree VALUES ('A1', NULL);
INSERT INTO codes_tree VALUES ('A101', 'A1');
---
INSERT INTO codes_tree VALUES ('A2', NULL);
INSERT INTO codes_tree VALUES ('A201', 'A2');
INSERT INTO codes_tree VALUES ('A20101', 'A201');
INSERT INTO codes_tree VALUES ('A201010001', 'A20101');

Any help will be much appreciated.

Thank you... Best regards

Published by: naive2Oracle on May 12, 2011 19:40

Published by: naive2Oracle on May 12, 2011 19:41

Hello

Of course, you can do it in SQL.
One way is to take the normal output of hierarchical and manipulate the result set so that the leaves are repeated as often as necessary to make all branches of the same length. I have Oracle 10.2 available right now, so here's a solution that will work in Oracle 10 (and more):

WITH     original_hierarchy     AS
(
     SELECT     node_id
     ,     LEVEL               AS lvl
     ,     CONNECT_BY_ISLEAF     AS isleaf
     ,     ROWNUM               AS rnum
     FROM     codes_tree
     START WITH     parent_node_id     IS NULL
     CONNECT BY     parent_node_id     = PRIOR node_id
)
,     got_max_lvl     AS
(
     SELECT     o.*
     ,     MAX (lvl) OVER ()     AS max_lvl
     FROM     original_hierarchy     o
)
SELECT       LPAD ( ' '
            , 3 * ( ( d.lvl
                 + NVL (c.rnum, 1)
                 - 1
                 )
               - 1
               )
            ) || CASE
               WHEN c.rnum > 1
               THEN '*' || d.node_id || '*'
               ELSE        d.node_id
              END          AS display_id
FROM            got_max_lvl     d
LEFT OUTER JOIN       got_max_lvl     c  ON     d.isleaf     = 1
                       AND     c.rnum          <= 1 + d.max_lvl - d.lvl
ORDER BY  d.rnum
,       c.rnum
;

With the help of Oracle 11.2, it would be preferable to generate original_hierarchy as above, but to manipulate using a WITH recursive clause.
Analytical functions often interfere with CONNECT BY, so I used a separate subquery to get max_lvl, do CONNECT BY in a sub-querry and analytic function in a separate subquery. I don't know what is needed on all versions.

Output of your sample data:

DISPLAY_ID
-------------------------------
A0
   A001
      A00101
         *A00101*
A1
   A101
      *A101*
         *A101*
A2
   A201
      A20101
         A201010001

Below is a generic version of the same query, which I used to test this on scott.emp:

DEFINE     table_name     = scott.emp
DEFINE     id_col          = empno
DEFINE     parent_id_col     = mgr
DEFINE     display_col     = ename

WITH     original_hierarchy     AS
(
     SELECT     &display_col          AS display_txt
     ,     LEVEL               AS lvl
     ,     CONNECT_BY_ISLEAF     AS isleaf
     ,     ROWNUM               AS rnum
     FROM     &table_name
     START WITH     &parent_id_col     IS NULL
     CONNECT BY     &parent_id_col     = PRIOR &id_col
)
,     got_max_lvl     AS
(
     SELECT     o.*
     ,     MAX (lvl) OVER ()     AS max_lvl
     FROM     original_hierarchy     o
)
SELECT       LPAD ( ' '
            , 3 * ( ( d.lvl
                 + NVL (c.rnum, 1)
                 - 1
                 )
               - 1
               )
            ) || CASE
               WHEN c.rnum > 1
               THEN '*' || d.display_txt || '*'
               ELSE        d.display_txt
              END          AS display_id
FROM            got_max_lvl     d
LEFT OUTER JOIN       got_max_lvl     c  ON     d.isleaf     = 1
                       AND     c.rnum          <= 1 + d.max_lvl - d.lvl
ORDER BY  d.rnum
,       c.rnum
;

Output:

DISPLAY_ID
-----------------------------
KING
   JONES
      SCOTT
         ADAMS
      FORD
         SMITH
   BLAKE
      ALLEN
         *ALLEN*
      WARD
         *WARD*
      MARTIN
         *MARTIN*
      TURNER
         *TURNER*
      JAMES
         *JAMES*
   CLARK
      MILLER
         *MILLER*

Published by: Frank Kulash, May 13, 2011 06:38
Adding the generic version

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I want to order the nodes of the tree using the column POSITION.

How can I do this?

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http://www.freeimagehosting.NET/uploads/7460540c52.gif
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I have this problem,
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TXS so you can help

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How to create multiple tables with SQL commands
Hi, first of all, is it possible to create a table and then one more both with SQL commands?
because I tried to do this in the following way
Example:
CREATE TABLE 'HB_Product')
"productId" number 4 NOT NULL.
"productName" VARCHAR2 (20) NOT NULL,
"prodPricePerTon" NUMBER (10,2).
"HB_Product_PK" PRIMARY KEY CONSTRAINT ("productId")
)
/
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"orderNo" NUMBER (5) NOT NULL,
'lineNo' number 4 NOT NULL.
"billNo" NUMBER (10) NOT NULL,
'actions' VARCHAR2 (15).
'place' VARCHAR2 (15).
"actDate" DATE NOT NULL,
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so you're saying I can't do several tables at the same time?

Yes... Well,... not this way

You can do this way
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create schema authorization alex
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"productId"    NUMBER(4) NOT NULL,
"productName"    VARCHAR2(20) NOT NULL,
"prodPricePerTon"    NUMBER(10,2),
CONSTRAINT "HB_Product_PK" PRIMARY KEY ("productId")
)
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"orderNo"    NUMBER(5) NOT NULL,
"lineNo"    NUMBER(4) NOT NULL,
"billNo"    NUMBER(10) NOT NULL,
"actions"    VARCHAR2(15),
"place"    VARCHAR2(15),
"actDate"    DATE NOT NULL,
"totWeight"    NUMBER(6,3),
CONSTRAINT "HB_Operations_PK" PRIMARY KEY ("lineNo")
)
/
```
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SQL query: How to match multiple names strictly with its appropriate values
Hi all

Assume there is a table of CODES (with 3 columns) as follows:...
------------------------------------------------
Name ID value
------------------------------------------------
SFO 100 080
200 NY 044
300. THE 040
------------------------------------------------

I would need to write a SQL to retrieve the rows that match few of names and values. For example: (1) name = "NY" and value = '044' and (2) name = "LA" and value = "040".

I wrote this sql...
SELECT id, name, value of CODES whose name IN ('NY', THE ') and the value IN ('044 ', ' 040')
This returns 2 rows as expected.

At the same time, even for the following query, the same 2 rows are returned (which is false)
SELECT id, name, value of CODES whose name in ('NY', THE ') and value in ('044 ', ' 444')
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My requirement is that the above query should return to the line for NY.

My question is: how to match multiple names strictly with its appropriate values... ? That is to say, unless '040' is specified in the part 'value', the corresponding line in 'LA' should not be returned...

I feel, nor the case AND neither the case of GOLD can be used in Where clause to check these 4 values.

Any ideas... ? Can you please let me know... ?

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as:
```
select id, name, value from CODES where (name, value) in (('NY', '044'), ('LA', '040'))
```
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How to order a tree balanced with SQL hierarchical queries

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