DB2 Common table expressions

Common table expressions

A common table expression is like a temporary view that is defined and used for the duration of an SQL statement.

You can define a common table expression wherever you can have a fullselect statement. For example, you can include a common table expression in a SELECT, INSERT, SELECT INTO, or CREATE VIEW statement.

Each common table expression must have a unique name and be defined only once. However, you can reference a common table expression many times in the same SQL statement. Unlike regular views or nested table expressions, which derive their result tables for each reference, all references to common table expressions in a given statement share the same result table.

You can use a common table expression in the following situations:

• When you want to avoid creating a view (when general use of the view is not required, and positioned updates or deletes are not used)
• When the desired result table is based on host variables
• When the same result table needs to be shared in a fullselect
• When the results need to be derived using recursion

Creating a common table expression

Creating a common table expression saves you the overhead of creating and dropping a regular view that you need to use only once. Also, during statement preparation, DB2® does not need to access the catalog for the view, which saves you additional overhead.

Use the WITH clause to create a common table expression.

You can use a common table expression in a SELECT statement by using the WITH clause at the beginning of the statement.

Example: WITH clause in a SELECT statement: The following statement finds the department with the highest total pay. The query involves two levels of aggregation. First, you need to determine the total pay for each department by using the SUM function and order the results by using the GROUP BY clause. You then need to find the department with highest total pay based on the total pay for each department.

WITH DTOTAL (deptno, totalpay) AS

               (SELECT deptno, sum(salary+bonus)

                   FROM  DSN8810.EMP

                   GROUP BY deptno)

        SELECT deptno

      FROM DTOTAL

           WHERE totalpay = (SELECT max(totalpay)

                                          FROM  DTOTAL);

The result table for the common table expression, DTOTAL, contains the department number and total pay for each department in the employee table. The fullselect in the previous example uses the result table for DTOTAL to find the department with the highest total pay. The result table for the entire statement looks similar to the following results:

DEPTNO

======

D11

Using common table expressions with views:

You can use common table expressions before a fullselect in a CREATE VIEW statement. This technique is useful if you need to use the results of a common table expression in more than one query.

Example: Using a WITH clause in a CREATE VIEW statement: The following statement finds the departments that have a greater-than-average total pay and saves the results as the view RICH_DEPT:

CREATE VIEW RICH_DEPT (deptno) AS

             WITH DTOTAL (deptno, totalpay) AS

                 (SELECT deptno, sum(salary+bonus)

                     FROM DSN81010.EMP

                     GROUP BY deptno)

             SELECT deptno

           FROM DTOTAL

                WHERE totalpay > (SELECT AVG(totalpay)

                                               FROM DTOTAL);

The fullselect in the previous example uses the result table for DTOTAL to find the departments that have a greater-than-average total pay. The result table is saved as the RICH_DEPT view and looks similar to the following results:

DEPTNO

======

A00

D11

D21

Using common table expressions when you use INSERT:

You can use common table expressions before a fullselect in an INSERT statement.

Example: Using a common table expression in an INSERT statement: The following statement uses the result table for VITALDEPT to find the manager's number for each department that has a greater-than-average number of senior engineers. Each manager's number is then inserted into the vital_mgr table.

INSERT INTO vital_mgr (mgrno)

          WITH VITALDEPT (deptno, se_count)  AS

                 (SELECT deptno, count(*)

                     FROM DSN81010.EMP

                     WHERE job = 'senior engineer'

                      GROUP BY deptno)

          SELECT d.manager

             FROM DSN81010.DEPT d, VITALDEPT s

             WHERE d.deptno = s.deptno

                 AND s.se_count  >  (SELECT  AVG(se_count)

                                                           FROM VITALDEPT);

Creating recursive SQL by using common table expressions

Queries that use recursion are useful in applications like bill-of-materials applications, network planning applications, and reservation systems.

You can use common table expressions to create recursive SQL If a fullselect of a common table expression contains a reference to itself in a FROM clause, the common table expression is a recursive common table expression.

Recursive common table expressions must follow these rules:

• The first fullselect of the first union (the initialization fullselect) must not include a reference to the common table expression.
• Each fullselect that is part of the recursion cycle must:
• Start with SELECT or SELECT ALL. SELECT DISTINCT is not allowed.
• Include only one reference to the common table expression that is part of the recursion cycle in its FROM clause.
• Not include aggregate functions, a GROUP BY clause, or a HAVING clause.
• The column names must be specified after the table name of the common table expression.
• The data type, length, and CCSID of each column from the common table expression must match the data type, length, and CCSID of each corresponding column in the iterative fullselect.
• If you use the UNION keyword, specify UNION ALL instead of UNION.
• You cannot specify INTERSECT or EXCEPT.
• Outer joins must not be part of any recursion cycle.
• A subquery must not be part of any recursion cycle.

Important: You should be careful to avoid an infinite loop when you use a recursive common table expression. DB2® issues a warning if one of the following items is not found in the iterative fullselect of a recursive common table expression:

• An integer column that increments by a constant
• A predicate in the WHERE clause in the form of counter_column < constant or counter_column < :host variable

Examples of recursive common table expressions

Recursive SQL is very useful in bill of materials (BOM) applications.

Consider a table of parts with associated subparts and the quantity of subparts required by each part

For the examples in this topic, create the following table:

CREATE TABLE PARTLIST

     (PART VARCHAR(8),

      SUBPART VARCHAR(8),

      QUANTITY INTEGER);

Assume that the PARTLIST table is populated with the values that are in the following table:

Table 1. PARTLIST table
PART	SUBPART	QUANTITY
00	01	5
00	05	3
01	02	2
01	03	3
01	04	4
01	06	3
02	05	7
02	06	6
03	07	6
04	08	10
04	09	11
05	10	10
05	11	10
06	12	10
06	13	10
07	14	8
07	12	8

Example 1: Single level explosion:

Single level explosion answers the question, "What parts are needed to build the part identified by '01'?". The list will include the direct subparts, subparts of the subparts and so on. However, if a part is used multiple times, its subparts are only listed once.

WITH RPL (PART, SUBPART, QUANTITY) AS

     (SELECT ROOT.PART, ROOT.SUBPART, ROOT.QUANTITY

        FROM PARTLIST ROOT

        WHERE ROOT.PART = '01'

     UNION ALL

        SELECT CHILD.PART, CHILD.SUBPART, CHILD.QUANTITY

           FROM RPL PARENT, PARTLIST CHILD

           WHERE PARENT.SUBPART = CHILD.PART)

SELECT DISTINCT PART, SUBPART, QUANTITY

    FROM RPL

    ORDER BY PART, SUBPART, QUANTITY;

The preceding query includes a common table expression, identified by the name RPL, that expresses the recursive part of this query. It illustrates the basic elements of a recursive common table expression.

The first operand (fullselect) of the UNION, referred to as the initialization fullselect, gets the direct subparts of part '01'. The FROM clause of this fullselect refers to the source table and will never refer to itself (RPL in this case). The result of this first fullselect goes into the common table expression RPL. As in this example, the UNION must always be a UNION ALL.

The second operand (fullselect) of the UNION uses RPL to compute subparts of subparts by using the FROM clause to refer to the common table expression RPL and the source table PARTLIST with a join of a part from the source table (child) to a subpart of the current result contained in RPL (parent). The result goes then back to RPL again. The second operand of UNION is used repeatedly until no more subparts exist.

The SELECT DISTINCT in the main fullselect of this query ensures the same part/subpart is not listed more than once.

The result of the query is shown in the following table:

Table 2. Result table for example 1
PART	SUBPART	QUANTITY
01	02	2
01	03	3
01	04	4
01	06	3
02	05	7
02	06	6
03	07	6
04	08	10
04	09	11
05	10	10
05	11	10
06	12	10
06	13	10
07	12	8
07	14	8

Observe in the result that part '01' contains subpart '02' which contains subpart '06' and so on. Further, notice that part '06' is reached twice, once through part '01' directly and another time through part '02'. In the output, however, the subparts of part '06' are listed only once (this is the result of using a SELECT DISTINCT).

Remember that with recursive common table expressions it is possible to introduce an infinite loop. In this example, an infinite loop would be created if the search condition of the second operand that joins the parent and child tables was coded as follows:

WHERE PARENT.SUBPART = CHILD.SUBPART

This infinite loop is created by not coding what is intended. You should carefully determining what to code so that there is a definite end of the recursion cycle.

The result produced by this example could be produced in an application program without using a recursive common table expression. However, such an application would require coding a different query for every level of recursion. Furthermore, the application would need to put all of the results back in the database to order the final result. This approach complicates the application logic and does not perform well. The application logic becomes more difficult and inefficient for other bill of material queries, such as summarized and indented explosion queries.

Example 2: Summarized explosion:

A summarized explosion answers the question, "What is the total quantity of each part required to build part '01'?" The main difference from a single level explosion is the need to aggregate the quantities. A single level explosion indicates the quantity of subparts required for the part whenever it is required. It does not indicate how many of each subpart is needed to build part '01'.

WITH RPL (PART, SUBPART, QUANTITY) AS

           (

              SELECT ROOT.PART, ROOT.SUBPART, ROOT.QUANTITY

                 FROM PARTLIST ROOT

                 WHERE ROOT.PART = '01'

             UNION ALL

              SELECT PARENT.PART, CHILD.SUBPART, 

                     PARENT.QUANTITY*CHILD.QUANTITY

                 FROM RPL PARENT, PARTLIST CHILD

                 WHERE PARENT.SUBPART = CHILD.PART

           )

SELECT PART, SUBPART, SUM(QUANTITY) AS "Total QTY Used"

   FROM RPL

   GROUP BY PART, SUBPART

   ORDER BY PART, SUBPART;

In the preceding query, the select list of the second operand of the UNION in the recursive common table expression, identified by the name RPL, shows the aggregation of the quantity. To determine how many of each subpart is used, the quantity of the parent is multiplied by the quantity per parent of a child. If a part is used multiple times in different places, it requires another final aggregation. This is done by the grouping the parts and subparts in the common table expression RPL and using the SUM column function in the select list of the main fullselect.

The result of the query is shown in the following table:

Table 3. Result table for example 2
PART	SUBPART	Total QTY Used
01	02	2
01	03	3
01	04	4
01	05	14
01	06	15
01	07	18
01	08	40
01	09	44
01	10	140
01	11	140
01	12	294
01	13	150
01	14	144

Consider the total quantity for subpart '06'. The value of 15 is derived from a quantity of 3 directly for part '01' and a quantity of 6 for part '02' which is needed two times by part '01'.

Example 3: Controlling depth:

You can control the depth of a recursive query to answer the question, "What are the first two levels of parts that are needed to build part '01'?" For the sake of clarity in this example, the level of each part is included in the result table.

WITH RPL (LEVEL, PART, SUBPART, QUANTITY) AS

       (

         SELECT 1, ROOT.PART, ROOT.SUBPART, ROOT.QUANTITY

           FROM PARTLIST ROOT

           WHERE ROOT.PART = '01'

     UNION ALL

         SELECT PARENT.LEVEL+1, CHILD.PART, CHILD.SUBPART, CHILD.QUANTITY

           FROM RPL PARENT, PARTLIST CHILD

           WHERE PARENT.SUBPART = CHILD.PART

             AND PARENT.LEVEL < 2

        )

SELECT PART, LEVEL, SUBPART, QUANTITY

  FROM RPL;

This query is similar to the query in example 1. The column LEVEL is introduced to count the level each subpart is from the original part. In the initialization fullselect, the value for the LEVEL column is initialized to 1. In the subsequent fullselect, the level from the parent table increments by 1. To control the number of levels in the result, the second fullselect includes the condition that the level of the parent must be less than 2. This ensures that the second fullselect only processes children to the second level.

The result of the query is shown in the following table:

Table 4. Result table for example 3
PART	LEVEL	SUBPART	QUANTITY
01	1	02	2
01	1	03	3
01	1	04	4
01	1	06	3
02	2	05	7
02	2	06	6
03	2	07	6
04	2	08	10
04	2	09	11
06	2	12	10
06	2	13	10