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8.10 Set Operators
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You can use the SQL set operators UNION, MINUS, and INTERSECT to combine the results of two independent query blocks into a single result. As you saw in 2, the set operators have the syntax shown in Figure 8-5.
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Figure 8-5. Set operators syntax diagram These SQL operators correspond with the union, minus, and intersect operators you know from mathematics. Don t we all have fond memories of our teachers drawing those Venn diagrams on the whiteboard (or blackboard, for you older readers) See also Figure 1-1 (in 1). The meanings of these set operators in SQL are listed in Table 8-4.
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RETRIEVAL: MULTIPLE TABLES AND AGGREGATION
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Table 8-4. Set Operators
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Q1 UNION Q2 Q1 UNION ALL Q2 Q1 MINUS Q2 Q1 INTERSECT Q2
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All rows occurring in Q1 or in Q2 (or in both) As UNION, retaining duplicate rows The rows from Q1, without the rows from Q2 The rows occurring in Q1 and in Q2
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By default, all three set operators suppress duplicaterows in the query result. The only exception to this rule is the UNION ALL operator, which does not eliminate duplicate rows. One important advantage of the UNION ALL operator is that the Oracle DBMS does not need to sort the rows. Sorting is needed for all other set operators to trace duplicate rows. The UNION, MINUS, and INTERSECT operators cannot be applied to any arbitrary set of two queries. The intermediate (separate) results of queries Q1 and Q2 must be compatible in order to use them as arguments to a set operator. In this context, compatibility means the following: Q1 and Q2 must select the same number of column expressions. The datatypes of those column expressions must match.
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Some other rules and guidelines for SQL set operators are the following: The result table inherits the column names (or aliases) from Q1. Q1 cannot contain an ORDER BY clause. If you specify an ORDER BY clause at the end of the query, it doesn t refer to Q2, but rather to the total result of the set operator.
Set operators are very convenient when building new queries by combining the multiple query blocks you wrote (and tested) before, without writing completely new SQL code. This simplifies testing, because you have more control over correctness. Listing 8-44 answers the following question: Which locations host course offerings without having a department Listing 8-44. MINUS Set Operator Example select o.location from offerings o MINUS select d.location from departments d; LOCATION -------SEATTLE You can also try to solve this problem without using the MINUS operator. See Listing 8-45 for a suggestion.
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Listing 8-45. Alternative Solution Without Using the MINUS Operator select DISTINCT o.location from offerings o where o.location not in (select d.location from departments d) Note that you must add a DISTINCT operator, to handle situations where you have multiple course offerings in the same location. As explained before, the MINUS operator automatically removes duplicate rows. Are the two queries in Listing 8-44 and 8-45 logically equivalent They appear to be logically the same, but they are not quite as identical logically as they first appear. The first query will return two rows. One is for Seattle. The other is a null, representing the one course offering with an unknown location. The MINUS operator does not remove the null value, whereas that same null value fails to pass the WHERE condition in Listing 8-45. This is just one more example of the subtle pitfalls inherent in dealing with nulls in your data. You can also produce outer join results by using the UNION operator. You will see how to do this in Listings 8-46 and 8-47. We start with a regular join in Listing 8-46. In Listing 8-47 you add the additional department(s) needed for the outer join with a UNION operator, while assigning the right number of employees for those departments: zero. Listing 8-46. Regular Join select , , from , where group by , DEPTNO -------10 20 30 d.deptno d.dname count(e.empno) as headcount employees e departments d e.deptno = d.deptno d.deptno d.dname; DNAME HEADCOUNT ---------- --------ACCOUNTING 3 TRAINING 5 SALES 6
Listing 8-47. Expansion to an Outer Join with a UNION Operator select , , from , where group by , union d.deptno d.dname count(e.empno) as headcount employees e departments d e.deptno = d.deptno d.deptno d.dname
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