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RETRIEVAL: MULTIPLE TABLES AND AGGREGATION
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COUNT(DEPTNO) COUNT(DISTINCTDEPTNO) AVG(COMM) AVG(COALESCE(COMM,0)) ------------- --------------------- --------- --------------------14 3 550 157.1429 Note that Listing 8-19 also shows that you can use group functions in the SELECT clause of a query without a GROUP BY clause. The absence of a GROUP BY clause in combination with the presence of group functions in the SELECT clause always results in a single-row result. In other words, the full table is aggregated into a single row. You can achieve precisely the same result by grouping on a constant expression. Try this yourself; for example, see what happens if you add GROUP BY 'x' to the query in Listing 8-19.
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Group Functions and Null Values
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The ANSI/ISO SQL standard postulates group functions to ignore null values completely. There is only one exception to this rule: the COUNT(*) function. This special case is discussed later in this section. This is a reasonable compromise. The only other consistent behavior for group functions would be to return a null value as soon as the input contains a null value. This would imply that all your SQL statements (containing group functions) should contain additional code to handle null values explicitly. So, ignoring null values completely is not a bad idea. Just make sure that you understand the consequences of this behavior. See Table 8-2 for some typical examples. Table 8-2. Behavior of Group Functions and Null Values
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Set X
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{1,2,3,NULL} {1,2,3,0} {1,2,3,2}
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AVG(X)
2 1.5 2
MAX(X)
3 3 3
The SUM function does not make any distinction between {1,2,3,NULL} and {1,2,3,0}. The MIN and AVG functions don t make any distinction between {1,2,3,NULL} and {1,2,3,2}. The MAX function gives the same result on all three sets. Looking back at Listing 8-19, you see an example of function nesting: the AVG function operates on the result of the COALESCE function. This is a typical method to handle null values explicitly. As you can see from Listing 8-19, the results of AVG(COMM) and AVG(COALESCE(COMM,0)) are obviously different. In this case, the Oracle DBMS replaces all null values by zeros before applying the AVG function, because the null values in the COMM column actually mean not applicable. The next query, shown in Listing 8-20, tells us how many different courses are scheduled for each trainer and the total number of scheduled courses. Listing 8-20. GROUP BY and DISTINCT select trainer , count(distinct course) , count(*)
RETRIEVAL: MULTIPLE TABLES AND AGGREGATION
from group
offerings by trainer;
TRAINER COUNT(DISTINCTCOURSE) COUNT(*) -------- --------------------- -------7369 2 3 7566 2 2 7788 2 2 7876 1 1 7902 2 2 3 3 Apparently, we have three course offerings without a trainer being assigned.
Grouping the Results of a Join
The query in Listing 8-21 shows the average evaluation ratings for each trainer, over all courses delivered. Listing 8-21. GROUP BY on a Join select o.trainer, avg(r.evaluation) from offerings o join registrations r using (course,begindate) group by o.trainer; TRAINER AVG(R.EVALUATION) -------- ----------------7369 4 7566 4.25 7788 7876 4 7902 4 Notice the USING clause in line 5, with the COURSE and BEGINDATE columns. This USING clause with two columns is needed to get the correct join results.
The COUNT(*) Function
As mentioned earlier, group functions operate on a set of values, with one important exception. Besides column names, you can specify the asterisk (*) as an argument to the COUNT function. This widens the scope of the COUNT function from a specific column to the full row level. COUNT(*) returns the number of rows in the entire group.
RETRIEVAL: MULTIPLE TABLES AND AGGREGATION
Note If you think that SELECT COUNT(1) is faster than SELECT COUNT(*), try a little experiment and prepare to be surprised you will find out that there is no difference. Don t trust opinions...
Listing 8-20 already showed an example of using the COUNT(*) function, to get the total number of scheduled courses for each trainer from the OFFERINGS table. Listing 8-22 shows another example of using the COUNT(*) function, this time applied against the EMPLOYEES table. Listing 8-22. Count Employees Per Department (First Attempt) select e.deptno, count(*) from employees e group by e.deptno; DEPTNO COUNT(*) -------- -------10 3 20 5 30 6 Obviously, department 40 is missing in this result. If you want to change the query into an outer join in order to show department 40 as well, you must be careful. What s wrong with the query in Listing 823 Apparently, we suddenly have one employee working for department 40. Listing 8-23. Count Employees Per Department (Second Attempt) select deptno, count(*) from employees e right outer join departments d using (deptno) group by deptno; DEPTNO COUNT(*) -------- -------10 3 20 5 30 6 40 1 Compare the results in Listing 8-23 with the results in Listing 8-24. The only difference is the argument of the COUNT function. Listing 8-24 obviously shows the correct result, because department 40 has no employees. By counting over the primary key e.EMPNO, you are sure that all real employees are counted, while the null value introduced by the outer join is correctly ignored. You could have used any other NOT NULL column as well.
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