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Combining aggregate and non-aggregate values in SQL using Joins and Over clause

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  • Difficulty Level : Medium
  • Last Updated : 31 Aug, 2021
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Prerequisite – Aggregate functions in SQL, Joins in SQL
Aggregate functions perform a calculation on a set of values and return a single value. Now, consider an employee table EMP and a department table DEPT with following structure:

Table – EMPLOYEE TABLE

Name Null Type
EMPNO NOT NULL NUMBER(4)
ENAME   VARCHAR2(10)
JOB   VARCHAR2(9)
MGR   NUMBER(4)
HIREDATE   DATE
SAL   NUMBER(7, 2)
COMM   NUMBER(7, 2)
DEPTNO   NUMBER(2)

Table – DEPARTMENT TABLE

Name Null Type
DEPTNO   NUMBER(2)
DNAME   VARCHAR2(14)
LOC   VARCHAR2(13)

And the following results are needed:

  1. DISPLAY NAME, SAL, JOB OF EMP ALONG WITH MAX, MIN, AVG, TOTAL SAL OF THE EMPS DOING THE SAME JOB.
  2. DISPLAY DEPTNAME WITH NUMBER OF EMP WORKING IN IT.

The aggregated values can’t be directly used with non-aggregated values to obtain a result. Thus one can use the following concepts:

1. Using Joins –

  1. Create a sub-table containing the result of aggregated values.
  2. Using Join, use the results from the sub-table to display them with non-aggregated values.

Solutions for problem 1 using JOIN:

SELECT ENAME, SAL, EMP.JOB, 
            SUBTABLE.MAXSAL, SUBTABLE.MINSAL, 
            SUBTABLE.AVGSAL, SUBTABLE.SUMSAL
FROM EMP
INNER JOIN
        (SELECT JOB, MAX(SAL) MAXSAL, MIN(SAL) 
                MINSAL, AVG(SAL) AVGSAL, SUM(SAL) SUMSAL 
         FROM EMP 
          GROUP BY JOB) SUBTABLE
                  ON EMP.JOB = SUBTABLE.JOB; 

Output for sample data:

Ename Sal Job MaxSal MinSal AvgSal SumSal
SCOTT 3300 ANALYST 3300 1925 2841.67 8525
HENRY 1925 ANALYST 3300 1925 2841.67 8525
FORD 3300 ANALYST 3300 1925 2841.67 8525
SMITH 3300 CLERK 3300 1045 1746.25 6985
MILLER 1430 CLERK 3300 1045 1746.25 6985

2. Using ‘Over’ clause –

  1. OVER CLAUSE ALONG WITH PARTITION BY IS USED TO BRAKE UP DATA INTO PARTITIONS.
  2. THE SPECIFIED FUNCTION OPERATES FOR EACH PARTITION.

Solutions for problem 2 using OVER Clause:

SELECT DISTINCT(DNAME),
COUNT(ENAME) OVER (PARTITION BY EMP.DEPTNO) EMP
FROM EMP
RIGHT OUTER JOIN DEPT
ON EMP.DEPTNO=DEPT.DEPTNO
ORDER BY EMP DESC; 
Dname Emp
SALES 6
RESEARCH 5
ACCOUNTING 3
OPERATIONS 0
OTHERS 0

 

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