Oracle PL/SQL Tutorial/Aggregate Functions/Introduction — различия между версиями
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Версия 13:45, 26 мая 2010
Содержание
List employee whose salary is higher than the average salary
SQL>
SQL>
SQL> create table employees(
2 empno NUMBER(4)
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , msal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) ) ;
Table created.
SQL>
SQL>
SQL> insert into employees values(1,"Jason", "N", "TRAINER", 2, date "1965-12-18", 800 , NULL, 10);
1 row created.
SQL> insert into employees values(2,"Jerry", "J", "SALESREP",3, date "1966-11-19", 1600, 300, 10);
1 row created.
SQL> insert into employees values(3,"Jord", "T" , "SALESREP",4, date "1967-10-21", 1700, 500, 20);
1 row created.
SQL> insert into employees values(4,"Mary", "J", "MANAGER", 5, date "1968-09-22", 1800, NULL, 20);
1 row created.
SQL> insert into employees values(5,"Joe", "P", "SALESREP",6, date "1969-08-23", 1900, 1400, 30);
1 row created.
SQL> insert into employees values(6,"Black", "R", "MANAGER", 7, date "1970-07-24", 2000, NULL, 30);
1 row created.
SQL> insert into employees values(7,"Red", "A", "MANAGER", 8, date "1971-06-25", 2100, NULL, 40);
1 row created.
SQL> insert into employees values(8,"White", "S", "TRAINER", 9, date "1972-05-26", 2200, NULL, 40);
1 row created.
SQL> insert into employees values(9,"Yellow", "C", "DIRECTOR",10, date "1973-04-27", 2300, NULL, 20);
1 row created.
SQL> insert into employees values(10,"Pink", "J", "SALESREP",null,date "1974-03-28", 2400, 0, 30);
1 row created.
SQL>
SQL> select empno
2 from employees
3 where msal > avg(msal);
where msal > avg(msal)
*
ERROR at line 3:
ORA-00934: group function is not allowed here
SQL>
SQL>
SQL> drop table employees;
Table dropped.
SQL>
SQL>
Use the aggregate functions with valid expression
The following query passes the expression salary + 2 to AVG().
This adds 2 to each row"s salary column value and then returns the average of those values.
SQL> -- create demo table
SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL,
3 First_Name VARCHAR2(10 BYTE),
4 Last_Name VARCHAR2(10 BYTE),
5 Start_Date DATE,
6 End_Date DATE,
7 Salary Number(8,2),
8 City VARCHAR2(10 BYTE),
9 Description VARCHAR2(15 BYTE)
10 )
11 /
Table created.
SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values ("01","Jason", "Martin", to_date("19960725","YYYYMMDD"), to_date("20060725","YYYYMMDD"), 1234.56, "Toronto", "Programmer")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("02","Alison", "Mathews", to_date("19760321","YYYYMMDD"), to_date("19860221","YYYYMMDD"), 6661.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("03","James", "Smith", to_date("19781212","YYYYMMDD"), to_date("19900315","YYYYMMDD"), 6544.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("04","Celia", "Rice", to_date("19821024","YYYYMMDD"), to_date("19990421","YYYYMMDD"), 2344.78, "Vancouver","Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("05","Robert", "Black", to_date("19840115","YYYYMMDD"), to_date("19980808","YYYYMMDD"), 2334.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("06","Linda", "Green", to_date("19870730","YYYYMMDD"), to_date("19960104","YYYYMMDD"), 4322.78,"New York", "Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("07","David", "Larry", to_date("19901231","YYYYMMDD"), to_date("19980212","YYYYMMDD"), 7897.78,"New York", "Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("08","James", "Cat", to_date("19960917","YYYYMMDD"), to_date("20020415","YYYYMMDD"), 1232.78,"Vancouver", "Tester")
3 /
1 row created.
SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01 Jason Martin 25-JUL-96 25-JUL-06 1234.56 Toronto Programmer
02 Alison Mathews 21-MAR-76 21-FEB-86 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL>
SQL> SELECT AVG(salary + 2) FROM employee;
AVG(SALARY+2)
-------------
4073.7525
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
Use the DISTINCT keyword to exclude identical values from a group computation
The following query uses the DISTICT keyword to exclude identical values in the salary column when computing the average using AVG():
SQL> -- create demo table
SQL> create table Employee(
2 ID VARCHAR2(4 BYTE) NOT NULL,
3 First_Name VARCHAR2(10 BYTE),
4 Last_Name VARCHAR2(10 BYTE),
5 Start_Date DATE,
6 End_Date DATE,
7 Salary Number(8,2),
8 City VARCHAR2(10 BYTE),
9 Description VARCHAR2(15 BYTE)
10 )
11 /
Table created.
SQL>
SQL> -- prepare data
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values ("01","Jason", "Martin", to_date("19960725","YYYYMMDD"), to_date("20060725","YYYYMMDD"), 1234.56, "Toronto", "Programmer")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("02","Alison", "Mathews", to_date("19760321","YYYYMMDD"), to_date("19860221","YYYYMMDD"), 6661.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("03","James", "Smith", to_date("19781212","YYYYMMDD"), to_date("19900315","YYYYMMDD"), 6544.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("04","Celia", "Rice", to_date("19821024","YYYYMMDD"), to_date("19990421","YYYYMMDD"), 2344.78, "Vancouver","Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("05","Robert", "Black", to_date("19840115","YYYYMMDD"), to_date("19980808","YYYYMMDD"), 2334.78, "Vancouver","Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("06","Linda", "Green", to_date("19870730","YYYYMMDD"), to_date("19960104","YYYYMMDD"), 4322.78,"New York", "Tester")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("07","David", "Larry", to_date("19901231","YYYYMMDD"), to_date("19980212","YYYYMMDD"), 7897.78,"New York", "Manager")
3 /
1 row created.
SQL> insert into Employee(ID, First_Name, Last_Name, Start_Date, End_Date, Salary, City, Description)
2 values("08","James", "Cat", to_date("19960917","YYYYMMDD"), to_date("20020415","YYYYMMDD"), 1232.78,"Vancouver", "Tester")
3 /
1 row created.
SQL>
SQL>
SQL>
SQL> -- display data in the table
SQL> select * from Employee
2 /
ID FIRST_NAME LAST_NAME START_DAT END_DATE SALARY CITY DESCRIPTION
---- ---------- ---------- --------- --------- ---------- ---------- ---------------
01 Jason Martin 25-JUL-96 25-JUL-06 1234.56 Toronto Programmer
02 Alison Mathews 21-MAR-76 21-FEB-86 6661.78 Vancouver Tester
03 James Smith 12-DEC-78 15-MAR-90 6544.78 Vancouver Tester
04 Celia Rice 24-OCT-82 21-APR-99 2344.78 Vancouver Manager
05 Robert Black 15-JAN-84 08-AUG-98 2334.78 Vancouver Tester
06 Linda Green 30-JUL-87 04-JAN-96 4322.78 New York Tester
07 David Larry 31-DEC-90 12-FEB-98 7897.78 New York Manager
08 James Cat 17-SEP-96 15-APR-02 1232.78 Vancouver Tester
8 rows selected.
SQL>
SQL>
SQL> SELECT AVG(DISTINCT salary) FROM employee;
AVG(DISTINCTSALARY)
-------------------
4071.7525
SQL>
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
Using Aggregate Functions
Aggregate functions operate on a group of rows at the same time.
Aggregate functions return one row of output for each group of rows.
For example, computing the average salary of employees and finding the maximum salary of employees.
Aggregate functions are also called group functions because they operate on groups of rows.
The aggregate functions are mainly numerical.
Function Description AVG(x) Returns the average value of x. COUNT(x) Returns the number of rows returned by a query involving x. GLC Greatest lower bound of an MLSLABEL. LUB Least upper bound of an MLSLABEL. MAX(x) Returns the maximum value of x. MEDIAN(x) Returns the median value of x. MIN(x) Returns the minimum value of x. STDDEV(x) Returns the standard deviation of x. SUM(x) Returns the sum of x. VARIANCE(x) Returns the variance of x.
Here are some points to remember when using aggregate functions
- COUNT(), MAX(), and MIN() functions operate numbers, strings, and datetimes.
- Null values are ignored by aggregate functions.
- DISTINCT keyword with an aggregate function excludes duplicate entries from the aggregate function"s calculation.
