Oracle PL/SQL Tutorial/Aggregate Functions/SUM
Содержание
Sum column for a certain time period
SQL> CREATE TABLE sales(
2 product_id NUMBER(6),
3 cid NUMBER,
4 time_id DATE,
5 channel_id CHAR(1),
6 promo_id NUMBER(6),
7 sold NUMBER(3),
8 amount NUMBER(10,2),
9 cost NUMBER(10,2)
10 );
Table created.
SQL>
SQL> select sum(amount)
2 from sales
3 where time_id between to_date("2006-02-01", "YYYY-MM-DD") and to_date("2006-02-28", "YYYY-MM-DD");
SUM(AMOUNT)
-----------
1 row selected.
SQL>
SQL> drop table sales;
Table dropped.
Sum() function and having clause
SQL>
SQL> CREATE TABLE server_usage (
2 pro_id NUMBER(4),
3 emp_id NUMBER,
4 time_log_date DATE,
5 hours_logged NUMBER(8,2),
6 dollars_charged NUMBER(8,2),
7 CONSTRAINT server_usage_pk PRIMARY KEY (pro_id, emp_id, time_log_date)
8 );
Table created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1001,101,to_date("4-Apr-2004","dd-mon-yyyy"),1123,222);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1002,102,to_date("4-Apr-2005","dd-mon-yyyy"),1124,223);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1003,103,to_date("4-Apr-2006","dd-mon-yyyy"),1125,224);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1004,104,to_date("4-Apr-2007","dd-mon-yyyy"),1126,225);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1005,105,to_date("4-Apr-2008","dd-mon-yyyy"),1127,226);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1001,106,to_date("4-Apr-2009","dd-mon-yyyy"),1128,227);
1 row created.
SQL> INSERT INTO server_usage(pro_id, emp_id, time_log_date, hours_logged, dollars_charged)
2 VALUES (1002,107,to_date("4-Apr-2010","dd-mon-yyyy"),1129,228);
1 row created.
SQL>
SQL> SET ECHO ON
SQL> SELECT emp_id, pro_id
2 FROM server_usage
3 WHERE pro_id = 1001 OR pro_id=1002
4 GROUP BY emp_id, pro_id
5 HAVING SUM(hours_logged) > 20;
101 1001
102 1002
106 1001
107 1002
4 rows selected.
SQL>
SQL>
SQL> drop table server_usage;
Table dropped.
Sum salary group by department number
SQL> create table emp
2 ( empno NUMBER(4) constraint E_PK primary key
3 , ename VARCHAR2(8)
4 , init VARCHAR2(5)
5 , job VARCHAR2(8)
6 , mgr NUMBER(4)
7 , bdate DATE
8 , sal NUMBER(6,2)
9 , comm NUMBER(6,2)
10 , deptno NUMBER(2) default 10
11 ) ;
Table created.
SQL> insert into emp values(1,"Tom","N", "TRAINER", 13,date "1965-12-17", 800 , NULL, 20);
1 row created.
SQL> insert into emp values(2,"Jack","JAM", "Tester",6,date "1961-02-20", 1600, 300, 30);
1 row created.
SQL> insert into emp values(3,"Wil","TF" , "Tester",6,date "1962-02-22", 1250, 500, 30);
1 row created.
SQL> insert into emp values(4,"Jane","JM", "Designer", 9,date "1967-04-02", 2975, NULL, 20);
1 row created.
SQL> insert into emp values(5,"Mary","P", "Tester",6,date "1956-09-28", 1250, 1400, 30);
1 row created.
SQL> insert into emp values(6,"Black","R", "Designer", 9,date "1963-11-01", 2850, NULL, 30);
1 row created.
SQL> insert into emp values(7,"Chris","AB", "Designer", 9,date "1965-06-09", 2450, NULL, 10);
1 row created.
SQL> insert into emp values(8,"Smart","SCJ", "TRAINER", 4,date "1959-11-26", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(9,"Peter","CC", "Designer",NULL,date "1952-11-17", 5000, NULL, 10);
1 row created.
SQL> insert into emp values(10,"Take","JJ", "Tester",6,date "1968-09-28", 1500, 0, 30);
1 row created.
SQL> insert into emp values(11,"Ana","AA", "TRAINER", 8,date "1966-12-30", 1100, NULL, 20);
1 row created.
SQL> insert into emp values(12,"Jane","R", "Manager", 6,date "1969-12-03", 800 , NULL, 30);
1 row created.
SQL> insert into emp values(13,"Fake","MG", "TRAINER", 4,date "1959-02-13", 3000, NULL, 20);
1 row created.
SQL> insert into emp values(14,"Mike","TJA","Manager", 7,date "1962-01-23", 1300, NULL, 10);
1 row created.
SQL>
SQL> select deptno
2 , sum(sal)
3 from emp
4 group by deptno;
DEPTNO SUM(SAL)
---------- ----------
10 8750
20 10875
30 9250
SQL>
SQL>
SQL>
SQL> drop table emp;
Table dropped.
Sum with "group by cude"
SQL>
SQL>
SQL> select owner,
2 tablespace_name,
3 sum(bytes)/(1024*1024) total_meg
4 from dba_extents
5 where rownum < 50
6 group by cube (owner,tablespace_name)
7 /
OWNER TABLESPACE_NAME TOTAL_MEG
------------------------------ ------------------------------ ----------
4
SYSTEM 4
SYS 4
SYS SYSTEM 4
Sum with null value
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", NULL, 0, 30);
1 row created.
SQL>
SQL> select
2 sum(msal)
3 from employees;
SQL>
SQL>
SQL> drop table employees;
Table dropped.
SQL>
SQL>
SUM(x) adds all the values in x and returns the total.
The following example displays the sum of the salary column from the employee table using SUM():
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 SUM(salary) FROM employee;
SUM(SALARY)
-----------
32574.02
SQL>
SQL>
SQL> -- clean the table
SQL> drop table Employee
2 /
Table dropped.
SQL>
SQL>