Oracle PL/SQL/Numeric Math Functions/FLOOR
Содержание
FLOOR: Returns the floor value (next lowest integer below number)
SQL>
SQL>
SQL> -- create demo table
SQL> create table TestTable(
2 ID VARCHAR2(4 BYTE) NOT NULL,
3 MyName VARCHAR2(10 BYTE),
4 MyDate DATE,
5 MyNumber Number(8,2)
6 )
7 /
Table created.
SQL>
SQL>
SQL> insert into TestTable (ID, MyName, MyDate, MyNumber) values("1","Alison",to_date("19960711","YYYYMMDD"),12.12);
1 row created.
SQL> insert into TestTable (ID, MyName, MyDate, MyNumber) values("2","Jason",to_date("19970622","YYYYMMDD"),-12.12);
1 row created.
SQL> insert into TestTable (ID, MyName, MyDate, MyNumber) values("3","Smith",to_date("19980513","YYYYMMDD"),22.1);
1 row created.
SQL> insert into TestTable (ID, MyName, MyDate, MyNumber) values("4","Tailor",to_date("19990624","YYYYMMDD"),-2.12);
1 row created.
SQL> insert into TestTable (ID, MyName, MyDate, MyNumber) values("5","Darlene",to_date("20000415","YYYYMMDD"),2.1);
1 row created.
SQL>
SQL>
SQL> select * from TestTable;
ID MYNAME MYDATE MYNUMBER
---- ---------- --------- ----------
1 Alison 11-JUL-96 12.12
2 Jason 22-JUN-97 -12.12
3 Smith 13-MAY-98 22.1
4 Tailor 24-JUN-99 -2.12
5 Darlene 15-APR-00 2.1
SQL>
SQL>
SQL>
SQL> -- FLOOR: Returns the floor value (next lowest integer below number).
SQL>
SQL> select MyNumber, FLOOR(MyNumber) from TestTable;
MYNUMBER FLOOR(MYNUMBER)
---------- ---------------
12.12 12
-12.12 -13
22.1 22
-2.12 -3
2.1 2
SQL>
SQL>
SQL>
SQL> drop table TestTable;
Table dropped.
SQL>
SQL>
FLOOR(x): Returns the largest integer less than or equal to x
SQL> --FLOOR(x): Returns the largest integer less than or equal to x.
SQL>
SQL> select FLOOR(-5.2) from dual;
FLOOR(-5.2)
-----------
-6
SQL>
round(345.678), ceil(345.678), floor(345.678)
SQL>
SQL> select round(345.678), ceil(345.678), floor(345.678)
2 from dual;
ROUND(345.678) CEIL(345.678) FLOOR(345.678)
-------------- ------------- --------------
346 346 345
Use floor(), mod() and date calculation to show how many weeks and days an employee has been working here
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", "Coder", 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", "Coder", 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", "Coder", 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", "Coder", 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 ename
2 , floor( (sysdate-bdate)/7) as weeks
3 , floor(mod(sysdate-bdate,7)) as days
4 from emp
5 where deptno = 10;
ENAME WEEKS DAYS
-------- ---------- ----------
Chris 2315 5
Peter 2971 0
Mike 2491 6
SQL>
SQL> drop table emp;
Table dropped.