INTERVAL ‘integer [- integer]‘ {YEAR | MONTH} [(precision)][TO {YEAR | MONTH}]
该数据类型常用来表示一段时间差, 注意时间差只精确到年和月. precision为年或月的精确域, 有效范围是0到9, 默认值为2.
eg:
INTERVAL ‘123-2‘ YEAR(3) TO MONTH
表示: 123年2个月, "YEAR(3)" 表示年的精度为3, 可见"123"刚好为3为有效数值, 如果该处YEAR(n), n<3就会出错, 注意默认是2.
INTERVAL ‘123‘ YEAR(3)
表示: 123年0个月
INTERVAL ‘300‘ MONTH(3)
表示: 300个月, 注意该处MONTH的精度是3啊.
INTERVAL ‘4‘ YEAR
表示: 4年, 同 INTERVAL ‘4-0‘ YEAR TO MONTH 是一样的
INTERVAL ‘50‘ MONTH
表示: 50个月, 同 INTERVAL ‘4-2‘ YEAR TO MONTH 是一样
INTERVAL ‘123‘ YEAR
表示: 该处表示有错误, 123精度是3了, 但系统默认是2, 所以该处应该写成 INTERVAL ‘123‘ YEAR(3) 或"3"改成大于3小于等于9的数值都可以的
INTERVAL ‘5-3‘ YEAR TO MONTH + INTERVAL ‘20‘ MONTH =
INTERVAL ‘6-11‘ YEAR TO MONTH
表示: 5年3个月 + 20个月 = 6年11个月
与该类型相关的函数:
NUMTODSINTERVAL(n, ‘interval_unit‘)
将n转换成interval_unit所指定的值, interval_unit可以为: DAY, HOUR, MINUTE, SECOND
注意该函数不可以转换成YEAR和MONTH的.
NUMTOYMINTERVAL(n, ‘interval_unit‘)
interval_unit可以为: YEAR, MONTH
eg: (Oracle Version 9204, RedHat Linux 9.0)
SQL> select numtodsinterval(100,‘DAY‘) from dual;
NUMTODSINTERVAL(100,‘DAY‘)
---------------------------------------------------------------------------
+000000100 00:00:00.000000000
SQL> c/DAY/SECOND
1* select numtodsinterval(100,‘SECOND‘) from dual
SQL> /
NUMTODSINTERVAL(100,‘SECOND‘)
---------------------------------------------------------------------------
+000000000 00:01:40.000000000
SQL> c/SECOND/MINUTE
1* select numtodsinterval(100,‘MINUTE‘) from dual
SQL> /
NUMTODSINTERVAL(100,‘MINUTE‘)
---------------------------------------------------------------------------
+000000000 01:40:00.000000000
SQL> c/MINUTE/HOUR
1* select numtodsinterval(100,‘HOUR‘) from dual
SQL> /
NUMTODSINTERVAL(100,‘HOUR‘)
---------------------------------------------------------------------------
+000000004 04:00:00.000000000
SQL> c/HOUR/YEAR
1* select numtodsinterval(100,‘YEAR‘) from dual
SQL> /
select numtodsinterval(100,‘YEAR‘) from dual
*
ERROR at line 1:
ORA-01760: illegal argument for function
SQL> select numtoyminterval(100,‘year‘) from dual;
NUMTOYMINTERVAL(100,‘YEAR‘)
---------------------------------------------------------------------------
+000000100-00
SQL> c/year/month
1* select numtoyminterval(100,‘month‘) from dual
SQL> /
NUMTOYMINTERVAL(100,‘MONTH‘)
---------------------------------------------------------------------------
+000000008-04
时间的计算:
SQL> select to_date(‘1999-12-12‘,‘yyyy-mm-dd‘) - to_date(‘1999-12-01‘,‘yyyy-mm-dd‘) from dual;
TO_DATE(‘1999-12-12‘,‘YYYY-MM-DD‘)-TO_DATE(‘1999-12-01‘,‘YYYY-MM-DD‘)
---------------------------------------------------------------------
11
-- 可以相减的结果为天.
SQL> c/1999-12-12/1999-01-12
1* select to_date(‘1999-01-12‘,‘yyyy-mm-dd‘) - to_date(‘1999-12-01‘,‘yyyy-mm-dd‘) from dual
SQL> /
TO_DATE(‘1999-01-12‘,‘YYYY-MM-DD‘)-TO_DATE(‘1999-12-01‘,‘YYYY-MM-DD‘)
---------------------------------------------------------------------
-323
-- 也可以为负数的
SQL> c/1999-01-12/2999-10-12
1* select to_date(‘2999-10-12‘,‘yyyy-mm-dd‘) - to_date(‘1999-12-01‘,‘yyyy-mm-dd‘) from dual
SQL> /
TO_DATE(‘2999-10-12‘,‘YYYY-MM-DD‘)-TO_DATE(‘1999-12-01‘,‘YYYY-MM-DD‘)
---------------------------------------------------------------------
365193
下面看看INTERVAL YEAR TO MONTH怎么用.
SQL> create table bb(a date, b date, c interval year(9) to month);
Table created.
SQL> desc bb;
Name Null? Type
----------------------------------------- -------- ----------------------------
A DATE
B DATE
C INTERVAL YEAR(9) TO MONTH
SQL> insert into bb values(to_date(‘1985-12-12‘, ‘yyyy-mm-dd‘), to_date(‘1984-12-01‘,‘yyyy-mm-dd‘), null)
1 row created.
SQL> select * from bb;
A B
--------- ---------
C
---------------------------------------------------------------------------
12-DEC-85 01-DEC-84
SQL> update bb set c = numtoyminterval(a-b, ‘year‘);
1 row updated.
SQL> select * from bb;
A B
--------- ---------
C
---------------------------------------------------------------------------
12-DEC-85 01-DEC-84
+000000376-00
-- 直接将相减的天变成年了, 因为我指定变成年的
SQL> select a-b, c from bb;
A-B
----------
C
---------------------------------------------------------------------------
376
+000000376-00
SQL> insert into bb values(null,null,numtoyminterval(376,‘month‘));
1 row created.
SQL> select * from bb;
A B C
--------- --------- --------------------------------------------
12-DEC-85 01-DEC-84 +000000376-00
+000000031-04
SQL> insert into bb values ( null,null, numtoyminterval(999999999,‘year‘));
1 row created.
SQL> select * from bb;
A B C
--------- --------- ---------------------------------------------------------------------
12-DEC-85 01-DEC-84 +000000376-00
+000000031-04
+999999999-00
感谢 einyboy 转自：http://www.cnblogs.com/einyboy/archive/2012/08/06/2624886.html