interval year to month数据类型
oracle语法:
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
========================
今天来添加点新的东西![2008-07-26] 这部分东东来源:
interval year to month类型2个timestamp类型的时间差别。内部类型是182,长度是5。其中4个字节存储年份差异,存储的时候在差异上加了一个0x80000000的偏移量。一个字节存储月份的差异,这个差异加了60的偏移量。
sql> alter table testtimestamp add e interval year to month;
sql> update testtimestamp set e=(select interval '5' year interval '10' month year from dual);
已更新3行。
sql> commit;
提交完成。
sql> select dump(e,16) from testtimestamp;
dump(e,16)
---------------------------------------------
typ=182 len=5: 80,0,0,5,46
typ=182 len=5: 80,0,0,5,46
typ=182 len=5: 80,0,0,5,46
年:0x80000005-0x80000000=5
月:0x46-60=10
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