oracle日期格式总结

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select to_char(sysdate,'yyyy')||'年'||to_char(sysdate,'mm')||'月'||to_char(sysdate,'dd')||'日'||to_char(sysdate,'hh24')||'时'||to_char(sysdate,'mi')||'分'||to_char(sysdate,'ss')||'秒’from dual

在oracle中处理日期大全

TO_DATE格式

Day: dd number 12

dy abbreviated fri

day spelled out friday

ddspth spelled out, ordinal twelfth

Month:

mm number 03

mon abbreviated mar

month spelled out march

Year:

yy two digits 98

yyyy four digits 1998

24小时格式下时间范围为: 0:00:00 - 23:59:59....

12小时格式下时间范围为: 1:00:00 - 12:59:59 ....

1. 日期和字符转换函数用法

(to_date,to_char)

2.

select to_char( to_date(222,'J'),'Jsp') from dual

显示Two Hundred Twenty-Two 3. 求某天是星期几和设置语言格式

select to_char(to_date('2002-08-26','yyyy-mm-dd'),'day')

from dual;

星期一

select

to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE = American') from dual;

monday

设置日期语言

ALTER SESSION SET NLS_DATE_LANGUAGE='AMERICAN';

也可以这样

TO_DATE ('2002-08-26', 'YYYY-mm-dd', 'NLS_DATE_LANGUAGE =

American')

4. 两个日期间的天数

select floor(sysdate - to_date('20020405','yyyymmdd')) from

dual;

5. 时间为null的用法

select id, active_date from table1

UNION

select 1, TO_DATE(null) from dual;

注意要用TO_DATE(null)

6. 时间的范围

a_date between to_date('20011201','yyyymmdd') and

to_date('20011231','yyyymmdd')

那么12月31号中午12点之后和12月1号的12点之前是不包含在这个范围之内的。

所以,当时间需要精确的时候,觉得to_char还是必要的

7. 日期格式冲突问题

输入的格式要看你安装的ORACLE字符集的类型, 比如: US7ASCII,

date格式的类型就是: '01-Jan-01'

alter system set NLS_DATE_LANGUAGE = American

alter session set NLS_DATE_LANGUAGE = American

或者在to_date中写

select

to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE = American') from dual;

注意我这只是举了NLS_DATE_LANGUAGE,当然还有很多,

可查看

select * from nls_session_parameters

select * from V$NLS_PARAMETERS

8. 查找2002-02-28至2002-02-01间除星期一和七的天数

select count(*) from ( select rownum-1 rnum

from all_objects where rownum <= to_date('2002-02-28','yyyy-mm-dd') - to_date('2002-

02-01','yyyy-mm-dd')+1 ) where

to_char( to_date('2002-02-01','yyyy-mm-dd')+rnum-1, 'D' )not

in ( '1', '7' )

在前后分别调用DBMS_UTILITY.GET_TIME, 让后将结果相减(得到的是1/100秒, 而不是毫秒).

9.

select months_between(to_date('01-31-1999','MM-DD-YYYY'),

to_date('12-31-1998','MM-DD-YYYY')) "MONTHS" FROM DUAL;

1

select months_between(to_date('02-01-1999','MM-DD-YYYY'),

to_date('12-31-1998','MM-DD-YYYY')) "MONTHS" FROM DUAL;

1.03225806451613

10. Next_day的用法

Next_day(date, day)

Monday-Sunday, for format code DAY

Mon-Sun, for format code DY

1-7, for format code D

11

select to_char(sysdate,'hh:mi:ss') TIME from all_objects

注意:第一条记录的TIME 与最后一行是一样的

可以建立一个函数来处理这个问题 create or replace function sys_date return date is

begin

return sysdate;

end;

select to_char(sys_date,'hh:mi:ss') from all_objects;

12. 获得小时数

SELECT EXTRACT(HOUR FROM TIMESTAMP '2001-02-16 2:38:40') from

offer

SQL> select sysdate ,to_char(sysdate,'hh') from dual;

SYSDATE TO_CHAR(SYSDATE,'HH')

-------------------- ---------------------

2003-10-13 19:35:21 07

select sysdate ,to_char(sysdate,'hh24') from dual;

SYSDATE TO_CHAR(SYSDATE,'HH24')

-------------------- -----------------------

2003-10-13 19:35:21 19

获取年月日与此类似

13. 年月日的处理

select older_date, newer_date, years, months,

abs( trunc( newer_date-

add_months( older_date,years*12+months ) ) ) days

from ( select trunc(months_between( newer_date, older_date )/12) YEARS, mod(trunc(months_between( newer_date,

older_date )), 12 ) MONTHS, newer_date, older_date from

( select hiredate

older_date,add_months(hiredate,rownum)+rownum newer_date

from emp ) )

14.处理月份天数不定的办法

select to_char(add_months(last_day(sysdate) +1, -2),

'yyyymmdd'),last_day(sysdate) from dual

16. 找出今年的天数

select add_months(trunc(sysdate,'year'), 12) -

trunc(sysdate,'year') from dual

闰年的处理方法

to_char( last_day( to_date('02' || :year,'mmyyyy') ), 'dd' )

如果是28就不是闰年

17. yyyy与rrrr的区别

'YYYY99 TO_C

------- ----

yyyy 99 0099

rrrr 99 1999

yyyy 01 0001

rrrr 01 2001

18.不同时区的处理 select to_char( NEW_TIME( sysdate, 'GMT','EST'), 'dd/mm/yyyy

hh:mi:ss') ,sysdate from dual;

19. 5秒钟一个间隔

Select TO_DATE(FLOOR(TO_CHAR(sysdate,'SSSSS')/300) *

300,'SSSSS') ,TO_CHAR(sysdate,'SSSSS')

from dual

2002-11-1 9:55:00 35786

SSSSS表示5位秒数

20. 一年的第几天

select TO_CHAR(SYSDATE,'DDD'),sysdate from dual

310 2002-11-6 10:03:51

21.计算小时,分,秒,毫秒

select Days, A, TRUNC(A*24) Hours, TRUNC(A*24*60 -

60*TRUNC(A*24)) Minutes, TRUNC(A*24*60*60 -

60*TRUNC(A*24*60)) Seconds, TRUNC(A*24*60*60*100 -

100*TRUNC(A*24*60*60)) mSeconds from ( select trunc(sysdate)

Days, sysdate - trunc(sysdate) A from dual )

select * from tabname order by

decode(mode,'FIFO',1,-1)*to_char(rq,'yyyymmddhh24miss');

// floor((date2-date1) /365) 作为年 floor((date2-date1, 365)