ORACLE日期时间函数大全(一)
TO_DATE格式(以时间:2007-11-02?? 13:45:25为例)
??
??????? Year:?????
??????? yy two digits 两位年??????????????? 显示值:07
??????? yyy three digits 三位年??????????????? 显示值:007
??????? yyyy four digits 四位年??????????????? 显示值:2007
???????????
??????? Month:?????
??????? mm??? number???? 两位月????????????? 显示值:11
??????? mon??? abbreviated 字符集表示????????? 显示值:11月,若是英文版,显示nov????
??????? month spelled out 字符集表示????????? 显示值:11月,若是英文版,显示november
?????????
??????? Day:?????
??????? dd??? number???????? 当月第几天??????? 显示值:02
??????? ddd??? number???????? 当年第几天??????? 显示值:02
??????? dy??? abbreviated 当周第几天简写??? 显示值:星期五,若是英文版,显示fri
??????? day??? spelled out?? 当周第几天全写??? 显示值:星期五,若是英文版,显示friday???????
??????? ddspth spelled out, ordinal twelfth
????????????
????????????? Hour:
????????????? hh??? two digits 12小时进制??????????? 显示值:01
????????????? hh24 two digits 24小时进制??????????? 显示值:13
?????????????
????????????? Minute:
????????????? mi??? two digits 60进制??????????????? 显示值:45
?????????????
????????????? Second:
????????????? ss??? two digits 60进制??????????????? 显示值:25
?????????????
????????????? 其它
????????????? Q???? digit???????? 季度????????????????? 显示值:4
????????????? WW??? digit???????? 当年第几周??????????? 显示值:44
????????????? W??? digit????????? 当月第几周??????????? 显示值:1
?????????????
??????? 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）
????????
select to_char(sysdate,'yyyy-mm-dd hh24:mi:ss') as nowTime from dual;?? //日期转化为字符串??
select to_char(sysdate,'yyyy') as nowYear?? from dual;?? //获取时间的年??
select to_char(sysdate,'mm')??? as nowMonth from dual;?? //获取时间的月??
select to_char(sysdate,'dd')??? as nowDay??? from dual;?? //获取时间的日??
select to_char(sysdate,'hh24') as nowHour?? from dual;?? //获取时间的时??
select to_char(sysdate,'mi')??? as nowMinute from dual;?? //获取时间的分??
select to_char(sysdate,'ss')??? as nowSecond from dual;?? //获取时间的秒
???
select to_date('2004-05-07 13:23:44','yyyy-mm-dd hh24:mi:ss')??? from dual//
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?????
?? 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.月份差??
?? 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.?????
?? 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' )?????
??
?? 查找2002-02-28至2002-02-01间除星期一和七的天数?????
?? 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. 查找月份????
??? 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?????
???? extract()找出日期或间隔值的字段值
??? 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?????
???
??? SQL> select sysdate ,to_char(sysdate,'hh24') from dual;?????
???
??? SYSDATE TO_CHAR(SYSDATE,'HH24')?????
??? -------------------- -----------------------?????
??? 2003-10-13 19:35:21 19????
??????
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) /30) 作为月?????
?? d(mod(date2-date1, 365), 30)作为日.
23.next_day函数????? 返回下个星期的日期,day为1-7或星期日-星期六,1表示星期日
?? next_day(sysdate,6)是从当前开始下一个星期五。后面的数字是从星期日开始算起。
?1 2 3 4 5 6 7?????
?? 日 一 二 三 四 五 六???
??
?? ---------------------------
??
?? select??? (sysdate-to_date('2003-12-03 12:55:45','yyyy-mm-dd hh24:mi:ss'))*24*60*60 from ddual
?? 日期 返回的是天 然后 转换为ss
?
