Oracle求日期的格式

Oracle中TO_DATE格式2009-04-14 10:53TO_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,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,显示fri

??????? day??? spelled out?? 当周第几天全写??? 显示值:星期五,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‘,‘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‘,‘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‘)) "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,‘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

????

24,round[舍入到最接近的日期](day:舍入到最接近的星期日)

?? select sysdate S1,

?? round(sysdate) S2,

?? round(sysdate,‘year‘) YEAR,‘month‘) MONTH,‘day‘) DAY from dual

25,trunc[截断到最接近的日期,单位为天],返回的是日期类型

?? select sysdate S1,????????????????????

???? trunc(sysdate) S2,???????????????? //返回当前日期,无时分秒

???? trunc(sysdate,??????? //返回当前年的1月1日,???? //返回当前月的1日,‘day‘) DAY?????????? //返回当前星期的星期天,无时分秒

?? from dual

26,返回日期列表中最晚日期

?? select greatest(‘01-1月-04‘,‘04-1月-04‘,‘10-2月-04‘) from dual

27.计算时间差

???? 注:oracle时间差是以天数为单位,所以换算成年月,日

????

????? select floor(to_number(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))/365) as spanYears from dual??????? //时间差-年

????? select ceil(moths_between(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))) as spanMonths from dual??????? //时间差-月

????? select floor(to_number(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))) as spanDays from dual???????????? //时间差-天

????? select floor(to_number(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))*24) as spanHours from dual???????? //时间差-时

????? select floor(to_number(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))*24*60) as spanMinutes from dual??? //时间差-分

????? select floor(to_number(sysdate-to_date(‘2007-11-02 15:55:03‘,‘yyyy-mm-dd hh24:mi:ss‘))*24*60*60) as spanSeconds from dual //时间差-秒

28.更新时间

???? 注:oracle时间加减是以天数为单位,设改变量为n,日

???? select to_char(sysdate,‘yyyy-mm-dd hh24:mi:ss‘),to_char(sysdate+n*365,‘yyyy-mm-dd hh24:mi:ss‘) as newTime from dual??????? //改变时间-年

???? select to_char(sysdate,add_months(sysdate,n) as newTime from dual???????????????????????????????? //改变时间-月

???? select to_char(sysdate,to_char(sysdate+n,‘yyyy-mm-dd hh24:mi:ss‘) as newTime from dual??????????? //改变时间-日

???? select to_char(sysdate,to_char(sysdate+n/24,‘yyyy-mm-dd hh24:mi:ss‘) as newTime from dual???????? //改变时间-时

???? select to_char(sysdate,to_char(sysdate+n/24/60,‘yyyy-mm-dd hh24:mi:ss‘) as newTime from dual????? //改变时间-分

???? select to_char(sysdate,to_char(sysdate+n/24/60/60,‘yyyy-mm-dd hh24:mi:ss‘) as newTime from dual?? //改变时间-秒

29.查找月的第一天,最后一天

???? SELECT Trunc(Trunc(SYSDATE,‘MONTH‘) - 1,‘MONTH‘) First_Day_Last_Month,

?????? Trunc(SYSDATE,‘MONTH‘) - 1 / 86400 Last_Day_Last_Month,‘MONTH‘) First_Day_Cur_Month,

?????? LAST_DAY(Trunc(SYSDATE,‘MONTH‘)) + 1 - 1 / 86400 Last_Day_Cur_Month

?? FROM dual;

30,查询一年12个月的信息

Select to_char(add_months(trunc(sysdate,Rownum - 1),‘yyyy-mm‘)
From dual Connect By Rownum <= 12

1 2009-01
2 2009-02
。。。
11 2009-11
12 2009-12

31、查询每月30天的信息

Select to_char(trunc(sysdate,‘month‘) + Rownum - 1,‘yyyy-mm-dd‘)?
From dual Connect By Rownum <= extract(Day From last_day(trunc(Sysdate,‘month‘)))

1 2009-07-01
2 2009-07-02
3 2009-07-03

。。。
30 2009-07-30
31 2009-07-31

32、目前时间日期信息

SELECT EXTRACT(YEAR FROM SYSDATE) FROM DUAL; return Current Year?? SELECT EXTRACT(MONTH FROM SYSDATE) FROM DUAL; return Current Month?? SELECT EXTRACT(DAY FROM SYSDATE) FROM DUAL; return Current Day

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