ORACLE时间相关问题

部分 说明 yyyy-mm-dd 年月日 hh24 时间为24小时格式，如果不需要可以换为hh mi 这个必须写mi写成mm会报错，表达式出现两次 ss 秒

如果时间要进行加减，和现在的时间进行比较。

to_date(字符串,‘yyyy-mm-dd hh24:mi:ss‘)+1

后面为1的时候就是加1天，2的时候就是加两天，如果加一个小时，就是为1/24。

二、常用时间转化方法以及函数使用

1.取得当前日期是本月的第几周?

select to_char(sysdate,‘YYYYMMDD W HH24:MI:SS‘) from dual;?

TO_CHAR(SYSDATE,‘YY‘)?

select to_char(sysdate,‘W‘) from dual;?

2.取得当前日期是一个星期中的第几天,注意星期日是第一天

select sysdate,to_char(sysdate,‘D‘) from dual;?

3.得到当天凌晨0点0分0秒的日期?

select trunc(sysdate) from dual;?

4.得到这天的最后一秒?
select trunc(sysdate) + 0.99999 from dual;?

5.得到小时的具体数值?
select trunc(sysdate) + 1/24 from dual;?
select trunc(sysdate) + 7/24 from dual;?

6.得到明天凌晨0点0分0秒的日期?

select trunc(sysdate+1) from dual;?
select trunc(sysdate)+1 from dual;

7.本月一日的日期?
select trunc(sysdate,‘mm‘) from dual;?

8.得到下月一日的日期?
select trunc(add_months(sysdate,1),‘mm‘) from dual;?

9.返回当前月的最后一天
select last_day(sysdate) from dual;?
select last_day(trunc(sysdate)) from dual;?
select trunc(last_day(sysdate)) from dual;?
select trunc(add_months(sysdate,‘mm‘) - 1 from dual;?

10.得到一年的每一天?
select trunc(sysdate,‘yyyy‘)+ rn -1 date0?
from?
(select rownum rn from all_objects?
where rownum<366);

11.今天是今年的第N天?
SELECT TO_CHAR(SYSDATE,‘DDD‘) FROM DUAL;?

12.如何在给现有的日期加上2年?
select add_months(sysdate,24) from dual;?

13.判断某一日子所在年分是否为润年?
select decode(to_char(last_day(trunc(sysdate,‘y‘)+31),‘dd‘),‘29‘,‘闰年‘,‘平年‘) from dual;?

14.判断两年后是否为润年?
select decode(to_char(last_day(trunc(add_months(sysdate,24),?
‘dd‘),‘平年‘) from dual;

17.得到日期的季度?
select ceil(to_number(to_char(sysdate,‘mm‘))/3) from dual;?
select to_char(sysdate,‘Q‘) from dual;?

三、TO_DATE函数的使用

TO_DATE格式(以时间:2018-11-02?? 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(‘2018-11-12 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,‘hh‘) from dual;??????
????
??? SYSDATE TO_CHAR(SYSDATE,‘HH‘)??????
? ???
????
??? SQL> select sysdate,‘hh24‘) from dual;??????
????
??? SYSDATE TO_CHAR(SYSDATE,‘HH24‘)??????

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;

（编辑：李大同）

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一、遇到一个情况数据库存入的时间格式为20180831154546，需要将其转化为日期时分秒格式的。 to_date(‘20180831154546‘,‘yyyy-mm-dd hh24:mi:ss‘); 也就是to_date(字符串,‘yyyy-mm-dd hh24:mi:ss‘); 说明：