Oracle时间处理大全
SQL Server和Oracle常用时间日期处理语句
在SQL语句中,常常用会对时间(或日期)进行一些处理,下面是比较通用的一些语句:
延迟:
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1.sysdate+(5/24/60/60) 在系统时间基础上延迟5秒
2.
3.sysdate+5/24/60 在系统时间基础上延迟5分钟
4.
5.sysdate+5/24 在系统时间基础上延迟5小时
6.
7.sysdate+5 在系统时间基础上延迟5天
8.
9.add_months(sysdate,-5) 在系统时间基础上延迟5月
10.
11.add_months(sysdate,-5*12) 在系统时间基础上延迟5年
上月末的日期:
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1.select last_day(add_months(sysdate, -1)) from dual;
本月的最后一秒:
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1.select trunc(add_months(sysdate,1),'MM') - 1/24/60/60 from dual
本周星期一的日期:
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1.select trunc(sysdate,'day')+1 from dual
年初至今的天数:
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1.select ceil(sysdate - trunc(sysdate, 'year')) from dual;
今天是今年的第几周 :
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1.select to_char(sysdate,'fmww') from dual
今天是本月的第几周:
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1.SELECT TO_CHAR(SYSDATE,'WW') - TO_CHAR(TRUNC(SYSDATE,'MM'),'WW') + 1 AS
"weekOfMon"FROM dual
本月的天数
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1.SELECT to_char(last_day(SYSDATE),'dd') days FROM dual
今年的天数
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1.select add_months(trunc(sysdate,'year'), 12) - trunc(sysdate,'year') from dual 下个星期一的日期
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1.SELECT Next_day(trunc(SYSDATE),'monday') FROM dual
================================
计算工作日方法
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1.create table t(s date,e date);
2.
3.alter session set nls_date_format = 'yyyy-mm-dd';
4.
5.insert into t values('2003-03-01','2003-03-03');
6.
7.insert into t values('2003-03-02','2003-03-03');
8.
9.insert into t values('2003-03-07','2003-03-08');
10.
11.insert into t values('2003-03-07','2003-03-09');
12.
13.insert into t values('2003-03-05','2003-03-07');
14.
15.insert into t values('2003-02-01','2003-03-31');
16.
17.-- 这里假定日期都是不带时间的,否则在所有日期前加trunc即可.
18.
19.select s,e,e-s+1 total_days, trunc((e-s+1)/7)*5 +
length(replace(substr('01111100111110',to_char(s,'d'),mod(e-s+1,7)),'0','')) work_days from t;
20.
21.-- drop table t;
====================================================
判断当前时间是上午下午还是晚上
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1.SELECT CASE
2.WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 6 AND 11 THEN'上午'
3.
4.WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 11 AND 17 THEN'下午'
5.
6.WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 17 AND 21 THEN'晚上'
7.
8.END
9.
10.FROM dual;
===================================================
Oracle 中的一些处理日期
将数字转换为任意时间格式.如秒:需要转换为天/小时
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1.SELECT to_char(floor(TRUNC(936000/(60*60))/24))||'天
'||to_char(mod(TRUNC(936000/(60*60)),24))||'小时'FROM DUAL
TO_DATE格式
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1.Day:
2.
3.dd number 12
4.
5.dy abbreviated fri
6.
7.day spelled out friday
8.
9.ddspth spelled out, ordinal twelfth
10.
11.Month:
12.
13.mm number 03
14.
15.mon abbreviated mar
16.
17.month spelled out march
18.
19.Year:
20.
21.yy two digits 98
22.
23.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.
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1.select to_char( to_date(222,'J'),'Jsp') from dual
显示Two Hundred Twenty-Two
3. 求某天是星期几
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1.select to_char(to_date('2002-08-26','yyyy-mm-dd'),'day') from dual;
2.
3.select to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE =
American') from dual;
设置日期语言
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1.ALTER SESSION SET NLS_DATE_LANGUAGE='AMERICAN';
也可以这样
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1.TO_DATE ('2002-08-26', 'YYYY-mm-dd', 'NLS_DATE_LANGUAGE = American')
4. 两个日期间的天数
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1.select floor(sysdate - to_date('20020405','yyyymmdd')) from dual;
5. 时间为null的用法
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1.select id, active_date from table1
2.UNION
3.select 1, TO_DATE(null) from dual;
注意要用TO_DATE(null)
6.
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1.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′
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1.alter system set NLS_DATE_LANGUAGE = American
2.
3.alter session set NLS_DATE_LANGUAGE = American
或者在to_date中写
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1.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
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1.select * from nls_session_parameters
2.select * from V$NLS_PARAMETERS
8.
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1.select count(*)
2.from ( select rownum-1 rnum
3.from all_objects
4.where rownum <= to_date('2002-02-28','yyyy-mm-dd') -
to_date('2002-02-01','yyyy-mm-dd')+1 )
5.where to_char( to_date('2002-02-01','yyyy-mm-dd')+rnum-1, 'D' )
6.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, 然后将结果相减(得到的是1/100秒, 而不是毫秒).
9.
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1.select months_between(to_date('01-31-1999','MM-DD-YYYY'),
2.to_date('12-31-1998','MM-DD-YYYY')) "MONTHS"FROM DUAL;
3. 1
4.select months_between(to_date('02-01-1999','MM-DD-YYYY'),
5.to_date('12-31-1998','MM-DD-YYYY')) "MONTHS"FROM DUAL;
6. 1.03225806451613
10. Next_day的用法
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1.Next_day(date, day)
2.
3.Monday-Sunday, for format code DAY
4.Mon-Sun, for format code DY
5.1-7, for format code D
11
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1.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
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1.create or replace function sys_date return date is
2.begin
3.return sysdate;
4.end;
5.select to_char(sys_date,'hh:mi:ss') from all_objects;
12. 获得小时数
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1.SQL> select sysdate ,to_char(sysdate,'hh') from dual;
2.SYSDATE TO_CHAR(SYSDATE,'HH')
——————–———————
2003-10-13 19:35:21 07
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1.SQL> select sysdate ,to_char(sysdate,'hh24') from dual;
2.
3.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
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1.select older_date,
2.newer_date,
3.years,
4.months,
5.
6.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,
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1.newer_date,
2.older_date
3.from ( select hiredate older_date,
4.add_months(hiredate,rownum)+rownum newer_date from emp ) )
14. 处理月份天数不定的办法
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1.select to_char(add_months(last_day(sysdate) +1, -2),
'yyyymmdd'),last_day(sysdate) from dual
16. 找出今年的天数
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1.select add_months(trunc(sysdate,'year'), 12) - trunc(sysdate,'year') from dual 闰年的处理方法
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1.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
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1.select to_char( NEW_TIME( sysdate, 'GMT','EST'), 'dd/mm/yyyy hh:mi:ss') ,sysdate
2.from dual;
19. 5秒钟一个间隔
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1.Select TO_DATE(FLOOR(TO_CHAR(sysdate,'SSSSS')/300) *
300,'SSSSS') ,TO_CHAR(sysdate,'SSSSS')
2.from dual
TO_DATE(FL TO_CH
———- —–
2007-01-01 60368
SSSSS表示5位秒数
20. 一年的第几天
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1.select TO_CHAR(SYSDATE,'DDD'),sysdate from dual
TO_ SYSDATE
————-
017 2007-01-17
21.计算小时,分,秒,毫秒
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1.select
2.Days,
3.A,
4.TRUNC(A*24) Hours,
5.TRUNC(A*24*60 - 60*TRUNC(A*24)) Minutes,
6.TRUNC(A*24*60*60 - 60*TRUNC(A*24*60)) Seconds,
7.TRUNC(A*24*60*60*100 - 100*TRUNC(A*24*60*60)) mSeconds
8.from
9.( select trunc(sysdate) Days, sysdate - trunc(sysdate) A
10.from dual
11.)
12.
13.select * from tabname
14.order by decode(mode,'FIFO',1,-1)*to_char(rq,'yyyymmddhh24miss');
15.
16.//
17.
18.floor((date2-date1) /365) 作为年
19.floor((date2-date1, 365) /30) 作为月
20.mod(mod(date2-date1, 365), 30) 作为日.
22.next_day函数
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—————————————————————
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1.select (sysdate-to_date('2003-12-03 12:55:45','yyyy-mm-dd
hh24:mi:ss'))*24*60*60 from dual
日期返回的是天然后转换为ss