Oracle時(shí)間日期操作
sysdate+(5/24/60/60) 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5秒
sysdate+5/24/60 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5分鐘
sysdate+5/24 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5小時(shí)
sysdate+5 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5天
add_months(sysdate,-5) 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5月
add_months(sysdate,-5*12) 在系統(tǒng)時(shí)間基礎(chǔ)上延遲5年
上月末的日期：select last_day(add_months(sysdate, -1)) from dual;
本月的最后一秒：select trunc(add_months(sysdate,1),'MM') - 1/24/60/60 from dual
本周星期一的日期：select trunc(sysdate,'day')+1 from dual
年初至今的天數(shù)：select ceil(sysdate - trunc(sysdate, 'year')) from dual;
今天是今年的第幾周 :select to_char(sysdate,'fmww') from dual
今天是本月的第幾周:SELECT TO_CHAR(SYSDATE,'WW') - TO_CHAR(TRUNC(SYSDATE,'MM'),'WW') + 1 AS "weekOfMon" FROM dual
本月的天數(shù)
SELECT to_char(last_day(SYSDATE),'dd') days FROM dual
今年的天數(shù)
select add_months(trunc(sysdate,'year'), 12) - trunc(sysdate,'year') from dual
下個(gè)星期一的日期
SELECT Next_day(SYSDATE,'monday') FROM dual
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
--計(jì)算工作日方法
create table t(s date,e date);
alter session set nls_date_format = 'yyyy-mm-dd';
insert into t values('2003-03-01','2003-03-03');
insert into t values('2003-03-02','2003-03-03');
insert into t values('2003-03-07','2003-03-08');
insert into t values('2003-03-07','2003-03-09');
insert into t values('2003-03-05','2003-03-07');
insert into t values('2003-02-01','2003-03-31');
-- 這里假定日期都是不帶時(shí)間的，否則在所有日期前加trunc即可。
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;
-- drop table t;
引此：http://www.itpub.net/showthread.php?s=1635506cd5f48b1bc3adbe4cde96f227&threadid=104060&perpage=15&pagenumber=1
================================================================================
判斷當(dāng)前時(shí)間是上午下午還是晚上
SELECT CASE
WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 6 AND 11 THEN '上午'
WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 11 AND 17 THEN '下午'
WHEN to_number(to_char(SYSDATE,'hh24')) BETWEEN 17 AND 21 THEN '晚上'
END
FROM dual;
================================================================================
Oracle 中的一些處理日期
將數(shù)字轉(zhuǎn)換為任意時(shí)間格式.如秒:需要轉(zhuǎn)換為天/小時(shí)
SELECT to_char(floor(TRUNC(936000/(60*60))/24))||'天'||to_char(mod(TRUNC(936000/(60*60)),24))||'小時(shí)'??FROM DUAL
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小時(shí)格式下時(shí)間范圍為：??0:00:00??-??23:59:59....? ?? ?
12小時(shí)格式下時(shí)間范圍為：??1:00:00??-??12:59:59??....? ?? ?
1.? ?? ?
日期和字符轉(zhuǎn)換函數(shù)用法（to_date,to_char）? ?? ?
2.? ?? ?
select??to_char(??to_date(222,'J'),'Jsp')??from??dual? ?? ?
顯示Two??Hundred??Twenty-Two? ?? ?
3.? ?? ?
求某天是星期幾? ?? ?
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? ?? ?
設(shè)置日期語言? ?? ?
ALTER??SESSION??SET??NLS_DATE_LANGUAGE='AMERICAN';? ?? ?
也可以這樣? ?? ?
TO_DATE??('2002-08-26',??'YYYY-mm-dd',??'NLS_DATE_LANGUAGE??=??American')? ?? ?
4.? ?? ?
兩個(gè)日期間的天數(shù)? ?? ?
select??floor(sysdate??-??to_date('20020405','yyyymmdd'))??from??dual;? ?? ?
5.??時(shí)間為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.? ?? ?
a_date??between??to_date('20011201','yyyymmdd')??and??to_date('20011231','yyyymmdd')? ?? ?
那么12月31號中午12點(diǎn)之后和12月1號的12點(diǎn)之前是不包含在這個(gè)范圍之內(nèi)的。? ?? ?
所以，當(dāng)時(shí)間需要精確的時(shí)候，覺得to_char還是必要的? ?? ?
7.??日期格式?jīng)_突問題? ?? ?
輸入的格式要看你安裝的ORACLE字符集的類型,??比如:??US7ASCII,??date格式的類型就是:??'01-Jan-01'? ?? ?
alter??system??set??NLS_DATE_LANGUAGE??=??American? ?? ?
alter??session??set??NLS_DATE_LANGUAGE??=??American? ?? ?
或者在to_date中寫? ?? ?
select??to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE??=??American')??from??dual;? ?? ?
注意我這只是舉了NLS_DATE_LANGUAGE，當(dāng)然還有很多，? ?? ?
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間除星期一和七的天數(shù)? ?? ?
在前后分別調(diào)用DBMS_UTILITY.GET_TIME,??讓后將結(jié)果相減(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：第一條記錄的TIME??與最后一行是一樣的? ?? ?
可以建立一個(gè)函數(shù)來處理這個(gè)問題? ?? ?
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.? ?? ?
獲得小時(shí)數(shù)? ?? ?
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? ?? ?
獲取年月日與此類似? ?? ?
13.? ?? ?
年月日的處理? ?? ?
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.? ?? ?
處理月份天數(shù)不定的辦法? ?? ?
select??to_char(add_months(last_day(sysdate)??+1,??-2),??'yyyymmdd'),last_day(sysdate)??from??dual? ?? ?
16.? ?? ?
找出今年的天數(shù)? ?? ?
select??add_months(trunc(sysdate,'year'),??12)??-??trunc(sysdate,'year')??from??dual? ?? ?
閏年的處理方法? ?? ?
to_char(??last_day(??to_date('02'? ? |??|??:year,'mmyyyy')??),??'dd'??)? ?? ?
如果是28就不是閏年? ?? ?
17.? ?? ?
yyyy與rrrr的區(qū)別? ?? ?
'YYYY99??TO_C? ?? ?
-------??----? ?? ?
yyyy??99??0099? ?? ?
rrrr??99??1999? ?? ?
yyyy??01??0001? ?? ?
rrrr??01??2001? ?? ?
18.不同時(shí)區(qū)的處理? ?? ?
select??to_char(??NEW_TIME(??sysdate,??'GMT','EST'),??'dd/mm/yyyy??hh:mi:ss')??,sysdate? ?? ?
from??dual;? ?? ?
19.? ?? ?
5秒鐘一個(gè)間隔? ?? ?
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位秒數(shù)? ?? ?
20.? ?? ?
一年的第幾天? ?? ?
select??TO_CHAR(SYSDATE,'DDD'),sysdate??from??dual? ?? ?
310??2002-11-6??10:03:51? ?? ?
21.計(jì)算小時(shí),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)??作為年? ?? ?
floor((date2-date1,??365)??/30)??作為月? ?? ?
mod(mod(date2-date1,??365),??30)作為日.? ?? ?
23.next_day函數(shù)? ?? ?
next_day(sysdate,6)是從當(dāng)前開始下一個(gè)星期五。后面的數(shù)字是從星期日開始算起。? ?? ?
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??dual??
日期??返回的是天??然后??轉(zhuǎn)換為ss
轉(zhuǎn)此：http://www.onlinedatabase.cn/leadbbs/Announce/Announce.asp?BoardID=42&ID=1769
將數(shù)字轉(zhuǎn)換為任意時(shí)間格式.如秒:需要轉(zhuǎn)換為天/小時(shí)
SELECT to_char(floor(TRUNC(936000/(60*60))/24))||'天'||to_char(mod(TRUNC(936000/(60*60)),24))||'小時(shí)'??FROM DUAL
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小時(shí)格式下時(shí)間范圍為：??0:00:00??-??23:59:59....? ?? ?
12小時(shí)格式下時(shí)間范圍為：??1:00:00??-??12:59:59??....? ?? ?
1.? ?? ?
日期和字符轉(zhuǎn)換函數(shù)用法（to_date,to_char）? ?? ?
2.? ?? ?
select??to_char(??to_date(222,'J'),'Jsp')??from??dual? ?? ?
顯示Two??Hundred??Twenty-Two? ?? ?
3.? ?? ?
求某天是星期幾? ?? ?
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? ?? ?
設(shè)置日期語言? ?? ?
ALTER??SESSION??SET??NLS_DATE_LANGUAGE='AMERICAN';? ?? ?
也可以這樣? ?? ?
TO_DATE??('2002-08-26',??'YYYY-mm-dd',??'NLS_DATE_LANGUAGE??=??American')? ?? ?
4.? ?? ?
兩個(gè)日期間的天數(shù)? ?? ?
select??floor(sysdate??-??to_date('20020405','yyyymmdd'))??from??dual;? ?? ?
5.??時(shí)間為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.? ?? ?
a_date??between??to_date('20011201','yyyymmdd')??and??to_date('20011231','yyyymmdd')? ?? ?
那么12月31號中午12點(diǎn)之后和12月1號的12點(diǎn)之前是不包含在這個(gè)范圍之內(nèi)的。? ?? ?
所以，當(dāng)時(shí)間需要精確的時(shí)候，覺得to_char還是必要的? ?? ?
7.??日期格式?jīng)_突問題? ?? ?
輸入的格式要看你安裝的ORACLE字符集的類型,??比如:??US7ASCII,??date格式的類型就是:??'01-Jan-01'? ?? ?
alter??system??set??NLS_DATE_LANGUAGE??=??American? ?? ?
alter??session??set??NLS_DATE_LANGUAGE??=??American? ?? ?
或者在to_date中寫? ?? ?
select??to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE??=??American')??from??dual;? ?? ?
注意我這只是舉了NLS_DATE_LANGUAGE，當(dāng)然還有很多，? ?? ?
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間除星期一和七的天數(shù)? ?? ?
在前后分別調(diào)用DBMS_UTILITY.GET_TIME,??讓后將結(jié)果相減(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：第一條記錄的TIME??與最后一行是一樣的? ?? ?
可以建立一個(gè)函數(shù)來處理這個(gè)問題? ?? ?
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.? ?? ?
獲得小時(shí)數(shù)? ?? ?
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? ?? ?
獲取年月日與此類似? ?? ?
13.? ?? ?
年月日的處理? ?? ?
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.? ?? ?
處理月份天數(shù)不定的辦法? ?? ?
select??to_char(add_months(last_day(sysdate)??+1,??-2),??'yyyymmdd'),last_day(sysdate)??from??dual? ?? ?
16.? ?? ?
找出今年的天數(shù)? ?? ?
select??add_months(trunc(sysdate,'year'),??12)??-??trunc(sysdate,'year')??from??dual? ?? ?
閏年的處理方法? ?? ?
to_char(??last_day(??to_date('02'? ? |??|??:year,'mmyyyy')??),??'dd'??)? ?? ?
如果是28就不是閏年? ?? ?
17.? ?? ?
yyyy與rrrr的區(qū)別? ?? ?
'YYYY99??TO_C? ?? ?
-------??----? ?? ?
yyyy??99??0099? ?? ?
rrrr??99??1999? ?? ?
yyyy??01??0001? ?? ?
rrrr??01??2001? ?? ?
18.不同時(shí)區(qū)的處理? ?? ?
select??to_char(??NEW_TIME(??sysdate,??'GMT','EST'),??'dd/mm/yyyy??hh:mi:ss')??,sysdate? ?? ?
from??dual;? ?? ?
19.? ?? ?
5秒鐘一個(gè)間隔? ?? ?
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位秒數(shù)? ?? ?
20.? ?? ?
一年的第幾天? ?? ?
select??TO_CHAR(SYSDATE,'DDD'),sysdate??from??dual? ?? ?
310??2002-11-6??10:03:51? ?? ?
21.計(jì)算小時(shí),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)??作為年? ?? ?
floor((date2-date1,??365)??/30)??作為月? ?? ?
mod(mod(date2-date1,??365),??30)作為日.? ?? ?
23.next_day函數(shù)? ?? ?
next_day(sysdate,6)是從當(dāng)前開始下一個(gè)星期五。后面的數(shù)字是從星期日開始算起。? ?? ?
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??dual??
日期??返回的是天??然后??轉(zhuǎn)換為ss
轉(zhuǎn)此：http://www.onlinedatabase.cn/leadbbs/Announce/Announce.asp?BoardID=42&ID=1769
將數(shù)字轉(zhuǎn)換為任意時(shí)間格式.如秒:需要轉(zhuǎn)換為天/小時(shí)
SELECT to_char(floor(TRUNC(936000/(60*60))/24))||'天'||to_char(mod(TRUNC(936000/(60*60)),24))||'小時(shí)'??FROM DUAL
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小時(shí)格式下時(shí)間范圍為：??0:00:00??-??23:59:59....? ?? ?
12小時(shí)格式下時(shí)間范圍為：??1:00:00??-??12:59:59??....? ?? ?
1.? ?? ?
日期和字符轉(zhuǎn)換函數(shù)用法（to_date,to_char）? ?? ?
2.? ?? ?
select??to_char(??to_date(222,'J'),'Jsp')??from??dual? ?? ?
顯示Two??Hundred??Twenty-Two? ?? ?
3.? ?? ?
求某天是星期幾? ?? ?
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? ?? ?
設(shè)置日期語言? ?? ?
ALTER??SESSION??SET??NLS_DATE_LANGUAGE='AMERICAN';? ?? ?
也可以這樣? ?? ?
TO_DATE??('2002-08-26',??'YYYY-mm-dd',??'NLS_DATE_LANGUAGE??=??American')? ?? ?
4.? ?? ?
兩個(gè)日期間的天數(shù)? ?? ?
select??floor(sysdate??-??to_date('20020405','yyyymmdd'))??from??dual;? ?? ?
5.??時(shí)間為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.? ?? ?
a_date??between??to_date('20011201','yyyymmdd')??and??to_date('20011231','yyyymmdd')? ?? ?
那么12月31號中午12點(diǎn)之后和12月1號的12點(diǎn)之前是不包含在這個(gè)范圍之內(nèi)的。? ?? ?
所以，當(dāng)時(shí)間需要精確的時(shí)候，覺得to_char還是必要的? ?? ?
7.??日期格式?jīng)_突問題? ?? ?
輸入的格式要看你安裝的ORACLE字符集的類型,??比如:??US7ASCII,??date格式的類型就是:??'01-Jan-01'? ?? ?
alter??system??set??NLS_DATE_LANGUAGE??=??American? ?? ?
alter??session??set??NLS_DATE_LANGUAGE??=??American? ?? ?
或者在to_date中寫? ?? ?
select??to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE??=??American')??from??dual;? ?? ?
注意我這只是舉了NLS_DATE_LANGUAGE，當(dāng)然還有很多，? ?? ?
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間除星期一和七的天數(shù)? ?? ?
在前后分別調(diào)用DBMS_UTILITY.GET_TIME,??讓后將結(jié)果相減(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：第一條記錄的TIME??與最后一行是一樣的? ?? ?
可以建立一個(gè)函數(shù)來處理這個(gè)問題? ?? ?
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.? ?? ?
獲得小時(shí)數(shù)? ?? ?
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? ?? ?
獲取年月日與此類似? ?? ?
13.? ?? ?
年月日的處理? ?? ?
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.? ?? ?
處理月份天數(shù)不定的辦法? ?? ?
select??to_char(add_months(last_day(sysdate)??+1,??-2),??'yyyymmdd'),last_day(sysdate)??from??dual? ?? ?
16.? ?? ?
找出今年的天數(shù)? ?? ?
select??add_months(trunc(sysdate,'year'),??12)??-??trunc(sysdate,'year')??from??dual? ?? ?
閏年的處理方法? ?? ?
to_char(??last_day(??to_date('02'? ? |??|??:year,'mmyyyy')??),??'dd'??)? ?? ?
如果是28就不是閏年? ?? ?
17.? ?? ?
yyyy與rrrr的區(qū)別? ?? ?
'YYYY99??TO_C? ?? ?
-------??----? ?? ?
yyyy??99??0099? ?? ?
rrrr??99??1999? ?? ?
yyyy??01??0001? ?? ?
rrrr??01??2001? ?? ?
18.不同時(shí)區(qū)的處理? ?? ?
select??to_char(??NEW_TIME(??sysdate,??'GMT','EST'),??'dd/mm/yyyy??hh:mi:ss')??,sysdate? ?? ?
from??dual;? ?? ?
19.? ?? ?
5秒鐘一個(gè)間隔? ?? ?
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位秒數(shù)? ?? ?
20.? ?? ?
一年的第幾天? ?? ?
select??TO_CHAR(SYSDATE,'DDD'),sysdate??from??dual? ?? ?
310??2002-11-6??10:03:51? ?? ?
21.計(jì)算小時(shí),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)??作為年? ?? ?
floor((date2-date1,??365)??/30)??作為月? ?? ?
mod(mod(date2-date1,??365),??30)作為日.? ?? ?
23.next_day函數(shù)? ?? ?
next_day(sysdate,6)是從當(dāng)前開始下一個(gè)星期五。后面的數(shù)字是從星期日開始算起。? ?? ?
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??dual??
日期??返回的是天??然后??轉(zhuǎn)換為ss
轉(zhuǎn)此：http://www.onlinedatabase.cn/leadbbs/Announce/Announce.asp?BoardID=42&ID=1769
將數(shù)字轉(zhuǎn)換為任意時(shí)間格式.如秒:需要轉(zhuǎn)換為天/小時(shí)
SELECT to_char(floor(TRUNC(936000/(60*60))/24))||'天'||to_char(mod(TRUNC(936000/(60*60)),24))||'小時(shí)'??FROM DUAL
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小時(shí)格式下時(shí)間范圍為：??0:00:00??-??23:59:59....? ?? ?
12小時(shí)格式下時(shí)間范圍為：??1:00:00??-??12:59:59??....? ?? ?
1.? ?? ?
日期和字符轉(zhuǎn)換函數(shù)用法（to_date,to_char）? ?? ?
2.? ?? ?
select??to_char(??to_date(222,'J'),'Jsp')??from??dual? ?? ?
顯示Two??Hundred??Twenty-Two? ?? ?
3.? ?? ?
求某天是星期幾? ?? ?
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? ?? ?
設(shè)置日期語言? ?? ?
ALTER??SESSION??SET??NLS_DATE_LANGUAGE='AMERICAN';? ?? ?
也可以這樣? ?? ?
TO_DATE??('2002-08-26',??'YYYY-mm-dd',??'NLS_DATE_LANGUAGE??=??American')? ?? ?
4.? ?? ?
兩個(gè)日期間的天數(shù)? ?? ?
select??floor(sysdate??-??to_date('20020405','yyyymmdd'))??from??dual;? ?? ?
5.??時(shí)間為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.? ?? ?
a_date??between??to_date('20011201','yyyymmdd')??and??to_date('20011231','yyyymmdd')? ?? ?
那么12月31號中午12點(diǎn)之后和12月1號的12點(diǎn)之前是不包含在這個(gè)范圍之內(nèi)的。? ?? ?
所以，當(dāng)時(shí)間需要精確的時(shí)候，覺得to_char還是必要的? ?? ?
7.??日期格式?jīng)_突問題? ?? ?
輸入的格式要看你安裝的ORACLE字符集的類型,??比如:??US7ASCII,??date格式的類型就是:??'01-Jan-01'? ?? ?
alter??system??set??NLS_DATE_LANGUAGE??=??American? ?? ?
alter??session??set??NLS_DATE_LANGUAGE??=??American? ?? ?
或者在to_date中寫? ?? ?
select??to_char(to_date('2002-08-26','yyyy-mm-dd'),'day','NLS_DATE_LANGUAGE??=??American')??from??dual;? ?? ?
注意我這只是舉了NLS_DATE_LANGUAGE，當(dāng)然還有很多，? ?? ?
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間除星期一和七的天數(shù)? ?? ?
在前后分別調(diào)用DBMS_UTILITY.GET_TIME,??讓后將結(jié)果相減(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：第一條記錄的TIME??與最后一行是一樣的? ?? ?
可以建立一個(gè)函數(shù)來處理這個(gè)問題? ?? ?
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.? ?? ?
獲得小時(shí)數(shù)? ?? ?
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? ?? ?
獲取年月日與此類似? ?? ?
13.? ?? ?
年月日的處理? ?? ?
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.? ?? ?
處理月份天數(shù)不定的辦法? ?? ?
select??to_char(add_months(last_day(sysdate)??+1,??-2),??'yyyymmdd'),last_day(sysdate)??from??dual? ?? ?
16.? ?? ?
找出今年的天數(shù)? ?? ?
select??add_months(trunc(sysdate,'year'),??12)??-??trunc(sysdate,'year')??from??dual? ?? ?
閏年的處理方法? ?? ?
to_char(??last_day(??to_date('02'? ? |??|??:year,'mmyyyy')??),??'dd'??)? ?? ?
如果是28就不是閏年? ?? ?
17.? ?? ?
yyyy與rrrr的區(qū)別? ?? ?
'YYYY99??TO_C? ?? ?
-------??----? ?? ?
yyyy??99??0099? ?? ?
rrrr??99??1999? ?? ?
yyyy??01??0001? ?? ?
rrrr??01??2001? ?? ?
18.不同時(shí)區(qū)的處理? ?? ?
select??to_char(??NEW_TIME(??sysdate,??'GMT','EST'),??'dd/mm/yyyy??hh:mi:ss')??,sysdate? ?? ?
from??dual;? ?? ?
19.? ?? ?
5秒鐘一個(gè)間隔? ?? ?
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位秒數(shù)? ?? ?
20.? ?? ?
一年的第幾天? ?? ?
select??TO_CHAR(SYSDATE,'DDD'),sysdate??from??dual? ?? ?
310??2002-11-6??10:03:51? ?? ?
21.計(jì)算小時(shí),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)??作為年? ?? ?
floor((date2-date1,??365)??/30)??作為月? ?? ?
mod(mod(date2-date1,??365),??30)作為日.? ?? ?
23.next_day函數(shù)? ?? ?
next_day(sysdate,6)是從當(dāng)前開始下一個(gè)星期五。后面的數(shù)字是從星期日開始算起。? ?? ?
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??dual??
日期??返回的是天??然后??轉(zhuǎn)換為ss
轉(zhuǎn)此：http://www.onlinedatabase.cn/leadbbs/Announce/Announce.asp?BoardID=42&ID=1769
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