oracle 取某個(gè)時(shí)間段的數(shù)據(jù)，具體代碼如下所示：
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select count(*),t.分組字段 from (select t.* ,to_char(t.時(shí)間,'HH24') stime,to_char(t.時(shí)間,'HH24mi') fz,to_char(時(shí)間,'d') from A twhere 時(shí)間>=to_date('2019-12-01','yyyy-MM-dd') and ghsj<=to_date('2019-12-31','yyyy-MM-dd') and to_char(時(shí)間,'d')='2') where stime in ('08','09','10','11','12') and fz>=0800 a group by T.分組字段to_char(時(shí)間,'d') 取當(dāng)前時(shí)間是星期幾 每星期第一天為周日to_char(t.時(shí)間,'HH24mi') 取當(dāng)前時(shí)間的小時(shí)分o_char(t.時(shí)間,'HH24') 取當(dāng)前時(shí)間的小時(shí) |
ps：oracle 同一個(gè)數(shù)據(jù)有多條記錄,根據(jù)條件取時(shí)間最大的那一條
1.第一種方式
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select max(t1.INVALID_TIME) from T_CUSTOMER t1 where t1.customer_code = '5101' |
1.第二種方式
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SELECT INVALID_TIME FROM (SELECT * FROM T_CUSTOMER WHERE customer_code='5101' ORDER BY INVALID_TIME desc) WHERE ROWNUM =1 |
1.第三種方式
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select INVALID_TIME from T_CUSTOMER t where INVALID_TIME=(select max(INVALID_TIME) from T_CUSTOMER where customer_code='5101') and ROWNUM =1 |
5.不過濾存在多條最大時(shí)間
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select INVALID_TIME from T_CUSTOMER t wheret.INVALID_TIME = (select max(t1.INVALID_TIME) from T_CUSTOMER t1 where t1.customer_code = '5101') |
總結(jié)
以上所述是小編給大家介紹的oracle 取某個(gè)時(shí)間段的數(shù)據(jù)（每周幾的上午幾點(diǎn)到幾點(diǎn)），希望對大家有所幫助！
原文鏈接：https://www.cnblogs.com/chenxiaoyao/archive/2020/01/03/12145813.html
