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解决时间 2021-02-28 23:35

提问者网友：人傍凄凉立暮秋
2021-02-28 05:59

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五星知识达人网友：酒醒三更
2021-02-28 07:18

【答案】 --迟到SELECt s.name 名字, w.daynumber 日期FROM tbl_staff s, tbl_worktime wWHERe s.id =w.idAND to_date(w.intime,'hh24-mi-ss')>to_date('9:00','hh24-mi-ss') ;--早退SELECt s.name 名字, w.daynumber 日期FROM tbl_staff s, tbl_worktime wWHERe s.id =w.idAND to_date(w.outtime,'hh24-mi-ss')<to_date('18:00','hh24-mi-ss');--缺勤,暂时没想到好的select name,1 daynumber from tbl_staff where id not in(select id from tbl_worktime where daynumber=1)union allselect name,2 daynumber from tbl_staff where id not in(select id from tbl_worktime where daynumber=2)union allselect name,3 daynumber from tbl_staff where id not in(select id from tbl_worktime where daynumber=3)...select name,30 daynumber from tbl_staff where id not in(select id from tbl_worktime where daynumber=30)--部门日均工作时间SELECt s.dept 部门, SUM(ceil((To_date(w.outtime , 'hh24-mi-ss') - To_date(w.intime , 'hh24-mi-ss')) * 24*60))/(30*COUNT(s.dept)) AS 按分钟计算FROM tbl_staff s, tbl_worktime wWHERe s.id=w.idGROUP BY s.dept;

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1楼网友：孤独的牧羊人
2021-02-28 08:02

谢谢解答

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