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im
V2EX  ›  MySQL

求一个考勤 SQL 语句,肯定有大神愿意帮我,多谢

  •  
  •   im · Nov 22, 2016 · 3368 views
    This topic created in 3606 days ago, the information mentioned may be changed or developed.
    按card_date和user_sn分组,取每天当中第一天数据和最后一条数据
    意思就是查询一个月当中每个人每天的第一次刷卡和最后一次刷卡

    card_time | user_sn | card_date
    2016-11-01 8:20:11 | 0012 | 2016-11-01
    2016-11-01 13:45:22 | 0012 | 2016-11-01
    2016-11-01 18:35:15 | 0012 | 2016-11-01
    2016-11-01 8:20:11 | 0013 | 2016-11-01
    2016-11-01 11:45:22 | 0013 | 2016-11-01
    2016-11-01 13:45:22 | 0013 | 2016-11-01
    2016-11-01 18:35:15 | 0013 | 2016-11-01
    2016-11-02 8:20:11 | 0012 | 2016-11-02
    2016-11-02 13:45:22 | 0012 | 2016-11-02
    2016-11-02 18:35:15 | 0012 | 2016-11-02
    2016-11-02 8:20:11 | 0013 | 2016-11-02
    2016-11-02 11:45:22 | 0013 | 2016-11-02
    2016-11-02 13:45:22 | 0013 | 2016-11-02
    2016-11-02 18:35:15 | 0013 | 2016-11-02
    2016-11-02 8:35:15 | 0014 | 2016-11-02
    2016-11-02 18:26:15 | 0014 | 2016-11-02
    2016-11-02 7:11:15 | 0015 | 2016-11-02

    取出结果:
    card_time | user_sn | card_date
    2016-11-01 8:20:11 | 0012 | 2016-11-01
    2016-11-01 18:35:15 | 0012 | 2016-11-01
    2016-11-01 8:20:11 | 0013 | 2016-11-01
    2016-11-01 18:35:15 | 0013 | 2016-11-01
    2016-11-02 8:20:11 | 0012 | 2016-11-02
    2016-11-02 18:35:15 | 0012 | 2016-11-02
    2016-11-02 8:20:11 | 0013 | 2016-11-02
    2016-11-02 18:35:15 | 0013 | 2016-11-02
    2016-11-02 8:35:15 | 0014 | 2016-11-02
    2016-11-02 18:26:15 | 0014 | 2016-11-02
    2016-11-02 7:11:15 | 0015 | 2016-11-02
    enenaaa
        1
    enenaaa  
       Nov 22, 2016   ❤️ 1
    SELECT MIN(card_time) AS mintime, MAX(card_time) AS maxtime , user_sn, card_date FROM testtbl GROUP BY user_sn, card_date
    im
        2
    im  
    OP
       Nov 22, 2016
    @enenaaa 多谢,你这样是可以的,但是我想要的是 min 和 max 分成二条数据显示
    enenaaa
        3
    enenaaa  
       Nov 22, 2016   ❤️ 1
    @im union all 一下就可以了
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