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  • 按天去除重复数据,为0则取0,否则取最大的那个值

    测试数据:
    mysql> select * from t2;
    +----+--------+---------------------+------------+
    | id | userid | inputDate           | infoStatus |
    +----+--------+---------------------+------------+
    |  1 |      1 | 2014-07-11 00:00:00 |      20013 | 
    |  2 |      1 | 2014-07-11 00:00:00 |          0 | 
    |  3 |      2 | 2014-07-12 00:00:11 |      20015 | 
    |  4 |      2 | 2014-07-12 00:00:22 |      20013 | 
    |  5 |      2 | 2014-07-14 00:00:00 |          0 | 
    |  6 |      2 | 2014-07-15 00:00:00 |      20013 | 
    |  7 |      2 | 2014-07-15 00:00:00 |          0 | 
    |  8 |      2 | 2014-07-16 00:00:00 |          0 | 
    |  9 |      2 | 2014-07-13 00:00:00 |      20013 | 
    | 10 |      2 | 2014-07-13 00:00:00 |          0 | 
    | 11 |      2 | 2014-07-14 00:00:00 |          0 | 
    +----+--------+---------------------+------------+
    
    查询sql:
    
    select id,userid,date(inputDate) as date,max(infoStatus) as m,min(infoStatus) as i,if(min(infoStatus)=0,0,max(infoStatus)) as max from t2 where infoStatus = 0 or infoStatus <= 99999 group by date;
    
    回执结果:
    +----+--------+------------+-------+-------+-------+
    | id | userid | date       | m     | i     | max   |
    +----+--------+------------+-------+-------+-------+
    |  1 |      1 | 2014-07-11 | 20013 |     0 |     0 | 
    |  3 |      2 | 2014-07-12 | 20015 | 20013 | 20015 | 
    |  9 |      2 | 2014-07-13 | 20013 |     0 |     0 | 
    |  5 |      2 | 2014-07-14 |     0 |     0 |     0 | 
    |  6 |      2 | 2014-07-15 | 20013 |     0 |     0 | 
    |  8 |      2 | 2014-07-16 |     0 |     0 |     0 | 
    +----+--------+------------+-------+-------+-------+
    
    我的思路是按照时间分组,并同时查询最小,跟最大的,如果最小的为0,那max字段就为0,否则就取最大的那个
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  • 原文地址:https://www.cnblogs.com/shibazi/p/3840774.html
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