Question

我在MySQL“mytable”中有一个表格，如下所示：

    id    car_num      date  
    -----  --------   -----------
    00001    BBB      2015-09-28
    00002    BBB      2015-10-04
    00003    DDD      2015-10-04
    00004    BBB      2015-10-04
    00005    AAA      2015-10-05
    00006    GGG      2015-10-05
    00007    GGG      2015-10-05
    00009    EEE      2015-10-06
    00010    AAA      2015-10-06
    00011    HHH      2015-10-10
    00012    FFF      2015-10-11
    00013    FFF      2015-10-11
    00014    CCC      2015-10-13

这是我的查询显示所有重复项（字段car_num）：

    SELECT  a.*
    FROM    mytable a
        INNER JOIN
        (
         SELECT  car_num
         FROM    mytable
         GROUP   BY car_num
         HAVING  COUNT(car_num) >= 2
        ) b ON a.car_num = b.car_num
    ORDER BY car_num ASC

结果是：

     id    car_num      date  
    -----  --------   -----------
    00005    AAA      2015-10-05
    00010    AAA      2015-10-06
    00001    BBB      2015-09-28
    00002    BBB      2015-10-04
    00004    BBB      2015-10-04
    00012    FFF      2015-10-11
    00013    FFF      2015-10-11
    00006    GGG      2015-10-05
    00007    GGG      2015-10-05

现在我只想在每个重复项的日期字段之间存在少于或等于1天的情况下显示重复项，并且我想按日期DESC订购它们

我的期望：

     id    car_num      date  
    -----  --------   -----------
    00012    FFF      2015-10-11
    00013    FFF      2015-10-11
    00010    AAA      2015-10-06
    00005    AAA      2015-10-05
    00006    GGG      2015-10-05
    00007    GGG      2015-10-05
    00002    BBB      2015-10-04
    00004    BBB      2015-10-04

是否可以直接使用MySQL查询（带有DATEDIFF的东西）？任何帮助表示赞赏

Answer 1

您可以这样做：

select t.*
from mytable t
where exists (select 1
              from mytable t2
              where t.car_num = t2.car_num and 
                    t.id <> t2.id and
                    abs(datediff(t2.date, t.date)) <= 1
             );

MySQL列出特定字段上的所有重复项，并且仅在另一个字段日期之间的日期<= 1

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