Question

希望您能够帮助我使用这个MYSQL语句。我有一张这样的桌子：

    |id            |duration     |start               |
    |1110460       |8.2          |20171211            |
    |2221104       |8.9          |20171112            |
    |1110460       |3.2          |20171113            |
    |1110460       |4.4          |20171214            |
    |3331938       |3.2          |20180115            |
    |3331722       |5.4          |20171216            |
    |1948212       |9.2          |20171217            |
    |9219302       |3.2          |20171218            |

我想要做的是按给定开始月份的降序排列前4个ID。

例如，对于201712的前4个ID：

    |id            |duration     |
    |1110460       |12.6         |
    |1948212       |9.2          |
    |3331722       |5.4          | 
    |9219302       |3.2          |

任何帮助将不胜感激。这是我到目前为止所做的，但它一直在返回错误的结果：

    SELECT id, sum(duration) FROM table WHERE 
    start LIKE '201712%' ORDER BY sum(duration) DESC LIMIT 4

Answer 1

您应该使用分组

SELECT id, sum(duration) 
FROM table 
WHERE start LIKE '201712%' 
GROUP BY id
ORDER BY sum(duration) DESC LIMIT 4

Answer 2

要获得4个不同的最大总和及其ID，您可以使用以下

SELECT a.id, a.sum_duration 
FROM (SELECT id, sum(duration) sum_duration
      FROM demo 
      WHERE `start` LIKE '201712%' 
      GROUP BY id
) a
JOIN (SELECT distinct sum(duration) max_durations
      FROM demo 
      WHERE `start` LIKE '201712%' 
      GROUP BY id
      ORDER BY sum(duration) DESC 
      LIMIT 4 
) b on a.sum_duration = b.max_durations

如果2个ID具有相同的总和

，则上述版本将返回超过4行

DEMO

MYSQL根据标准选择前4个总和

2 个答案: