每个客户过去 3 天的 mysql select 行

Question

我想根据每个客户的最大时间戳在 MySQL 过去三天的行中 select，但我不确定如何执行此操作。

示例table：

id     | timestamp           | cust_ID
899900 | 2016-04-09 12:00:00 | 500219
899901 | 2016-04-12 16:00:00 | 500219
899902 | 2016-04-14 11:00:00 | 500219
899903 | 2016-04-15 12:00:00 | 500219
909901 | 2016-04-08 16:00:00 | 500230
909902 | 2016-04-14 11:00:00 | 500230
909903 | 2016-04-15 12:00:00 | 500230
939905 | 2016-05-02 19:00:00 | 500240

到目前为止我有以下内容：

SELECT * 
FROM table 
WHERE timestamp BETWEEN max(timestamp) AND (max(timestamp) - INTERVAL 3 DAY) 
GROUP BY cust_ID

但是报错：

1111 'invalid use of group function'

Answer 1

我希望能正确理解您要查找的内容，但您可能需要聚合函数而不是 where

SELECT cust_id, max(timestamp) 
from table 
GROUP BY cust_ID
having max(timestamp)  BETWEEN  date_sub((max(timestamp) , INTERVAL 3 DAY) AND   max(timestamp)  ;

Answer 2

将 table 加入自身的聚合，为每个客户找到最大时间戳：

SELECT t.*
FROM mytable t
JOIN (SELECT cust_ID, MAX(timestamp) max_timestamp    -- each cust_ID's max timestamp
      FROM mytable
      GROUP BY cust_ID) m
  ON timestamp BETWEEN max_timestamp - INTERVAL 3 DAY AND max_timestamp
  AND t.cust_ID = m.cust_ID           -- join on the specific cust_ID's max timestamp

另请注意，您必须在 BETWEEN 中使用较小的表达式 first，即

WHERE myColumn BETWEEN smaller AND larger

如果你反过来编码：

WHERE myColumn BETWEEN larger AND smaller  -- don't do this

它仍会编译和执行，但您不会返回任何行。

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