Question

我有这个表结构

leadid agentid datetime            ip
     1       6 2016-06-17 12:55:48 127.0.0.1
     5       6 2016-06-17 12:56:26 127.0.0.1
     9       6 2016-06-17 12:58:18 127.0.0.].
    13       6 2016-06-17 12:58:19 127.0.0.1
    17       6 2016-06-17 12:58:20 127.0.0.1
     2       7 2016-06-17 12:55:54 127.0.0.1
     6       7 2016-06-17 12:56:32 127.0.0.1
    10       7 2016-06-17 12:58:18 127.0.0.1
    14       7 2016-06-17 12:58:19 127.0.0.1
    18       7 2016-06-17 12:58:20 127.0.0.1
     3       8 2016-06-17 12:55:56 127.0.0.].
     7       8 2016-06-17 12:58:18 127.0.0.1
    11       8 2016-06-17 12:58:19 127.0.0.1
    15       8 2016-06-17 12:58:20 127.0.0.1
    19       8 2016-06-17 12:58:21 127.0.0.1
     4       9 2016-06-17 12:56:22 127.0.0.1
     8       9 2016-06-17 12:58:18 127.0.0.1
    12       9 2016-06-17 12:58:19 127.0.0.1
    16       9 2016-06-17 12:58:20 127.0.0.1
    20       9 2016-06-17 12:58:21 127.0.0.1

我想为每个agentid选择一条记录，其中datetime按升序排序例如agentid 6 with datetime 2016-06-17 12:55:48应选择agentid 7 datetime 2016-06-17 12:55:54应该被选中

是否可以通过一个查询来完成？

我知道查询没有意义，只是为了解释我想做什么

SELECT COUNT(agentid)
     , `agentid`
     , `datetime` 
  FROM leadassignment 
 GROUP 
    BY agentid 
 ORDER 
    BY datetime ASC

Expression #3 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'timeshareleads.leadassignment.datetime' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by

Answer 1

您可以使用MIN功能，该功能会为您提供每个年龄段的最短日期时间。

SELECT count(agentid), `agentid`, min(`datetime`) as 'datetime'
FROM leadassignment 
GROUP BY agentid

mysql记录排序问题

1 个答案: