Question

在我的数据库中，我有两个表：

动作

+--------------+--------------+------+-----+---------+----------------+
| Field        | Type         | Null | Key | Default | Extra          |
+--------------+--------------+------+-----+---------+----------------+
| id           | int(11)      | NO   | PRI | NULL    | auto_increment |
| lead_id      | int(11)      | YES  | UNI | NULL    |                |
| type         | varchar(255) | NO   |     | NULL    |                |
+--------------+--------------+------+-----+---------+----------------+

铅

+---------+--------------+------+-----+---------+----------------+
| Field   | Type         | Null | Key | Default | Extra          |
+---------+--------------+------+-----+---------+----------------+
| id      | int(11)      | NO   | PRI | NULL    | auto_increment |
| status  | varchar(255) | YES  |     | NULL    |                |
| created | datetime     | NO   |     | NULL    |                |
| lead_id | int(11)      | NO   | MUL | NULL    |                |
+---------+--------------+------+-----+---------+----------------+

排行榜有点不寻常，因为潜在客户是非唯一的，并由lead_id标识，因此可能会有多个潜在客户具有相同的lead_id。

我想要做的是检索所有操作，然后将它们加入条件它们是最新的（MAX(created)？）。

This回答部分解释了如何处理引导表，但是我无法找到如何以正确的方式在子查询上进行连接。

Answer 1

有多种方法可以做到这一点。以下内容将条件放在on子句中：

select a.*
from lead l join
     action a
     on l.lead_id = a.lead_id and
        l.created = (select MAX(created) from leads l2 where l2.lead_id = l.lead_id)

您也可以使用显式聚合子查询执行此操作：

select a.*
from lead l join
     action a
     on l.lead_id = a.lead_id  join
     (select lead_id, max(created) as maxcreated
      from lead l
      group by lead_id
    ) lmax
    on lmax.lead_id = l.lead_id and lmax.maxcreated = l.created;

Answer 2

未经验证，但我认为您需要以下内容：

select action.*, max_leads.* from action left join
  (select lead_id, max(created) from lead group by lead_id) max_leads
  on action.lead_id = max_leads.lead_id

SQL - 加入子查询条件

2 个答案: