Question

我正在尝试按clientid和m.id分组。如何将这2个查询转换为1个查询？

查询1

select cs.clientid, count(distinct(m.id)) as cp
into #temp
from call_table cs 
join mem_table m 
on cs.CLIENTID = m.clientid
join activity_table a 
on a.memid = m.id 
where a.activity = 'abc'
group by cs.clientid, m.id

查询2

select clientid, sum(cp) from #temp
group by ClientId

更新

更具体地说，我希望在不使用临时表的情况下编写此查询。

Answer 1

您只能按客户ID分组

并且由于没有选择来自activity_table的字段，因此可以使用EXISTS代替。

select cs.clientid, count(distinct m.id) as cp
from call_table cs 
join mem_table m on m.clientid = cs.clientid
where exists 
(
    select 1
    from activity_table a 
    where a.memid = m.id 
      and a.activity = 'abc'
)
group by cs.clientid

Answer 2

为什么不这样写：

with temp as (
      select cs.clientid, count(distinct m.id) as cp
      from call_table cs join
           mem_table m 
           on cs.CLIENTID = m.clientid join
           activity_table a 
           on a.memid = m.id 
      where a.activity = 'abc'
      group by cs.clientid, m.id 
     )
select clientid, sum(cp)
from temp
group by ClientId;

我不知道逻辑应该是什么，但这似乎是组合查询的最简单方法。

我推测您只是想要此查询：

      select cs.clientid, count(distinct m.id) as cp
      from call_table cs join
           mem_table m 
           on cs.CLIENTID = m.clientid join
           activity_table a 
           on a.memid = m.id 
      where a.activity = 'abc'
      group by cs.clientid;

也就是说，从m.id中删除group by。而这又可以简化为：

      select m.clientid, count(distinct m.id) as cp
      from mem_table m join
           activity_table a 
           on a.memid = m.id 
      where a.activity = 'abc'
      group by cs.clientid;

假设m.id是唯一的而非非NULL，则可以进一步简化为：

      select m.clientid, count(*) as cp
      from mem_table m join
           activity_table a 
           on a.memid = m.id 
      where a.activity = 'abc'
      group by cs.clientid;

将多个查询合并为一个查询

2 个答案: