Question

我有两张表accounts和subs 第一个表格包含id，第二个表格包含字段id，requester_account_id，grabber_account_id

我想要计算一个帐户制作了多少个子请求，以及他做了多少抓取（基于requester_account_id/grabber_account_id是否填充了id）

在一个查询中

输出：

+------------+----------------+-----------------+
| account_id | subs_requested | subs_grabbed    |
+------------+----------------+-----------------+
|  1         | 4              | 3               |
|  3         | 2              | 1               |
+------------+----------------+-----------------+

Answer 1

使用类似

的内容

select accounts.id, 
    count(distinct s1.id) as num_req, 
    count(distinct s2.id) as num_grab
from accounts left join subs as s1 on accounts.id = s1.requester_account_id
    left join subs as s2 accounts.id = s2.grabber_account_id
group by accounts.id

诀窍是使用表格subs两次：subs as s1和subs as s2，每次都加入不同的字段....

关于效率的注意事项：我不确定，但我相信这个解决方案比子查询解决方案更快，但没有经过测试（至少它不会慢）。我总是喜欢left join而不是子查询。

Answer 2

这应该有效：

SELECT a.account_id,
(SELECT COUNT(*) FROM subs WHERE requester_account_id=a.account_id) AS subs_requested,
(SELECT COUNT(*) FROM subs WHERE grabber_account_id=a.account_id) AS subs_grabbed
FROM accounts a

Answer 3

据我了解你在寻找什么，它会是这样的：

select
    a.id as account_id,
    count(distinct sr.id) as requests,
    count(distinct gr.id) as grabs
from
    accounts a
left outer join subs sr
    on sr.requester_account_id = a.id
left outer join subs gr
    on gr.grabber_account_id = a.id

有时候，我在使用左边连接上的聚合函数时遇到了奇怪的问题，所以这是另一种方式：

select
    a.id as account_id,
    (select count(distinct sr.id) from subs sr where sr.requester_account_id = a.id) as requests,
    (select count(distinct gr.id) from subs sr where gr.grabber_account_id = a.id) as grabs
from 
    accounts a

Answer 4

如果需要，您只能加入子表一次（不确定这表现得更好）

select 
acc.id
sum(if(sub.requester_account_id = acc.id, 1, 0)) as req,
sum(if(sub.grabber_account_id = acc.id, 1, 0)) as grab
from subs sub, accounts acc
where sub.requester_account_id = acc.id or sub.grabber_account_id = acc.id
group by acc.id

在不同的密钥上连接两次相同的表

4 个答案: