mysql - 解码查询结果

Question

如果两个请求都返回相同的数据，那么它们将被“组合”到联合中。

query1 return X 
query2 return X

query1 union query2 renurn only X

But i need to get
+---+
| X |
| X |
+---+

如果每个请求返回不相等的数据，union 的结果将由两行组成。这是正常的。

mysql> select count(pid) from posts 
where uid_posts=8890 and postacc=1 and postcomacc=1 
union
select count(comid) from coms join posts on pid_coms=pid 
where uid_posts!=8890 and uid_coms=8890 and postacc=1 and postcomacc=1;
+------------+
| count(pid) |
+------------+
|          1 |
|          2 |
+------------+

我想实现这个。

---

但是使用联合的常见查询“分组”了这些相同的值。

mysql> select count(pid) from posts
where uid_posts=8890 and postacc=1 and postcomacc=1 ;
+------------+
| count(pid) |
+------------+
|          2 |
+------------+


mysql>  select count(comid) from coms join posts on pid_coms=pid 
where uid_posts!=8890 and uid_coms=8890 and postacc=1 and postcomacc=1;
+--------------+
| count(comid) |
+--------------+
|            2 |
+--------------+



//thay are "grouped" :( below

mysql> select count(pid) from posts
where uid_posts=8890 and postacc=1 and postcomacc=1 
union
select count(comid) from coms join posts on pid_coms=pid 
where uid_posts!=8890 and uid_coms=8890 and postacc=1 and postcomacc=1;
+------------+
| count(pid) |
+------------+
|          2 |
+------------+

如何取消对这些结果的分组？

或者我需要重新处理整个查询吗？

停下! UNION ALL 必须帮助我。

Answer 1

使用union all:

select count(pid)
from posts 
where uid_posts = 8890 and postacc = 1 and postcomacc = 1 
union all
select count(comid)
from coms join
     posts
     on pid_coms = pid 
where uid_posts <> 8890 and uid_coms = 8890 and postacc = 1 and postcomacc = 1

我还会添加第二列来指定哪个计数与哪个查询相关。