mysql—表中两列的计数，以及一列与另一列的计数

我希望将这些查询纳入其中：

select count(first_column) as first from table1 where user_id = $id
select count(second_column) as second from table1 where user_id = $id
select count(first_column) as third from table2 where user_id = $id

我有这个：

select COUNT(table1.first_column) AS first, 
       COUNT(table1.second_column) AS second, 
       COUNT(table2.first_column) AS third 
from table2 inner join
        table2 on table1.user_id = table2.user_id 
where table1.user_id = 1;

但它为表2返回了错误的值。

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如何解决此问题？

一种简单的方法是使用子查询：

select (select count(first_column) as first from table1 where user_id = $id) as first,
       (select count(second_column) as second from table1 where user_id = $id) as second,
       (sselect count(first_column) as third from table2 where user_id = $id) as third;

因为前两个来自同一个表，您可以考虑：

select count(t1.first_column), count(t1.second_column), t2.third_column
from table1 t1 cross join
     (select count(third_column as third from table2) t2
where t1.user_id = $id;

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这将在MySQL中工作。在其他数据库中，第三列需要一个聚合函数。

您能给出示例数据、输出和所需输出吗？我已经从表1和表2中了解了。我想计算用户关注度，关注度和帖子数量。数据如下：user_id:1，follows:11，follows:null-user_id:1，follows:24，follows:null-user_id:1，follows:null，follows:54和posts表中的post_id:22，user_id:1-post_id:26，user_id:1。当我添加group by table2.post_id时，它返回表1的正确值，但不返回表2的正确值。下面和后面是目标用户的id，而不是此处的计数。