Question

给出如下查询：

SELECT 
  id, 
  (SELECT COUNT(*) 
   FROM members 
   WHERE members.network_id = networks.id) AS mem_count
FROM 
  networks
WHERE mem_count > 2

使用此查询，where子句会中断，因为它不知道mem_count是什么...为什么我不能在where子句中使用as var？

由于

Answer 1

虽然bernie建议正确回答问题，但您的查询可以简化为：

SELECT
  network_id as id,
  count(*)
FROM
  members
GROUP BY
  network_id 
HAVING
  count(*) > 2

作为额外的奖励，可以更快。

Answer 2

你已经把这个概念搞定了。你只需要正确的语法。您可以像这样重写，并且可以使您的查询符合ANSI标准：

SELECT 
  id, 
  m.mem_count
FROM 
  networks n 
  JOIN (
    SELECT   m.network_id, 
             COUNT(*) AS mem_count
    FROM     members 
    GROUP BY m.network_id
  ) m 
  ON m.network_id = n.id 
  AND m.mem_count > 2;

Answer 3

尝试：

SELECT 
  id, 
  (SELECT COUNT(*) as mem_count
   FROM members 
   WHERE members.network_id = networks.id)
FROM 
  networks
WHERE mem_count > 2

Answer 4

一种方式是。

Select * From (
SELECT 
  id, 
  (SELECT COUNT(*) 
   FROM members 
   WHERE members.network_id = networks.id) AS mem_count
FROM 
  networks)) mem_counts
WHERE mem_count > 2

伯尼建议的加入会更好。基本上你混淆了解析器。当您使用AS来别名列名时，您会遇到group by或order by的同类问题。

如何在查询WHERE子句中使用AS名称？

4 个答案: