在IN查询中使用IN

Question

我无法合并此查询，但如果单独执行两者都可以正常工作。

select name 
   from zone 
   where id IN (select proj_tbl_id from project where proj_name = '$project2')

第二次查询的结果

select proj_tbl_id from project where proj_name = '$project2') = (13,14)

如果运行如下，结果为真。

select name from zone where id IN (13,14)

Answer 1

您可以使用exists运算符：

SELECT name
FROM   zone z
WHERE  EXISTS (SELECT *
               FROM   project p
               WHERE  z.id = p.proj_tbl_id AND
                      proj_name = '$project2')