SELECT查询中的IF和CASE语句，其中包含子查询

Question

如果menu_items.parent不为0 ，则下面的查询返回NULL，这是错误的。

我在这里要做的是，如果menu_items.parent行的值为0，则将其作为原始值返回，但如果该值不为0，则返回该值的相应行的名称（varchar）。

知道为什么我一直变为NULL吗？

注意：所有其他选定的列都正常，因此查询本身很好。

谢谢

例1：

SELECT
...
...
(
   CASE
      WHEN menu_items.parent = '0' THEN menu_items.parent
      ELSE (SELECT name FROM menu_items WHERE id = menu_items.parent)
   END
) AS ParentID
...
...
FROM menus
INNER JOIN menu_items ...
...
...

例2：

SELECT
...
...
IF
(
   menu_items.parent = '0',
      menu_items.parent,
      (SELECT name FROM menu_items WHERE id = menu_items.parent)
) ParentID
...
...
FROM menus
INNER JOIN menu_items ...
...
...

实际查询：

SELECT
menus.name AS MenuName,
menu_items.id AS MenuItemsId,
menu_items.name AS MenuItemsName

/*
Sub section should go here
*/


FROM users
INNER JOIN assigned_menus ON assigned_menus.fk_users_id = users.id
INNER JOIN menu_items ON menu_items.id = assigned_menus.fk_menu_items_id
INNER JOIN menus ON menus.id = menu_items.fk_menus_id
WHERE
users.id = ? AND
users.is_active = 'YES' AND
menu_items.is_active = 'YES' AND
menus.is_active = 'YES'
ORDER BY
MenuName,
MenuItemsName

menu_items表

ID      name                    parent       fk_menu_items_id
1   Menus           0       1
2   Roles           0       1
3   Permissions     0       1
4   Users           0       1
5   Files           0       1
6   File Uploads        0       1
7   University      6       1
8   Details         6       1
9   Progress        6       1
10  Assg            6       1
11  Applications        0       2
12  New             11      2
13  Edit            11      2
14  Rejected        11      2
15  Approved        11      2
16  Exs         0       2
17  Assm            16      2
18  Stf         0       3
19  Std         0       3
20  Prg         19      3
21  Comm            0       4
22  Sms         21      4
23  Sms2            21      4
24  New2            0       4
25  Act         0       4

Answer 1

 SELECT
...
...
(
   CASE
      WHEN menu_items.parent = '0' THEN menu_items.parent
      ELSE (SELECT mi.name FROM menu_items mi WHERE mi.id = menu_items.parent)
   END
) AS ParentID
...
...
FROM menus
INNER JOIN menu_items ...
...
...