MySQL SELECT 查询中的 IF 和 CASE 语句,其中包含子查询

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时间:2020-08-31 16:48:10  来源:igfitidea点击:

IF and CASE statements in SELECT query with a subquery in them

mysqlsubqueryinner-join

提问by BentCoder

Queries below return NULL if menu_items.parentis not 0which is wrong.

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

What I'm trying to do here is, if value of menu_items.parentrow is 0then return it as original value but if the value is not 0then return the name (varchar) of the corresponding row of what ever the value is.

我在这里尝试做的是,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,则返回该值对应的行的名称(varchar)。

Any idea why I keep getting NULL?

知道为什么我一直得到 NULL 吗?

Note: All the other selected columns comes as normal so the query itself is fine.

注意:所有其他选定的列都正常,因此查询本身没问题。

Thanks

谢谢

EXAMPLE 1:

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 ...
...
...

EXAMPLE 2:

示例 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 ...
...
...

ACTUAL QUERY:

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 TABLE

菜单项表

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

回答by Romil Kumar Jain

 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 ...
...
...