1

我正在尝试在下面运行脚本,但总是为 name 字段获取 NULL 值。

SELECT u.name AS _user_name, s.name AS _school_name
FROM fwg_files AS f
LEFT JOIN users AS u ON u.id = f.user_id
LEFT JOIN user_profiles AS up ON up.user_id = u.id
LEFT JOIN school AS s ON s.id = up.profile_value

在我看来,这个问题在 JOIN ON school 表中,我尝试 SELECT s.id 并且它也返回 NULL 值。

表 fwg_files

id  | user_id
240 | 414
241 | 436

表用户

id  | name
414 | Name 1
436 | Name 2

表 user_profiles

user_id | profile_value
414     | "6"
436     | "14"

表学校

id | name

 6 | School 1
14 | School 2

谢谢

4

1 回答 1

0

不确定您的数据和架构,但如果 up.profile_value 类似于“123”,您可以试试这个:

SELECT u.name AS _user_name, s.name AS _school_name
FROM fwg_files AS f
LEFT JOIN users AS u ON u.id = f.user_id
LEFT JOIN user_profiles AS up ON up.user_id = u.id
LEFT JOIN school AS s ON CONCAT('"', s.id, '"') = up.profile_value
于 2013-02-26T11:12:44.013 回答