mysql - 使用 mysql join 查询选择数据

Question

此查询有效。

表用户

userid   firstname    lastname
1         JOHN         DEO
2         JANE         DEO

TABLE msg

msg_id    msg_from     msg_to      received   
1         userid(1)    userid(2)   null       


SELECT SND.userid, SND.firstname, SND.lastname
FROM msg as M
JOIN users as SND 
  ON SND.userid = M.msg_from
WHERE M.msg_to = 'userid -1'
  AND M.received IS NULL

但是这个查询不起作用

TABLE users

userid   firstname    lastname
1         JOHN         DEO
2         JANE         DEO

TABLE msg

msg_id    msg_from     msg_to      received   age  city  country
1         userid(1)    userid(2)   null       26   any   any


SELECT SND.userid, SND.firstname, SND.lastname, SND.age, SND.city, SND.country
FROM msg as M
JOIN users as SND 
  ON SND.userid = M.msg_from
WHERE M.msg_to = 'userid -1'
  AND M.received IS NULL

我想从表 msg 上的数据库中选择年龄城市国家，但看起来我的查询有问题，从表 msg 中选择年龄城市和国家

Answer 1

您需要将 SND.age,SND.city,SND.country 更改为 M.age,M.city,M.country，因为它们来自表 消息

SELECT SND.userid, SND.firstname, SND.lastname, M.age, M.city, M.country
FROM msg as M
JOIN users as SND 
  ON SND.userid = M.msg_from
WHERE M.msg_to = 'userid -1'
  AND M.received IS NULL