mysql - 如何返回连接表中所有不匹配的行，如果匹配则检查条件？

Question

User 时如何返回表的所有行

1. a row has not any match with join table ex **UserDetails**
2. a row has a match in **UserDetails**, check some condition ex: user location is **IND**

示例表

     User
     -----------------------
     id   name
     -----------------------
     1    Hearaman
     2    Ramse
     3    Temmy
     4    Robert

     UserDetails
     -----------------------------
     id   user_id     location
     1    3           USA
     2    4           IND

     Expected results
     --------------------------
     id    name        location
     --------------------------
     1     Hearaman    null
     2     Ramse       null
     4     Robert      IND

Answer 1

您可以使用left join 并进行过滤

select u.id, u.name, ud.location
from User u 
left join UserDetails ud on ud.user_id = u.id 
where ud.user_id is null or 
      ud.location = 'IND';