在多个连接中引用相同的行

Question

我有一个名为relOwner的mySQL数据库，它有两列： OwnerID，RelationshipOwner

我正在编写一个带有引用db：

的连接的查询

$query = "SELECT b.Contact, b.ContactB, relOwner.OwnerID, relOwner.RelationshipOwner 
    FROM b 
    Left JOIN relOwner
    ON b.Contact = relOwner.OwnerID
    Left JOIN relOwner
    ON b.ContactB = relOwner.OwnerID
";

如何在我的php中单独引用RelationshipOwner的值？

$RelationshipOwner = $row['RelationshipOwner'];
$RelationshipOwnerB = $row['RelationshipOwner']; <--- Get value from second JOIN

提前致谢。

Answer 1

您似乎在表b上有两个外键列到表relOwner（即Contact和ContactB）。

根据Sverri的评论，您需要为表使用不同的别名（我使用过ro1和ro2），并从不同的表列中设置不同的名称（例如前缀带有ro2）的第二个表格列：

SELECT b.Contact, b.ContactB, ro1.OwnerID, ro1.RelationshipOwner, 
   ro2.OwnerID as ro2OwnerId, ro2.RelationshipOwner as ro2RelationshipOwner
FROM b -- Is this table Contact? If so then "Contact b"
  Left JOIN relOwner ro1
  ON b.Contact = ro1.OwnerID
  Left JOIN relOwner ro2
  ON b.ContactB = ro2.OwnerID;

然后您可以参考：

$row['ro2RelationshipOwner'];