Question

我想从连接构造中删除，我必须为其提供别名（“映射”），因为我还必须在连接中使用EXISTS子句。所以整个事情看起来像那样：

DELETE a
FROM (TableA a INNER JOIN 
        (SELECT * FROM TableX x INNER JOIN TableY y ON x.id = y.id) map
      ON a.key = map.key) mapped
WHERE EXISTS
      (SELECT * 
       FROM  LookUp l 
       WHERE l.key1 = mapped.TableAKey 
       AND   l.key2 = mapped.TableXKey
       AND   l.key3 = mapped.TableYKey)

问题似乎与括号有关，因为我收到错误：

Incorrect syntax near 'mapped'.

任何帮助都将不胜感激。

Answer 1

我认为DELETE语句需要引用FROM子句中使用的别名。由于您的别名是“已映射”，请尝试按以下方式更改DELETE：

DELETE mapped
FROM (TableA a INNER JOIN 
        (SELECT * FROM TableX x INNER JOIN TableY y ON x.id = y.id) map
      ON a.key = map.key) mapped
WHERE EXISTS
      (SELECT * 
       FROM  LookUp l 
       WHERE l.key1 = mapped.TableAKey 
       AND   l.key2 = mapped.TableXKey
       AND   l.key3 = mapped.TableYKey)

Answer 2

只需重写它，这样你就可以从显式表中删除并处理where子句中的所有条件。

DELETE TableA
WHERE
    EXISTS (
        SELECT * 
        FROM 
            TableX x 
            INNER JOIN TableY y ON x.id = y.id
        WHERE
            x.key = TableA.key and
            EXISTS (
                SELECT * 
                FROM  LookUp l 
                WHERE l.key1 = TableA.TableAKey 
                AND   l.key2 = x.TableXKey
                AND   l.key3 = y.TableYKey
            )
    )

另请注意，在您处理where子句时，将DELETE TableA替换为SELECT * FROM TableA可能会有所帮助，以确切了解哪些记录将被删除。

从别名连接构造中删除

2 个答案: