即使某些值不匹配，也显示连接表的结果

Question

访问数据库

table contacts
--------------
id
surname
name


table relations
---------------
contact_id
relation_id

contact_id 和relation_id 都是引用表contacts id 的外键

我想执行查询以获取联系人的姓/名和关系的姓/名如果当前联系人的关系存在。如果它不存在，我想获取联系人的姓/名和关系字段的空白值。

所有这些都在一个查询中

编辑：

我使用左连接。我正在使用 VB.NET 运行查询：

Dim myOleDbDataReader As OleDbDataReader = _
    New OleDbCommand( _
    "SELECT c.id           AS contact_id " & _
    "     , c.surname      AS contact_surname " & _
    "     , c.name         AS contact_name " & _
    "     , c2.id          AS related_id " & _
    "     , c2.surname     AS related_surname " & _
    "     , c2.name        AS related_name " & _
    "FROM ((contacts c " & _
    "LEFT JOIN relations r " & _
    "ON c.id = r.contact_id) " & _
    "INNER JOIN contacts c2 " & _
    "ON c2.id = r.relation_id)" _
    , connection).ExecuteReader()

我得到 OleDbException: Join expression not supported.

他们在另一篇文章中说： “当您在 FROM 子句中使用 LEFT/RIGHT/INNER JOINS 时，Access 不允许您在 where 子句中使用常规连接。这可能是故意让您购买更昂贵的软件。” - (Is the join expression not supported by MS Access?)

不完全是这样。从我尝试的一些示例中，我得出的结论是：

Access 不允许您将外部联接（左/右）与一个或多个内部联接一起使用。 以约翰·卡马克的名义，我能做什么？ 我想避免单独的选择查询。 请帮忙...

Answer 1

SELECT c.id       AS contact_id
     , c.surname  AS contact_surname
     , c.name     AS contact_name
     , c2.id      AS related_id
     , c2.surname AS related_surname
     , c2.name    AS related_name
FROM contacts c
  LEFT JOIN relations r
    ON c.id = r.contact_id
  JOIN contacts c2
    ON r.relation_id = c2.id

上述方法在 MS-Access 中不起作用。

这略有不同（两个左连接），但它有效：

SELECT c.id       AS contact_id
     , c.surname  AS contact_surname
     , c.name     AS contact_name
     , c2.id      AS related_id
     , c2.surname AS related_surname
     , c2.name    AS related_name
FROM contacts c
  LEFT JOIN
    ( relations r
      LEFT  JOIN contacts AS c2
        ON r.relation_id = c2.id
    )  
    ON c.id = r.contact_id

尽管有第二个LEFT JOIN，但它会给出相同的结果集，因为第二个 LEFT JOIN 涉及外键关系（方向从多 -> 一）。

要使用 INNER JOIN 进行 LEFT JOIN，您可以使用：

SELECT c.id       AS contact_id
     , c.surname  AS contact_surname
     , c.name     AS contact_name
     , g.id       AS related_id
     , g.surname  AS related_surname
     , g.name     AS related_name
FROM contacts c
  LEFT JOIN
    ( SELECT r.contact_id
           , c2.id      
           , c2.surname
           , c2.name 
      FROM relations r
        INNER JOIN contacts AS c2
          ON r.relation_id = c2.id
    )  AS g
    ON c.id = g.contact_id