即使某些值不匹配，也显示已连接表的结果

Question

访问数据库

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


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

contact_id和relation_id都是引用表联系人id

的外键

如果存在当前联系人的关系，我想执行查询以获取联系人的姓氏/姓名和关系的姓氏/名称。如果它不存在，我想获得联系人的姓氏/姓名和关系字段的空白值。

所有这一切都在一个查询中

编辑：

我用左连接。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：不支持连接表达式。

他们在另一篇文章中说： “在FROM子句中使用LEFT / RIGHT / INNER JOINS时，访问不会让你在where子句中使用常规连接。这可能是为了让你购买更昂贵的软件。” - （Is the join expression not supported by MS Access?）

并非如此。从一些例子我尝试了我得出的结论：

Access不允许您将外部联接（LEFT / RIGHT）与一个或多个INNER JOINS一起使用。 约翰卡马克的名字我能做什么？ 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

以上在MS-Access中执行无法正常工作。

略有不同（两个左连接），但有效：

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

尽管第二个LEFT JOIN，它将给出相同的结果集，因为第二个LEFT JOIN涉及外键关系（从许多方向 - >一个）。

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