连接表的显示结果，即使某些值不匹配

Question

Access数据库

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


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

两个CONTACT_ID和relation_id的外键参照表联系人的ID

我想执行一个查询来获取如果当前联系人的关系存在，则联系人的姓/名和关系的姓/名称。如果它不存在，我想获取联系人的姓氏/姓名和关系字段的空白值。

所有这一切都在一个查询

编辑：

我用左加入。我正在使用VB.NET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：不支持加入表达。

他们说，在另一篇文章中指出： “访问不会让你使用WHERE子句中的常规联接当您使用左/右/ INNER在FROM子句连接这可能是故意让你买。更昂贵的软件。“ - （Is the join expression not supported by MS Access?）

这不完全是。从我试过的一些例子中得出结论：

访问不会让你使用外连接（左/右）连同一个或多个INNER JOINS。 John Carmack的名字我能做什么？ 我想避免单独的选择查询。 请帮助...

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 

上述声明不起作用在MS-访问。

这是稍有不同（两个左联接），但它工作：

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LEFT JOIN，它会给相同的结果集，因为第二LEFT JOIN涉及外键关系（以从很多 - >一个方向）。

为了有一个LEFT JOIN与INNER JOIN你可以使用：

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