Question

我正在尝试联接2个表并收到此错误。

关于模式：PK是a.app_guid，s.space_guid，o.org_guid。

无法绑定多部分标识符“ apps.space_guid”

SELECT 
    a.app_guid,
    a.name,
    a.state,
    a.created_at,
    a.updated_at,
    deleted_at,
    a.space_guid,
    a.foundation,
    a.timestamp, 

    s.space_guid,
    s.name,
    s.created_at,
    s.updated_at,
    s.timestamp,
    s.foundation,
    s.org_guid, 

    o.org_guid,
    o.created_at,
    o.updated_at,
    o.name,
    o.timestamp,
    o.foundation
FROM 
    apps a, spaces s, organizations o
INNER JOIN 
    [spaces] ON [apps].[space_guid] = [spaces].[space_guid]
INNER JOIN 
    [organizations] ON [spaces].[org_guid] = [organizations].[org_guid]

预期结果将包括一个表格，其中所有表格均根据space_guid和org_guid

Answer 1

我认为您不想交叉连接和内部连接表格。
您只是弄乱了语法。
使用以下正确的联接语法：

SELECT 
    a.app_guid,
    a.name,
    a.state,
    a.created_at,
    a.updated_at,
    deleted_at,
    a.space_guid,
    a.foundation,
    a.timestamp, 

    s.space_guid,
    s.name,
    s.created_at,
    s.updated_at,
    s.timestamp,
    s.foundation,
    s.org_guid, 

    o.org_guid,
    o.created_at,
    o.updated_at,
    o.name,
    o.timestamp,
    o.foundation
FROM 
    apps a
INNER JOIN 
    spaces s ON a.space_guid = s.space_guid
INNER JOIN 
    organizations o ON s.org_guid = o.org_guid

错误：无法绑定“多部分”标识符

1 个答案: