无法在 SQL Server 2008 上绑定多部分标识符

Posted 2023-02-24

【中文标题】无法在 SQL Server 2008 上绑定多部分标识符

【英文标题】：The multi-part identifier could not be bound on SQL Server 2008

【发布时间】：2010-11-07 23:35:03

【问题描述】：

我有两张桌子

requests (ID, company_id, amount)

companies (ID, name)

带有 FK 约束 (requests.company_id -> companies.id)

requests.company 可以是NULL

我需要获取所有请求并将company_id 替换为适当的公司name，如果没有指定公司，则将其留空。

我有下一个问题：

SELECT R.[ID], C.[name] AS [company], R.[amount], ...
FROM [requests] AS R, [companies] AS C, ...
WHERE R.[company_id] = C.[ID]

在 company 字段中输入 NULL 之前它工作正常。

我试着做下一步：

SELECT R.[ID], C.[name] AS [company], ...
FROM [requests] AS R, ...
LEFT OUTER JOIN [companies] AS C
ON R.[company_id] = S.ID

得到了

无法绑定多部分标识符“R.company_id”

ON 子句转换中的字段也会出现相同的错误。我做错了什么？

【参考方案1】：

我想你想要：

SELECT R.[ID], ISNULL(C.[name], '') AS [company]
FROM [requests] AS R
    LEFT OUTER JOIN [companies] AS C
ON R.[company_id] = C.ID

编辑：见 cmets，需要左连接 ...

在我看来也不需要左连接，所以你可以重写为：

SELECT R.[ID], C.[name] AS [company]
FROM [requests] AS R
    JOIN [companies] AS C
ON R.[company_id] = C.ID

【讨论】：

他确实需要左连接。 Requests.Company_ID 可以为空。如果没有外部连接，Company_ID 为 null 的请求将从结果中删除。

@Shannon，你说得对，我错过了这个：如果没有指定公司，则将其留空。

【参考方案2】：

您展示的代码示例有省略号，我相信是省略号中的内容导致了问题。

你有：

SELECT R.[ID], C.[name] AS [company], ...
FROM [requests] AS R, ...
LEFT OUTER JOIN [companies] AS C
ON R.[company_id] = S.ID

假设是这样的：

SELECT R.[ID], C.[name] AS [company], X.Field
FROM [requests] AS R, [eXample] as X 
LEFT OUTER JOIN [companies] AS C
ON R.[company_id] = S.ID
WHERE X.[request_id] = R.ID

换句话说，ANSI 92 之前的内连接语法与 ANSI 92 外连接语法的混合。在 SQL Server 2005 上进行测试时，在您的示例中，在将 R 与 ... 分隔开的逗号之后，似乎看不到请求的别名 R，而在我的示例中，[eXample] 为 X。然而，以下确实有效：

SELECT R.[ID], C.[name] AS [company], X.Field
FROM [eXample] as X, [requests] AS R 
-- Requests and companies on the same side of the comma
LEFT OUTER JOIN [companies] AS C
ON R.[company_id] = S.ID
WHERE X.[request_id] = R.ID

或

SELECT R.[ID], C.[name] AS [company], X.Field
FROM [requests] AS R LEFT OUTER JOIN [companies] AS C
    ON R.[company_id] = S.ID, [eXample] as X 
WHERE X.[request_id] = R.ID
-- Yuck, I would hate to find this. Not at all sure from reading
-- the code how it would work.

还是我最喜欢的，因为我喜欢 ANSI 92 连接语法：

SELECT R.[ID], C.[name] AS [company], X.Field
FROM [requests] AS R
INNER JOIN [eXample] as X ON X.[request_id] = R.ID
LEFT OUTER JOIN [companies] AS C ON R.[company_id] = S.ID

【讨论】：

表在哪里别名为“S”以便约束“= S.ID”有效？

S.ID 在 OP 中，可能在椭圆中。 ？？？既然它帮助了 OP，我不想编辑，因为我担心我只会扼杀解释。处理丢失的代码很困难。

【参考方案3】：

用逗号更改表顺序使其在 sql server 2005,2008 和 2012 中工作

例如

假设您有如下查询，但在 sql 2005 及更高版本中将失败：

SELECT t1.*,t2.*, t3.*
FROM table1 AS t1, table2 as t2 
LEFT OUTER JOIN table3 AS t3
ON t1.id = t3.id
WHERE t1.id = t2.id

如果您重写查询并在连接前以逗号更改表的顺序，则该查询将在 sql 2005 及更高版本中工作。例如

SELECT t1.*,t2.*, t3.*
FROM table2 as t2, table1 as t1 
LEFT OUTER JOIN table3 AS t3
ON t1.id = t3.id
WHERE t1.id = t2.id