Question

任何人都可以帮助我解决这个问题，我曾多次尝试但仍然没有找到解决方案这是我在数据库中的原始视图，但现在我在数据库中进行了更改，并且视图也需要更改。

以下是视图：

SELECT 
    [tableN].*, 
    [tabB].[att1] AS tabB_email, [tabB].[name] AS tabB_name,
    [tabC].[name] AS tabC_name
FROM  
    [tabC] 
INNER JOIN 
    ([tableN] 
INNER JOIN [tabB] ON [tableN].[sender_id] = [tabB].[ID]) 
     ON [tabC].[ID] = [tableN].[recipient_id]

这对我来说有什么难点。现在我没有这两个表tabB和tabC

它们现在位于一个表tabX中，并且具有标识符字段roleId。我设法获取除最后一列[tabC].[name] AS tabC_name

之外的所有列

有什么想法吗？

Answer 1

  SELECT [tableN].*, [Tabx] .[att1] AS tabB_email, [Tabx] .[name] AS tabB_name  

  FROM  [Tabx] A 

  INNER JOIN [TABLEN] B

  ON A.ROLEID=B.RECIPIENT_ID

Answer 2

试试这个

SELECT [tableN].*, [tabX].[att1] AS tabB_email, [tabX].[name] AS tabB_name, 
t1.[name] AS tabC_name 
FROM [tabX] as t INNER JOIN ([tableN] INNER JOIN [tabX] 
ON [tableN].[sender_id] = [tabX].[roleid]) 
ON t.[roleid] = [tableN].[recipient_id]

Answer 3

 SELECT [TableN].*, Bd.[email] AS bd_email, Bd.[showname] AS bd_name,
        Pc.[showname] AS pc_name
   FROM 
      [TABLE_X] AS Pc 
         INNER JOIN ([TableN] 
         INNER JOIN [TABLE_X]  AS Bd
         ON [TableN].[sender_id] = Bd.[ID] AND Bd.roleID = 1) 
         ON Pc.[ID] = [TableN].[recipient_id] AND Pc.roleID = 2

我终于找到了根据需要运行的代码

带有2个INNER JOINS的SQL View语句

3 个答案: