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在尝试找出问题后，我自己做了一些调试

我发现 @cols 和 @cols2 变量没有带来结果，我有 PRINT

PRINT('INSERT INTO [' + @Destination_Database_Name + '].[dbo].[' + @tablename + '] (' + @cols2 + ']' + ') SELECT [' + @cols2 + ']' + ' FROM [' + @Source_Database_Name + '].[dbo].[' + @tablename + ']');

而且该语句不会显示我得到的所有输出

（1 行受影响）

（1 行受影响）我在这里2

c365online_script1 我在这里3 tCompany

这是我认为是问题的代码部分

   Print 'I am here2'
                        SET IDENTITY_INSERT c365online_script1.dbo.tCompany ON
                        declare @cols2 varchar(max)
                        PRINT @cols2
                        select @cols2 = (Select Stuff((Select '],[' + C.COLUMN_NAME From INFORMATION_SCHEMA.COLUMNS As C Where C.TABLE_SCHEMA = T.TABLE_SCHEMA And C.TABLE_NAME = T.TABLE_NAME Order By C.ORDINAL_POSITION For Xml Path('')), 1, 2, '') As Columns From INFORMATION_SCHEMA.TABLES As T WHERE T.TABLE_NAME = @tablename)
                        PRINT('INSERT INTO [' + @Destination_Database_Name + '].[dbo].[' + @tablename + '] (' + @cols2 + ']' + ') SELECT [' + @cols2 + ']' + ' FROM [' + @Source_Database_Name + '].[dbo].[' + @tablename + ']');
                        PRINT @Destination_Database_Name
                         Print 'I am here3'
                         Print @tablename
                    END

我可以根据要求发布完整代码

4

2 回答 2

0

尝试

SET @cols2 = (Select....)

代替

SELECT @cols2 = (Select....)
于 2013-11-01T16:55:17.090 回答
0

这是一个 NULL 连接问题。尝试将您的变量初始化为空字符串，如下所示，并查看您对查询结果的了解：

    declare @tablename varchar(128) = '', 
            @Destination_Database_Name varchar(128) = '', 
            @Source_Database_Name varchar(128) = '';
于 2013-11-01T22:37:49.620 回答