Question

首先，我已经看到了这个问题的答案，但它们并不适用于我。我的SELECT和INSERT语句中的列数相同。请帮忙，我已经花了半天时间。当我运行它时，它完美地工作但是在最后一次迭代后它给了我一个错误。

消息121，级别15，状态1，行12 INSERT的选择列表语句包含的项目多于插入列表。的数量 SELECT值必须与INSERT列的数量匹配。

proc [dbo].[uspPOP_WBC] 

--params
(@ptblname  varchar(MAX)        
,@pID_Sub varchar               
,@pSubCodeShort  varchar(2)     
,@pdbname  varchar(10)  )       
AS 
BEGIN
SET nocount ON; 
DECLARE @strSQL AS nvarchar(MAX)

SET @strSQL = 'INSERT INTO dbo.WBC 
( ID
, ID_SUB
, wellcompl
, id_wellcompl
, id_type_wellcompl
, type_wellcompl
, id_wellstring
, wellstring
, type_wellstring
, start_date
, end_date
, comment)'
SET @strSQL = @strSQL + 
' SELECT WC.id
,'+convert(varchar(1),@pID_Sub)+'   
, WC.name
, WC.id 
, WC.id_type_wellcompl
, TWC.name
, WC.id_wellstring
, WS.name
, TWS.name
, WC.start_date
, WC.end_date
, WC.comment
 FROM (
 ['+ @pdbname+ @pSubCodeShort +'].[dbo].[wellbore_compl] as WC 
INNER JOIN ['+@pdbname+ @pSubCodeShort+'].[dbo].[wellstring] as WS 
ON WC.id_wellstring = WS.id 
INNER JOIN ['+@pdbname+ @pSubCodeShort+'].[dbo].[type_wellcompl] as TWC 
ON WC.id_type_wellcompl = TWC.id
INNER JOIN ['+@pdbname+ @pSubCodeShort+'].[dbo].[type_wellstring] as TWS 
ON WS.id_type_wellstring = TWS.id)
where WC.deletedAt is null'
;

EXEC sp_executesql @strSQL
SET nocount OFF; 
END

打印@strSQL

    INSERT INTO dbo.WBC ( ID, ID_SUB, wellcompl, id_wellcompl,  
id_type_wellcompl, type_wellcompl, id_wellstring, wellstring, 
type_wellstring, start_date, end_date, comment) 
SELECT 
WC.id
, 1 
, WC.name
, WC.id 
, WC.id_type_wellcompl
, TWC.name
, WC.id_wellstring
, WS.name
, TWS.name
, WC.start_date
, WC.end_date
, WC.comment
 FROM (
 [PDS_SYNC_CM].[dbo].[wellbore_compl] as WC INNER JOIN [PDS_SYNC_CM].[dbo].[wellstring] as WS ON WC.id_wellstring = WS.id 
 INNER JOIN [PDS_SYNC_CM].[dbo].[type_wellcompl] as TWC ON WC.id_type_wellcompl = TWC.id
 INNER JOIN [PDS_SYNC_CM].[dbo].[type_wellstring] as TWS ON WS.id_type_wellstring = TWS.id)
 where WC.deletedAt is null

Answer 1

如果我的计数是正确的，那么代码在INSERT子句中有12列，在SELECT子句中只有11列。缺少一个，所以请检查您的列以找出哪个缺失并添加它。

为什么我得到消息121，级别15，状态1，第12行

1 个答案: