Question

我创建了以下存储过程，其中用户传递输入参数以获取数据。输入参数是可选的，即仅传递一个参数应该会产生一些数据。

这是我的代码。

ALTER PROC [dbo].[SPR_ECRM_CASE_INFORMATION] 

@ID_APPLICATION_NO NVARCHAR(20),
@TX_ID_NO NVARCHAR(20)

AS 
    BEGIN
        SET NOCOUNT ON

        IF (@ID_APPLICATION_NO IS NOT NULL ) OR (LEN(@ID_APPLICATION_NO) > 0) AND ((@TX_ID_NO IS NOT NULL) OR (LEN(@TX_ID_NO) > 0)) 
            BEGIN
                SELECT DISTINCT
                 [A].[ID_APPLICATION_NO]
                ,[A].[TX_APPLICATION_STATUS]
     
                FROM [VW_T_APPLICATION] As [A]
                INNER JOIN [CT_L1_APPLICATION_STATUS] [L] ON [L].[TX_L1_APPLICATION_STATUS_CODE] = [A].[TX_APPLICATION_STATUS_CODE]
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [I] WITH(NOLOCK) ON [I].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                [II].[ID_APPLICATION_APPLICANT]
                FROM  
                    [ST_APPLICATION_APPLICANT] [II] WITH(NOLOCK)
                WHERE [II].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                    AND [II].[CD_APPLICANT_TYPE] = 81
                    AND [II].[IN_ACTIVE] = 1
                ORDER BY 
                    [II].[ID_APPLICATION_APPLICANT] DESC)
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [J] WITH(NOLOCK) ON [J].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                    [JJ].[ID_APPLICATION_APPLICANT]
                FROM  
                    [ST_APPLICATION_APPLICANT] [JJ] WITH(NOLOCK)
                WHERE [JJ].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                    AND [JJ].[CD_APPLICANT_TYPE] = 82
                    AND [JJ].[IN_ACTIVE] = 1
                ORDER BY 
                    [JJ].[ID_APPLICATION_APPLICANT] DESC)
 
                WHERE
                [I].[TX_ID_NO] = @TX_ID_NO AND [J].[TX_ID_NO] = @TX_ID_NO
                AND
                [A].[TX_APPLICANT_ID_NO] = @ID_APPLICATION_NO
                AND
                [A].[IN_ACTIVE] = 1

            END

         ELSE IF (@ID_APPLICATION_NO IS NOT NULL) OR (LEN(@ID_APPLICATION_NO) > 0)
            BEGIN
                SELECT DISTINCT
                 [A].[ID_APPLICATION_NO]
                ,[A].[TX_APPLICATION_STATUS]
     
                FROM [VW_T_APPLICATION] As [A]
                INNER JOIN [CT_L1_APPLICATION_STATUS] [L] ON [L].[TX_L1_APPLICATION_STATUS_CODE] = [A].[TX_APPLICATION_STATUS_CODE]
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [I] WITH(NOLOCK) ON [I].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                        [II].[ID_APPLICATION_APPLICANT]
                    FROM  
                        [ST_APPLICATION_APPLICANT] [II] WITH(NOLOCK)
                    WHERE [II].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                        AND [II].[CD_APPLICANT_TYPE] = 81
                        AND [II].[IN_ACTIVE] = 1
                    ORDER BY 
                        [II].[ID_APPLICATION_APPLICANT] DESC)
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [J] WITH(NOLOCK) ON [J].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                        [JJ].[ID_APPLICATION_APPLICANT]
                    FROM  
                        [ST_APPLICATION_APPLICANT] [JJ] WITH(NOLOCK)
                    WHERE [JJ].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                        AND [JJ].[CD_APPLICANT_TYPE] = 82
                        AND [JJ].[IN_ACTIVE] = 1
                    ORDER BY 
                        [JJ].[ID_APPLICATION_APPLICANT] DESC)
 
                WHERE
                [A].[TX_APPLICANT_ID_NO] = @ID_APPLICATION_NO
                AND 
                [A].[IN_ACTIVE] = 1
                

            END

          ELSE IF (@TX_ID_NO IS NOT NULL) OR (LEN(@TX_ID_NO) > 0)
            BEGIN
                SELECT DISTINCT
                 [A].[ID_APPLICATION_NO]
                ,[A].[TX_APPLICATION_STATUS]
     
                FROM [VW_T_APPLICATION] As [A]
                INNER JOIN [CT_L1_APPLICATION_STATUS] [L] ON [L].[TX_L1_APPLICATION_STATUS_CODE] = [A].[TX_APPLICATION_STATUS_CODE]
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [I] WITH(NOLOCK) ON [I].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                    [II].[ID_APPLICATION_APPLICANT]
                FROM  
                    [ST_APPLICATION_APPLICANT] [II] WITH(NOLOCK)
                WHERE [II].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                    AND [II].[CD_APPLICANT_TYPE] = 81
                    AND [II].[IN_ACTIVE] = 1
                ORDER BY 
                    [II].[ID_APPLICATION_APPLICANT] DESC)
                LEFT JOIN [VW_ST_APPLICATION_APPLICANT] [J] WITH(NOLOCK) ON [J].[ID_APPLICATION_APPLICANT] = (SELECT TOP 1   
                    [JJ].[ID_APPLICATION_APPLICANT]
                FROM  
                    [ST_APPLICATION_APPLICANT] [JJ] WITH(NOLOCK)
                WHERE [JJ].[ID_APPLICATION_GUID] = [A].[ID_APPLICATION_GUID]
                    AND [JJ].[CD_APPLICANT_TYPE] = 82
                    AND [JJ].[IN_ACTIVE] = 1
                ORDER BY 
                    [JJ].[ID_APPLICATION_APPLICANT] DESC)
 
                WHERE
                [I].[TX_ID_NO] = @TX_ID_NO OR [J].[TX_ID_NO] = @TX_ID_NO
                AND
                [A].[IN_ACTIVE] = 1

            END
        ELSE
        BEGIN
            RETURN NULL
        END
        END

我已经使用过if if else条件，但是它无法返回数据。

DECLARE @A NVARCHAR(30)
DECLARE @B NVARCHAR(20)

SET @A = 'C192'
SET @B = 'ABC' 

EXEC SPR_ECRM_CASE_INFORMATION @A, ''
EXEC SPR_ECRM_CASE_INFORMATION '', @B
EXEC SPR_ECRM_CASE_INFORMATION '', ''

首先，else if条件失败并且我的代码在这里是多余的。有没有更好的方法来优化这一点。感谢您的帮助。

Answer 1

如果要优化性能，则可以为每个查询创建一个过程。每个人都会得到自己的查询计划。您可以在此过程中分别调用。

查看每个查询，我在ON子句中看到了子查询。我还看到了NOLOCK，我通常认为这是一个不好的信号。另请参见DISTINCT。这些都是可以简化查询的线索。这样，SQL Server将更有可能优化查询。（您可以通过多种方式要求同一件事。逻辑越复杂，SQL查找最佳计划的可能性就越小。）

如果[VW_T_APPLICATION]中的数据已经不同，则摆脱DISTINCT并使用EXISTS代替LEFT JOINS-一个EXISTS代表81，一个EXISTS代表82。（我假设您不需要这些列）

“ VW”的使用表示视图的使用。该视图可以隐藏很多妨碍简单查询的逻辑。

Answer 2

这部分（您的第一个IF）在包围式方面是否正确？如果@ID_Application_No不为NULL，它将运行，无论@TX_ID_No是否具有值。

IF (@ID_APPLICATION_NO IS NOT NULL ) OR (LEN(@ID_APPLICATION_NO) > 0) AND ((@TX_ID_NO IS NOT NULL) OR (LEN(@TX_ID_NO) > 0))

我猜你也想要

a）ID_Application_No部分周围的方括号，例如

IF ((@ID_APPLICATION_NO IS NOT NULL ) OR (LEN(@ID_APPLICATION_NO) > 0)) AND ((@TX_ID_NO IS NOT NULL) OR (LEN(@TX_ID_NO) > 0))

或b）希望它为AND（这是我最可能的猜测），例如，

IF (@ID_APPLICATION_NO IS NOT NULL ) AND (LEN(@ID_APPLICATION_NO) > 0) AND ((@TX_ID_NO IS NOT NULL) OR (LEN(@TX_ID_NO) > 0))

请注意，您可以简化一些操作，例如，当您拥有IF (@ID_APPLICATION_NO IS NOT NULL) OR (LEN(@ID_APPLICATION_NO) > 0)时，如果第二部分为true，则第一部分将始终为true。在这种情况下，您只能使用IF (@ID_APPLICATION_NO IS NOT NULL)，它将得到相同的结果。

在存储过程中如何使用If Else if和Else SQL Server

2 个答案: