Question

我目前有以下存储过程;

CREATE PROCEDURE web.insertNewCampaign
   (
   @tmp_Id BIGINT,
   @tmp_Title VARCHAR(100),
   @tmp_Content VARCHAR(8000),
   @tmp_Pledge DECIMAL(7,2),
   --@tmp_Recipients BIGINT,
   @tmp_Date DATETIME,
   @tmp_Private BIT,
   @tmp_Template BIGINT,
   @tmp_AddyBook BIGINT
   )
AS
   declare @recipients BIGINT
   declare @tmp_IDENTITY BIGINT
   declare @fave BIGINT
   declare @allocation VARCHAR(50)

   --insert campaign data
   BEGIN TRAN
   SELECT @recipients = addMaster_NoRecipients FROM tbl_AddressBookMaster 
   WHERE addMaster_UserId = @tmp_Id AND addMaster_Key = @tmp_AddyBook;
   INSERT INTO TBL_CAMPAIGNS ([campaign_MemberId], [campaign_Title], [campaign_Content], [campaign_Pledge], [campaign_Date], [campaign_Private], [campaign_Template], [campaign_AddressBook], [campaign_Recipients]) 
   VALUES (@tmp_Id, @tmp_Title, @tmp_Content, @tmp_Pledge, @tmp_Date, @tmp_Private, @tmp_Template, @tmp_AddyBook, @recipients)
   SELECT @tmp_IDENTITY = SCOPE_IDENTITY() --this returns the newly added IDENTITY ID
   COMMIT
......

所以我有两个问题：

1）如何将@tmp_Pledge除以@recipients给@allocation，例如：（@ allocation = @tmp_Pledge / @recipients）

2）是否有可能将这些语句复合到一个更有效的语句中，并将@allocation有效地作为值插入到[campaign_RecipShare]列中，并减少对这些声明变量的需求？

非常感谢您为这两个问题提供的任何帮助。

- ）

Answer 1

在第一次选择后，您可以执行此操作来设置@allocation：

set @allocation = @tmp_pledge / @recepients

至于提高效率，它已经相当有效 - 你不会经历任何更少的步骤，但你可以稍微压缩代码：

INSERT INTO TBL_CAMPAIGNS (
    [campaign_MemberId], [campaign_Title], [campaign_Content], 
    [campaign_Pledge], [campaign_Date], [campaign_Private], 
    [campaign_Template], [campaign_AddressBook], [campaign_Recipients],
    [capmain_RecipShare]) 
SELECT 
    @tmp_Id, @tmp_Title, @tmp_Content, 
    @tmp_Pledge, @tmp_Date, @tmp_Private, 
    @tmp_Template, @tmp_AddyBook, addMaster_NoRecipients,
    @tmp_Pledge / addMaster_NoReceipients as Allocation
FROM
    tbl_AddressBookMaster
WHERE
    addMaster_UserId = @tmp_Id
    AND addMaster_Key = @tmp_AddyBook

SELECT @tmp_IDENTITY = SCOPE_IDENTITY() --this returns the newly added IDENTITY ID

这也使您无需计算@allocation语句之外的insert成员。

Answer 2

1）@tmp_pledge / @recepients - 我假设分配是TBL_CAMPAIGNS中某个形式的数字字段，在varchar中保存一个数字并不是一个好主意。

2）你只需要构建一个select，它返回另一个表中的所有值以及与要插入的列匹配的参数。

insert into TBL_CAMPAIGNS ([campaign_MemberId], [campaign_Title], [campaign_Content],    [campaign_Pledge], [campaign_Date], [campaign_Private], [campaign_Template], [campaign_AddressBook], [campaign_Recipients], [campaign_allocation) 

select @tmp_Id, @tmp_Title, @tmp_Content, @tmp_Pledge, @tmp_Date, @tmp_Private, @tmp_Template, @tmp_AddyBook, addMaster_NoRecipients, @tmp_pledge / addMaster_NoRecipients

FROM FROM tbl_AddressBookMaster WHERE addMaster_UserId = @tmp_Id AND addMaster_Key = @tmp_AddyBook;

SELECT @tmp_IDENTITY = SCOPE_IDENTITY（） - 这会返回新添加的IDENTITY ID

Answer 3

set @allocation = @tmp_pledge / (@recepients* 1.0)

你想这样做，否则你将遇到整数数学，结果将舍入为整数。

存储过程计算＆amp;性能改进

3 个答案: