在單個存儲過程中插入和更新不起作用

Question

當用戶更新他的個人資料圖片時，我想將其插入到表中並更新另一個表中的兩列。它工作正常，只是一個入口和停止更新第二個表..不知道什麼是我創建了一個小提琴錯誤

ALTER PROCEDURE [dbo].[UploadProfilePic] 
    (@UserName Nvarchar(50), 
     @ImageName Nvarchar(max), 
     @OrgImageName Nvarchar(max), 
     @CommentImage Nvarchar(max)) 
AS 
BEGIN 
    IF NOT EXISTS (SELECT Username FROM ProfilePic WHERE UserName = @UserName) 
    Begin 
      insert into ProfilePic ([UserName], [ImageName], [OrgImageName], [IsActive], [CommentImage]) 
      values(@UserName, @ImageName, @OrgImageName, 'Y', @CommentImage) 

      Update UserProfile 
      set ProfileImg = @ImageName, 
       PostImage = @CommentImage 
      where UserName = @UserName 
    END 
    ELSE 
    BEGIN 
     UPDATE ProfilePic 
     SET IsActive = 'N' 
     WHERE UserName = @UserName 

     INSERT INTO ProfilePic ([UserName], [ImageName], [OrgImageName], [IsActive],[CommentImage]) 
     VALUES(@UserName, @ImageName, @OrgImageName, 'Y', @CommentImage) 
    END 
END 

Answer 1

UPDATE

，檢查此請：

http://sqlfiddle.com/#!3/73aff0/6

我認爲問題在於你錯過了在else情況下添加update語句，所以對於新用戶它一直在工作 - 但不是已經存在的那些用戶。

BEGIN 
     IF NOT EXISTS (SELECT Username FROM ProfilePic WHERE UserName = @UserName) 
     BEGIN 
      INSERT INTO ProfilePic ([UserName], [ImageName], [OrgImageName], [IsActive],  [CommentImage]) 
      VALUES (@UserName, @ImageName, @OrgImageName, 'Y', @CommentImage) 

      UPDATE UserProfile 
      SET ProfileImg = @ImageName, 
       PostImage = @CommentImage 
      WHERE UserName = @UserName 
     END 
    ELSE 
     BEGIN 
      UPDATE ProfilePic 
      SET IsActive = 'N' 
      WHERE UserName = @UserName 

      INSERT INTO ProfilePic ([UserName], [ImageName], [OrgImageName], [IsActive], [CommentImage]) 
      VALUES(@UserName, @ImageName, @OrgImageName, 'Y', @CommentImage) 

      UPDATE UserProfile 
      SET ProfileImg = @ImageName, 
       PostImage = @CommentImage 
      WHERE UserName = @UserName 
     END 
    END 

Answer 2

你的問題實際上帶來了一個非常有趣和非常常見的情況。這是一種使用合併來解決這種情況的技術。這並不完全正是我所要求的，因爲我已經修改了模式並移除了一些內容，但這是如何解決這種情況的一個很好的例子。因爲`@`變量的

-- 
------------------------------------------------- 
if schema_id(N'profile') is null 
    execute (N'create schema profile'); 
go 
if object_id(N'[profile].[picture]', N'U') is not null 
    drop table [profile].[picture]; 
go 
if object_id(N'[profile].[user]', N'U') is not null 
    drop table [profile].[user]; 
go 
create table [profile].[user] 
    (
    [id]   [int] not null identity(1, 1) constraint [profile.user.id.clustered_primary_key] primary key clustered 
    , [user_name] [sysname] 
); 
go 
create table [profile].[picture] 
    (
    [id]    [int] not null identity(1, 1) constraint [profile.picture.id.clustered_primary_key] primary key clustered 
    , [image_name] [sysname] 
    , [image_comment] [nvarchar](max) 
    , [is_active]  [bit] 
    , [user]   [int] constraint [profile.picture.user] references [profile].[user] ([id]) 
); 
go 
-- 
if object_id(N'[profile].[set_picture]', N'P') is not null 
    drop procedure [profile].[set_picture]; 
go 
create procedure [profile].[set_picture] 
    @user_id   [int] 
    , @image_name [nvarchar](max) 
    , @image_comment [nvarchar](max) 
as 
    begin 
     declare @output table 
     (
      [action] [sysname] 
      , [id] [int] 
     ); 
     -- 
     merge into [profile].[picture] as target 
     using (values(@user_id 
      , @image_name 
      , @image_comment)) as source ([user], [image_name], [image_comment]) 
     on source.[user] = target.[user] 
     and source.[image_name] = target.[image_name] 
     when matched then 
     update set target.[is_active] = 1 
        , target.[image_name] = source.[image_name] 
        , target.[image_comment] = source.[image_comment] 
     when not matched by target then 
     insert ([user] 
       , [image_name] 
       , [image_comment] 
       , [is_active]) 
     values ([user] 
       , [image_name] 
       , [image_comment] 
       , 0) 
     output $action 
      , coalesce(inserted.[id], deleted.[id]) 
     into @output([action], [id]); 
    end; 
go 
-- 
-- note that you cannot insert a picture for a user that doesn't exist 
------------------------------------------------- 
insert into [profile].[user] 
      ([user_name]) 
    values (N'sally'), 
      (N'kevin'); 
go 
select * 
    from [profile].[user]; 
-- 
-- and the first insert gets an [is_active] value of 0 
-- so you set it a second time to activate it 
------------------------------------------------- 
execute [profile].[set_picture] 
    @user_id   =1 
    , @image_name =N'at the park' 
    , @image_comment =N'she looks good in that dress'; 
go 
execute [profile].[set_picture] 
    @user_id   =1 
    , @image_name =N'at church' 
    , @image_comment =N'but I like the blue dress better'; 
go 
-- 
------------------------------------------------- 
execute [profile].[set_picture] 
    @user_id   =2 
    , @image_name =N'at work' 
    , @image_comment =N'nice smile'; 
go 
execute [profile].[set_picture] 
    @user_id   =2 
    , @image_name =N'at work' 
    , @image_comment =N'nice smile'; 
go 
-- 
------------------------------------------------- 
select * 
    from [profile].[picture]; 
go 
--