什么意思? 没太看明白.小问题弄得好复杂. 要修改SUB就修改SUB,要修改其它就修改其它.Create Project UP_1
@id int,
@Subject nvarchar(2000),
@succ bool output=false
...
if(exits(select id from VoteSubject ))
begin
update ...
set @output=true;
end2 CREATE PROCEDURE UP_VoteSubject_Update
@id int,
@VEID int,
@Order int,
@SubjectType char(1),
@IsShow bit
@id int,
@Subject nvarchar(2000),
@succ bool output=false
...
if(exits(select id from VoteSubject ))
begin
update ...
set @output=true;
end2 CREATE PROCEDURE UP_VoteSubject_Update
@id int,
@VEID int,
@Order int,
@SubjectType char(1),
@IsShow bit
解决方案 »
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/*--Author:xiaojie.shi--*/
/*--CreateTime:07-4-6--*/
CREATE PROCEDURE UP_VoteSubject_Update
@id int,
@VEID int,
@Subject nvarchar(200),
@Order int,
@SubjectType char(1),
@IsShow bit
AS
declare @r int,@r1 int;
select @r = count(subject) from VoteSubject where Subject = @Subject and [order] = @Order and SubjectType = @SubjectType and isshow = @IsShow
if @r = '1'
return '2';
--print '您未作任何修改该';
select @r1 = count(subject) from VoteSubject where [id] not in (select([id]) from votesubject where subject= @subject) and subject = @Subject
if @r1 = '1'
return '3';
--print '您修改该的标题已经存在';
else if @r1 = '0'
UPDATE VoteSubject SET
[VEID] = @VEID,[Subject] = @Subject,[Order] = @Order,[SubjectType] = @SubjectType,[IsShow] = @IsShow
WHERE [id] = @id
return '1';
--print'修该成功'
GO
我自己解决
@id int,
@VEID int,
@Subject nvarchar(200),
@Order int,
@SubjectType char(1),
@IsShow bit
AS
--标题已经存在且不是自己
IF EXISTS(SELECT * FROM VoteSubject WHERE Subject=@Subject AND [id]<>@id)
RETURN 3
ELSE
BEGIN
IF EXISTS (SELECT * FROM VoteSubject WHERE [id]=@id AND [VEID]=@VEID AND Subject=@Subject AND [order]=@Order AND SubjectType=@SubjectType AND isshow=@IsShow)
--未作任何修改,其实没必要。
RETURN 2
ELSE
UPDATE VoteSubject SET [VEID]=@VEID,[Subject]=@Subject,[Order]=@Order,[SubjectType]=@SubjectType,[IsShow]=@IsShow
WHERE [id]=@id
END
--修该成功
RETURN 1