游标中带变量，怎么写？

[fyqwork]

DECLARE Friend_Cursor CURSOR FOR
SELECT memberId FROM memberTb WHERE memberId in (@receiveIds);

sql server中这么写好像不得行？

[GoAwayZ]

DECLARE Friend_Cursor CURSOR FOR
SELECT memberId FROM memberTb WHERE charindex(memberId,@receiveIds)>0

[--小F--]

DECLARE Friend_Cursor CURSOR FOR  
  SELECT memberId FROM memberTb WHERE memberId in ('+@receiveIds+');

[cd731107]

declare @id int
declare @name varchar(50)
declare cursor1 cursor for --定义游标cursor1
select * from table1 where name in (@name) --使用游标的对象(跟据需要填入select文)
open cursor1 --打开游标

fetch next from cursor1 into @id,@name --将游标向下移1行，获取的数据放入之前定义的变量@id,@name中

while @@fetch_status=0 --判断是否成功获取数据
begin
update table1 set name=name+'1'
where id=@id --进行相应处理(跟据需要填入SQL文)

fetch next from cursor1 into @id,@name --将游标向下移1行
end

close cursor1 --关闭游标
deallocate cursor1

[Mr_Nice]

DECLARE @i INT 

SET @i = 1

DECLARE Friend_Cursor CURSOR FOR  
  SELECT name FROM master..spt_values WHERE number =@i;
OPEN Friend_Cursor
FETCH NEXT FROM Friend_Cursor;

try

[fyqwork]

[Quote=引用 2 楼 fredrickhu 的回复:]
PHP code
DECLARE Friend_Cursor CURSOR FOR
SELECT memberId FROM memberTb WHERE memberId in ('+@receiveIds+');
[/Quote]

这么写，我传什么呢
我传
1,3和
'1,3'都不行啊

[fyqwork]

[Quote=引用 3 楼 cd731107 的回复:]
declare @id int
declare @name varchar(50)
declare cursor1 cursor for --定义游标cursor1
select * from table1 where name in (@name) --使用游标的对象(跟据需要填入select文)
open cursor1 --打开游标

fetch next from cursor……
[/Quote]

我就是这么写的啊，我该怎么传值进去呢

[cd731107]

select * from table1 where name in (@name)
--改为下面
select * from table1 where @name like '%'+name+'%'

[AcHerat]

select * from table1 where charindex(','+name+',',','+@name+',')>0
--传  1,2,3

[fyqwork]

谢谢各位，问题已经解决
我是这么做的
exec('Declare Friend_Cursor Cursor For Select memberId,nickName From memberTb where memberId in ('+@receiveIds+')')

[cd731107]

恭喜楼主