2010-10-11 36 views
0

我正在尝试使用游标移动来创建一个包含与其之前的字段相同的所有部分的列。在示例Sql - 关于游标和变量的问题

| column1 | column 2 | 
|  1  |  a  | 
|  2  |  b  | 
|  3  |  c  | 

would to go... 

| column1 | column2 | column3 | 
|  1  |  a |  b  | 
|  2  |  b |  c  | 
|  3  |  c | NULL | 

所以为了做到这一点我使用光标和更新语句基于如下的最后一个fetch语句尝试:

DECLARE myCursor1 CURSOR READ_ONLY 
FOR 
SELECT lname AS 'lnamerecoff' 
FROM testingThis 
ORDER BY lname 

OPEN myCursor1 
DECLARE @previous char(15) 
DECLARE @new char(15) 

SET @previous = FETCH NEXT FROM myCursor1 

IF NOT EXISTS(SELECT * FROM sys.columns WHERE name = 'lnamerecoff' 
         AND object_id = OBJECT_ID('testingThis')) 
    ALTER TABLE testingThis ADD lnamerecoff int 

WHILE @@FETCH_STATUS = 0 
BEGIN 
SET @new = FETCH NEXT FROM myCursor1 
UPDATE testingThis 
SET lnamerecoff = @new 
SET @previous = @new 
END 

这是抛出一个语法错误靠近我的fetch语句。谁能帮我这个?谢谢!

+1

LEAD和LAG函数会做了这个如此简单:/ – 2010-10-11 17:41:38

+0

这些都不是在SQL Server 2005提供,虽然,是吗? – user416516 2010-10-11 17:48:13

回答

1

它应该是:

fetch next from myCursor1 into @previous