TSQL RAND()问题

Question

我正在尝试生成一个格式为:99-88-77的字符串,其中随机生成三个2位数.

我的TSQL有效:

declare @result nvarchar(50)
    DECLARE @counter smallint, @ci smallint, @cu smallint, @dc smallint

      SET @ci=RAND()*100
      SET @cu=RAND()*100
      SET @dc=RAND()*100

      --SET @counter = @counter + 1

   set @result = CAST(@ci AS varchar(2)) +'-'+CAST(@cu AS varchar(2))+'-'+CAST(@dc AS varchar(2))
   print @result

产生(这次):16-37-30

我需要为插入表中的每个记录获取此字符串.

现在我想把它包装成一个函数,但显然我不能在UDF中使用RAND().

如何在使用insert语句时将其包装为call？

Answer 1

您可以通过两个步骤完成此操作.

首先,创建一个视图来生成随机数:

CREATE VIEW vRandNumber
AS
SELECT RAND() as RandNumber

其次,创建您UDF从视图中拉出:

CREATE FUNCTION dbo.udfTest
(
)
RETURNS nvarchar(50)
AS
BEGIN

    DECLARE @result  nvarchar(50)
    DECLARE @counter smallint, @ci smallint, @cu smallint, @dc smallint

    SET @ci=(SELECT RandNumber FROM vRandNumber)*100
    SET @cu=(SELECT RandNumber FROM vRandNumber)*100
    SET @dc=(SELECT RandNumber FROM vRandNumber)*100

    set @result = CAST(@ci AS varchar(2)) +'-'+CAST(@cu AS varchar(2))+'-'+CAST(@dc AS varchar(2))

RETURN @result
END

这将返回您刚才要求的价值.然后当你想要你的价值时,你会得到你的随机答案:

SELECT dbo.udfTest()

要么

INSERT INTO yourTable 
(
    randNumber
)
SELECT dbo.udfTest()

我刚刚在sql server 2005中对它进行了测试,但它确实有效.