Question

DECLARE @tmp char(1) = '';
SET @tmp = CAST(RAND()*26 AS int)+97;

此代码不起作用。 select的相同表达式效果很好：

SELECT CAST(RAND()*26 AS int)+97;

可能是我的@tmp声明了错误的数据类型？

Answer 1

尝试将@tmp声明为int而不是char(1)：

DECLARE @tmp int = '';
SET @tmp = CAST(RAND()*26 AS int)+97;
SELECT @tmp

Answer 2

我认为您需要CHAR功能：

CHAR（integer_expression）

将int ASCII代码转换为字符。

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Answer 3

这可能会帮助你：

DECLARE @tmp nvarchar(38) = '';
SET @tmp = newid()
select @tmp

Answer 4

试试这个：

DECLARE @tmp varchar(3)= CAST(RAND()*26 AS int)+97;
select @tmp;

或者您也可以尝试将数据类型更改为int

DECLARE @tmp int= CAST(RAND()*26 AS int)+97;
select @tmp;

使用T-SQL生成随机字符串

4 个答案: