我正在尝试确定在SQL中截断或删除额外的小数位而不舍入的最佳方法。例如:

declare @value decimal(18,2)

set @value = 123.456

这将自动将@value舍入为123.46,这在大多数情况下都很好。然而,对于这个项目,我不需要那样。有没有一种简单的方法来截断我不需要的小数?我知道我可以使用left()函数并将其转换回小数。还有别的办法吗?


当前回答

你到底要不要小数点?

如果没有,请使用

select ceiling(@value),floor(@value)

如果你用0做,那么做一轮:

select round(@value,2)

其他回答

另一种方法是ODBC的TRUNCATE函数:

DECLARE @value DECIMAL(18,3) =123.456;

SELECT @value AS val, {fn TRUNCATE(@value, 2)} AS result

现场演示

输出:

╔═════════╦═════════╗
║   val   ║ result  ║
╠═════════╬═════════╣
║ 123,456 ║ 123,450 ║
╚═════════╩═════════╝

备注:

我建议使用内置的ROUND函数,第3个参数设置为1。

Round有一个可选参数

Select round(123.456, 2, 1)  will = 123.45
Select round(123.456, 2, 0)  will = 123.46
ROUND ( 123.456 , 2 , 1 )

当第三个参数!= 0时,它截断而不是舍入。

语法

ROUND ( numeric_expression , length [ ,function ] )

参数

numeric_expression Is an expression of the exact numeric or approximate numeric data type category, except for the bit data type. length Is the precision to which numeric_expression is to be rounded. length must be an expression of type tinyint, smallint, or int. When length is a positive number, numeric_expression is rounded to the number of decimal positions specified by length. When length is a negative number, numeric_expression is rounded on the left side of the decimal point, as specified by length. function Is the type of operation to perform. function must be tinyint, smallint, or int. When function is omitted or has a value of 0 (default), numeric_expression is rounded. When a value other than 0 is specified, numeric_expression is truncated.

这将删除任何数字的小数部分

SELECT ROUND(@val,0,1)

请尝试使用此代码将一个点后的3个十进制值转换为2位小数:

declare @val decimal (8, 2)
select @val = 123.456
select @val =  @val

select @val

输出值为123.46