在SQL Server 2008中舍入十进制数

I read all rounding functions of T-SQL like Round, Floor, and Ceil, but none of them has rounded down decimal numbers correctly for me.

我读了T-SQL的所有舍入函数，如Round，Floor和Ceil，但它们都没有正确地向下舍入十进制数。

I have 2 questions:

我有两个问题：

How to round down a decimal number (3.69 ==> 3.5)?
如何舍入十进制数（3.69 ==> 3.5）？
How to round up the last 3 digits of an integer (e.g. 142600 ==> 143000)?
如何舍入整数的最后3位数（例如142600 ==> 143000）？

3 个解决方案

#1

1) select CAST(FLOOR(2 * 3.69) / 2 AS decimal(2, 1)) handles the first case - courtesy of an answer to a similar question on SQL Server Forums, which I adapted and quickly checked.

1）选择CAST（FLOOR（2 * 3.69）/ 2 AS十进制（2,1））处理第一种情况 - 礼貌地回答SQL Server论坛上的类似问题，我对其进行了调整并快速检查。

Note that if the numbers you are rounding to the nearest 0.5 could be bigger (e.g. 333.69 => 333.5), be sure to specify more decimal precision when you cast (e.g. select CAST(FLOOR(2 * 3.69) / 2 AS decimal(10, 1))), or you could get an overflow error:

请注意，如果要舍入到最接近的0.5的数字可能更大（例如333.69 => 333.5），请确保在转换时指定更多小数精度（例如，选择CAST（FLOOR（2 * 3.69）/ 2 AS十进制（10），1））），或者您可能会收到溢出错误：

Msg 8115, Level 16, State 8, Line 1
Arithmetic overflow error converting numeric to data type numeric.

Extra precision will not affect the bottom-line result (i.e. select CAST(FLOOR(2 * 3.69) / 2 AS decimal(10, 1)) and select CAST(FLOOR(2 * 3.69) / 2 AS decimal(2, 1)) both yield 3.5); but it is wasteful if the numbers you are rounding will always be smaller.

额外的精度不会影响底线结果（即选择CAST（FLOOR（2 * 3.69）/ 2 AS十进制（10,1））并选择CAST（FLOOR（2 * 3.69）/ 2 AS十进制（2,1））两者产量3.5）;但如果你要四舍五入的数字总是更小，那就太浪费了。

Online references with examples are available for T-SQL FLOOR, CAST, and decimal to help.

带有示例的在线参考可用于T-SQL FLOOR，CAST和十进制帮助。

2) select ROUND(142600, -3) handles the second case.

2）选择ROUND（142600，-3）处理第二种情况。

A similar online reference is available for T-SQL ROUND.

类似的在线参考可用于T-SQL ROUND。

#2

As per @J0e3gan 's anwser, Sql Server's Round allows rounding to the nearest powers of 10 using the length parameter, where length is 10^(-length), e.g.

根据@ J0e3gan的anwser，Sql Server的Round允许使用length参数舍入到最接近的10的幂，其中length是10 ^（ - length），例如，

length = 0 : 10 ^ 0 = nearest 1
length = 3 : 10 ^ -3 = nearest .001
length = -3 : 10 ^ 3 = nearest 1000

etc

等等

However, in general, with a simple 1-based rounding function - e.g. (Sql Round with Length=0) to round to an arbitrary value of "nearest N" - with the formula:

然而，通常，使用简单的基于1的舍入函数 - 例如，（Sql Round with Length = 0）舍入到“最接近的N”的任意值 - 使用以下公式：

round(X / N) * N

e.g. nearest 100

例如最近的100

select round(12345 / 100.0, 0) * 100.0 -- 12300
select round(-9876 / 100.0, 0) * 100.0 -- -9900
select round(-9849 / 100.0, 0) * 100.0 -- -9800

... Nearest 0.5

......最近的0.5

select round(5.123 / 0.5, 0) * 0.5 -- 5.000
select round(6.499 / 0.5, 0) * 0.5 -- 6.500
select round(-4.499 / 0.5, 0) * 0.5 -- -4.50

... Nearest 0.02

......最近0.02

select round(5.123 / .02, 0) * .02 -- 5.12
select round(-9.871 / .02, 0) * .02 -- -9.88

etc

等等

Remember that the type used for the divisors must be numeric / decimal or float.

请记住，用于除数的类型必须是数字/小数或浮点数。

#3

The Oracle/PLSQL FLOOR function returns the largest integer value that is equal to or less than a number. eg:

Oracle / PLSQL FLOOR函数返回等于或小于数字的最大整数值。例如：

FLOOR(7.9)
Result: 7

FLOOR(30.29)
Result: 30

FLOOR(-7.9)
Result: -8

#1