我希望能够在SQL中将数字舍入为n个有效数字。所以:
123.456 rounded to 2sf would give 120
0.00123 rounded to 2sf would give 0.0012
我知道ROUND()函数,它舍入到n个小数位而不是有效数字。
答案 0 :(得分:20)
select round(@number,@sf-1- floor(log10(abs(@number))))
应该做到这一点!
成功测试了两个例子。
编辑:在@ number = 0上调用此函数将不起作用。在使用此代码之前,您应该为此添加测试。
create function sfround(@number float, @sf int) returns float as
begin
declare @r float
select @r = case when @number = 0 then 0 else round(@number ,@sf -1-floor(log10(abs(@number )))) end
return (@r)
end
答案 1 :(得分:1)
将Brann最流行的答案改编成MySQL,让那些看起来像我的人。
CREATE FUNCTION `sfround`(num FLOAT, sf INT) # creates the function
RETURNS float # defines output type
DETERMINISTIC # given input, will return same output
BEGIN
DECLARE r FLOAT; # make a variable called r, defined as a float
IF( num IS NULL OR num = 0) THEN # ensure the number exists, and isn't 0
SET r = num; # if it is; leave alone
ELSE
SET r = ROUND(num, sf - 1 - FLOOR(LOG10(ABS(num))));
/* see below*/
END IF;
RETURN (r);
END
/ *感觉太长,无法发表评论* /
ROUND(num,sf - 1 - FLOOR(LOG10(ABS(num))))
有效,因为ROUND(num,-ve num)向小数点左侧舍入
仅一次性,ROUND(123.456,-1)和ROUND(0.00123,4) 返回所要求的答案((120,0.0012)
答案 2 :(得分:0)
你可以在四舍五入之前除以100然后再乘以100 ......
答案 3 :(得分:0)
我想我已经成功了。
CREATE FUNCTION RoundSigFig(@Number float, @Figures int)
RETURNS float
AS
BEGIN
DECLARE @Answer float;
SET @Answer = (
SELECT
CASE WHEN intPower IS NULL THEN 0
ELSE FLOOR(fltNumber * POWER(CAST(10 AS float), intPower) + 0.5)
* POWER(CAST(10 AS float), -intPower)
END AS ans
FROM (
SELECT
@Number AS fltNumber,
CASE WHEN @Number > 0
THEN -((CEILING(LOG10(@Number)) - @Figures))
WHEN @Number < 0
THEN -((FLOOR(LOG10(@Number)) - @Figures))
ELSE NULL END AS intPower
) t
);
RETURN @Answer;
END