SQL中的小时计算 [英] Hours Calculation in SQL

问题描述

我这样做.. !!!



DECLARE @minutes AS INT



SET @minutes = 1.80 * 100



SELECT @minutes / 60 +（@minutes％60）/ 100.0 AS Hours



它给了我3.000但结果应该是2.20小时



我怎么能实现那个......？



请帮我解决

解决方案

整数。有趣的小恶魔不是他们......

什么是1.8 * 100？

答案：180

什么是180/60？

答案：3

什么是180％60？

答案：0



所以首先答案是3，而不是2-and-bit。



其次，当你将整数除以小于100（任何数字％60将始终是）100？

答案：0，因为整数算术按定义丢弃了数字的一小部分...



所以，你可能想要考虑在整个浮点数工作...

  DECLARE  @minutes INT 
  DECLARE  @hours INT 
  DECLARE  @time  decimal （ 5 ， 2 ）=  1 . 80  
 
  SET  @hours = floor（@time） 
  SET  @minutes =（@ time  -  @hours）*  100  
 < span class =code-keyword> SET  @minutes = @minutes +（@hours *  60 ）
 
  SELECT  @minutes /  60  +（@min utes％ 60 ）/  100 。 0   AS 小时

这个怎么样.......！ !!!

I'm doing like this..!!!

DECLARE @minutes AS INT

SET @minutes = 1.80*100

SELECT @minutes / 60 + (@minutes % 60) / 100.0 AS Hours

it gives me 3.000 but the result should be 2.20 hrs

how can i acheive that..?

pls help me out

解决方案

Integers. Interesting little devils aren't they...
What is 1.8 * 100?
Answer: 180
What is 180 / 60?
Answer: 3
What is 180 % 60?
Answer: 0

So first off the answer is 3, not 2-and-a-bit at all.

Secondly, What do you get when you integer divide a number less than 60 (as any number % 60 will always be) by 100?
Answer: 0, because integer arithmetic throws away fraction parts of numbers by definition...

So, probably, you want to consider working in floating point numbers throughout...

DECLARE @minutes INT
DECLARE @hours INT
DECLARE @time decimal(5,2) = 1.80

SET @hours = floor(@time)
SET @minutes = (@time - @hours) * 100
SET @minutes = @minutes + (@hours * 60)

SELECT  @minutes / 60 + (@minutes % 60) / 100.0 AS Hours

How about this.......!!!!

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