在SQL中将datediff分钟转换为100种基本格式

Question

I want to calculate date time difference between two dates but the minutes should be in .100 scale ie if date time difference is 2.30 (2 Hours 30 Minutes) i want it in format 2.50

30 minutes = 0.50

i wrote a query for it but it does not work well when minutes are in range of 01 - 09

Case 1 : Wrong Output

Declare @Start DateTime='02-03-2014 14:25:00'
Declare @End DateTime='02-03-2014 20:29:46'

Select STR(DateDiff(MINUTE,@Start,@End)/60)+'.'+STR(DateDiff(MINUTE,@Start,@End)%60/0.6)

DateTime Difference : 6.04 Expected Output : 6.10 Actual Output : 6.7

Case 2 : Correct Output

Declare @Start DateTime='02-03-2014 13:55:02'
Declare @End DateTime='02-03-2014 17:33:31'

Select STR(DateDiff(MINUTE,@Start,@End)/60)+'.'+STR(DateDiff(MINUTE,@Start,@End)%60/0.6)

DateTime Difference : 3.38 Expected Output : 6.63 Actual Output : 6.63

what i am missing in case, when minutes are less than 10 ??

DB : SQL Server 2008

Answer 1

It seems in STR(DateDiff(MINUTE,@Start,@End)%60/0.6) 0.6 should be replaced with 6.0

The above solution would round off upto 1 place after decimal.

If you want to ROUND off to 2 places after decimal you can try the below snippet:

SELECT CAST(CAST(ROUND(DateDiff(MINUTE,@Start,@End)%60/0.6) AS NUMERIC(12,2)) AS VARCHAR(50))

Answer 2

60 minutes = 60/60 = 1.0

30 minutes = 30/60 = 0.5

4 minutes = 4/60 = 0.066 , not 0.10

0.10 = 6 minutes, not 4 minutes difference as in your first example.

If we temporarily remove the seconds to reduce complexity, then you simply divide the seconds by 3600.00:

Declare @Start DateTime='02-03-2014 14:25:00'
Declare @End DateTime='02-03-2014 20:29:00'

SELECT DATEDIFF(s,@Start,@End)/3600.00

This returns 6.066 which is the correct portion of hours to return 6 hours 4 minutes difference and a far simpler expression.

Answer 3

Try this, it should give a very exact result(after rounding):

Declare @Start DateTime='02-03-2014 13:55:02'
Declare @End DateTime='02-03-2014 17:33:31'

Select round(cast(@end-@start as float)*24, 2)

use this if you want to round down

Select floor(cast(@end-@start as float)*2400) / 100

Answer 4

My friend following is tabular presentation of you requirement.

1   1.666666667
2   3.333333333
3   5
4   6.666666667
5   8.333333333
6   10
7   11.66666667
8   13.33333333
9   15
10  16.66666667
11  18.33333333
12  20
13  21.66666667
14  23.33333333
15  25
16  26.66666667
17  28.33333333
18  30
19  31.66666667
20  33.33333333
21  35
22  36.66666667
23  38.33333333
24  40
25  41.66666667
26  43.33333333
27  45
28  46.66666667
29  48.33333333
30  50
31  51.66666667
32  53.33333333
33  55
34  56.66666667
35  58.33333333
36  60
37  61.66666667
38  63.33333333
39  65
40  66.66666667
41  68.33333333
42  70
43  71.66666667
44  73.33333333
45  75
46  76.66666667
47  78.33333333
48  80
49  81.66666667
50  83.33333333
51  85
52  86.66666667
53  88.33333333
54  90
55  91.66666667
56  93.33333333
57  95
58  96.66666667
59  98.33333333
60  100

In First Case you actual difference is 6 hours & 4 Minutes. So as per requirement ans of 6.07 is correct how you are saying it is wrong?