Re: DateDiff() and Minutes Wierdness

2010-04-14 Thread Dan G. Switzer, II

Brad,

On Tuesday, April 13, 2010, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart yields 
 nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

There's only 13 seconds between those two dates, so zero is the
correct response.

-Dan

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Re: DateDiff() and Minutes Wierdness

2010-04-14 Thread James Holmes

Can't you just use seconds, since the difference is under a minute?

cfoutput
#DateDiff(s, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
/cfoutput

mxAjax / CFAjax docs and other useful articles:
http://www.bifrost.com.au/blog/


On 14 April 2010 00:08, Bradley Stone b...@nb9m.com wrote:


 Thanks guys. I think the problem is that that I was thinking that the
 function could function when the arguments are essentially the same day.
  Here's what I did to solve the problem: when the function returns 0, do
 this:

 cfoutput
 cfset holdTotalHours = ABS(DatePart(h, arguments.HighDate) -
 DatePart(h, arguments.LowDate))

 cfset holdTotalMinutes = ABS(DatePart(n, arguments.HighDate) -
 DatePart(n, arguments.LowDate))

 Results in (HH:MM) Format: #holdTotalHours#:#holdTotalMinutes#
 /cfoutput

 Using DatePart here seems to work fine!



 

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Re: DateDiff() and Minutes Wierdness

2010-04-14 Thread Andrew Scott

Dan,

There is actually 48 seconds...

@Brad - I am not sure what you are trying to do, but you might consider
doing the diff on seconds and converting it to the number of days, minutes
seconds etc by calculating this. Plenty of examples how to turn seconds into
mins, hours, days etc.


On Thu, Apr 15, 2010 at 1:49 PM, Dan G. Switzer, II dswit...@pengoworks.com
 wrote:


 Brad,

 On Tuesday, April 13, 2010, Bradley Stone b...@nb9m.com wrote:
 
  All:
 
  Feeding DateDiff() dates which are more than 24 hours apart yields
 minutes:
 
  cfoutput
  #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
  /cfoutput
 
  Result: 5728
 
  However, feeding DateDiff() dates which are LESS than 24 hours apart
 yields nothing:
 
  cfoutput
  #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
  /cfoutput
 
  Result: 0

 There's only 13 seconds between those two dates, so zero is the
 correct response.

 -Dan

 

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Re: DateDiff() and Minutes Wierdness

2010-04-13 Thread Raymond Camden

Um, aren't those times like 50 seconds or so apart? Isn't 50 seconds 
1 minute? Therefore 0 full minutes?

On Tue, Apr 13, 2010 at 7:32 AM, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart yields 
 nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

 What am I missing?  Any help appreciated!
 Brad Stone
 Benchmark Technologies Inc.



 

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Re: DateDiff() and Minutes Wierdness

2010-04-13 Thread Brian Thornton

teah change your evaluator to seconds and do the multiplication.

On Tue, Apr 13, 2010 at 9:09 AM, Raymond Camden rcam...@gmail.com wrote:

 Um, aren't those times like 50 seconds or so apart? Isn't 50 seconds 
 1 minute? Therefore 0 full minutes?

 On Tue, Apr 13, 2010 at 7:32 AM, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart yields 
 nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

 What am I missing?  Any help appreciated!
 Brad Stone
 Benchmark Technologies Inc.





 

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RE: DateDiff() and Minutes Wierdness

2010-04-13 Thread Rick Faircloth

Isn't that:

April  8th, 2010 / 19 hours 41 minutes 36 seconds
April 12th, 2010 / 19 hours 10 minutes 34 seconds

Around 3 days, 28 minutes, 58 seconds apart?
(No coffee yet, so results may be skewed!)

And what does the first decimal place in 36.0 represent, anyway?
A tenth of a second?  Never worked with time and decimal places.


-Original Message-
From: Raymond Camden [mailto:rcam...@gmail.com] 
Sent: Tuesday, April 13, 2010 9:09 AM
To: cf-talk
Subject: Re: DateDiff() and Minutes Wierdness


Um, aren't those times like 50 seconds or so apart? Isn't 50 seconds 
1 minute? Therefore 0 full minutes?

On Tue, Apr 13, 2010 at 7:32 AM, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields
minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart
yields nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

 What am I missing?  Any help appreciated!
 Brad Stone
 Benchmark Technologies Inc.



 



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RE: DateDiff() and Minutes Wierdness

2010-04-13 Thread Jason Fisher

Yes, the long datetime in SQL Server tracks out to fractions of seconds.



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Re: DateDiff() and Minutes Wierdness

2010-04-13 Thread Raymond Camden

That was his first example, which worked for him. His second example
was less then 60 seconds apart.


On Tue, Apr 13, 2010 at 9:37 AM, Rick Faircloth
r...@whitestonemedia.com wrote:

 Isn't that:

 April  8th, 2010 / 19 hours 41 minutes 36 seconds
 April 12th, 2010 / 19 hours 10 minutes 34 seconds

 Around 3 days, 28 minutes, 58 seconds apart?
 (No coffee yet, so results may be skewed!)

 And what does the first decimal place in 36.0 represent, anyway?
 A tenth of a second?  Never worked with time and decimal places.


 -Original Message-
 From: Raymond Camden [mailto:rcam...@gmail.com]
 Sent: Tuesday, April 13, 2010 9:09 AM
 To: cf-talk
 Subject: Re: DateDiff() and Minutes Wierdness


 Um, aren't those times like 50 seconds or so apart? Isn't 50 seconds 
 1 minute? Therefore 0 full minutes?

 On Tue, Apr 13, 2010 at 7:32 AM, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields
 minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart
 yields nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

 What am I missing?  Any help appreciated!
 Brad Stone
 Benchmark Technologies Inc.







 

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RE: DateDiff() and Minutes Wierdness

2010-04-13 Thread Rick Faircloth

Gotcha...I was thinking his examples were reversed...

-Original Message-
From: Raymond Camden [mailto:rcam...@gmail.com] 
Sent: Tuesday, April 13, 2010 11:55 AM
To: cf-talk
Subject: Re: DateDiff() and Minutes Wierdness


That was his first example, which worked for him. His second example
was less then 60 seconds apart.


On Tue, Apr 13, 2010 at 9:37 AM, Rick Faircloth
r...@whitestonemedia.com wrote:

 Isn't that:

 April  8th, 2010 / 19 hours 41 minutes 36 seconds
 April 12th, 2010 / 19 hours 10 minutes 34 seconds

 Around 3 days, 28 minutes, 58 seconds apart?
 (No coffee yet, so results may be skewed!)

 And what does the first decimal place in 36.0 represent, anyway?
 A tenth of a second?  Never worked with time and decimal places.


 -Original Message-
 From: Raymond Camden [mailto:rcam...@gmail.com]
 Sent: Tuesday, April 13, 2010 9:09 AM
 To: cf-talk
 Subject: Re: DateDiff() and Minutes Wierdness


 Um, aren't those times like 50 seconds or so apart? Isn't 50 seconds 
 1 minute? Therefore 0 full minutes?

 On Tue, Apr 13, 2010 at 7:32 AM, Bradley Stone b...@nb9m.com wrote:

 All:

 Feeding DateDiff() dates which are more than 24 hours apart yields
 minutes:

 cfoutput
 #DateDiff(n, 2010-04-08 19:41:36.0, 2010-04-12 19:10:34.0)#
 /cfoutput

 Result: 5728

 However, feeding DateDiff() dates which are LESS than 24 hours apart
 yields nothing:

 cfoutput
 #DateDiff(n, 2010-04-08 19:34:47.0, 2010-04-08 19:35:35.0)#
 /cfoutput

 Result: 0

 What am I missing?  Any help appreciated!
 Brad Stone
 Benchmark Technologies Inc.







 



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Re: DateDiff() and Minutes Wierdness

2010-04-13 Thread Bradley Stone

Thanks guys. I think the problem is that that I was thinking that the function 
could function when the arguments are essentially the same day.  Here's what I 
did to solve the problem: when the function returns 0, do this:

cfoutput
cfset holdTotalHours = ABS(DatePart(h, arguments.HighDate) - DatePart(h, 
arguments.LowDate))

cfset holdTotalMinutes = ABS(DatePart(n, arguments.HighDate) - DatePart(n, 
arguments.LowDate))

Results in (HH:MM) Format: #holdTotalHours#:#holdTotalMinutes#
/cfoutput

Using DatePart here seems to work fine!



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RE: DateDiff() and Minutes Wierdness

2010-04-13 Thread brad

 I was thinking that the function could function when the arguments are 
 essentially the same day.

I'm a little confused.  The dateDiff() function DOES work when the
arguments are the same day.  Have we led you to believe otherwise, or
are you talking about a different function functioning?

 Here's what I did to solve the problem: when the function returns 0, do this:

Exactly what combination of inputs to dateDiff() is returning zero? If
it is your second example in your original post, then it SHOULD return
zero because the two dates were less than one minute apart.  Your code
below is asking whether or not the two times are in the same hour and
minute which is a different question than asking how many full hours
and minutes exists between the times.  (It's also worth noting your code
below will ONLY work for times on the same day and with a 24 hour clock)

If your objective is to decide if the times are in the same minute
then you should probably always be using datePart() starting with the
largest unit (year) and working in and NOT using dateDiff() at all. 
Alternatively you could continue to use dateDiff() to the second level,
divide by 60, and ceiling() the result up to the next full minute
which is essentially what you are doing below.

Perhaps you can explain exactly what you were trying to do.

~Brad

cfoutput
cfset holdTotalHours = ABS(DatePart(h, arguments.HighDate) -
DatePart(h, arguments.LowDate))

cfset holdTotalMinutes = ABS(DatePart(n, arguments.HighDate) -
DatePart(n, arguments.LowDate))

Results in (HH:MM) Format: #holdTotalHours#:#holdTotalMinutes#
/cfoutput



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