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We call "delay" of a lottery number the number of draws that have elapsed without that number coming out. A number has a delay of 1 if it just came out, a delay of 30 if it last came out 30 draws ago, and so on.

... is slowing things down, it is all your other code. In order to find which section or sections are slowing it down the most, you need to log the elapsed time it takes to run each section of code. Then for the slowest code, you have to either decide to eliminate it or reprogram it another faster ...

... * para comparar datas com hora inicial e final * $DOC$ * $FUNCNAME$ * FT_ELAPSED() * $CATEGORY$ * Date/Time * $ONELINER$ * Return elapsed time between two days and/or times * $SYNTAX$ * FT_ELAPSED([ <dStart> ], [ <dEnd> ], ; * <cTimeStart>, <cTimeEnd>) -> ...

Hey Rick! Thank you posting and thank you for giving me something to smile about today. I'm here reading your text and smiling at the fact that you were -for a minute- confused about 12:30am being 00:30. Many won't understand that in the US you never think of time in "military" terms which...

For computing purposes, please use 24 hour format

Rao

Both your examples work great ... I just inserted the wrong value ( as you mentioned ) for 12:30am should have been 0030 .. and NOT 12:30

Rick Lipkin

Rao

Let me take a deep breath ... you are correct ... I cannot focus this morning due to all the interruptions .. Let me re-test.

Thanks
Rick Lipkn

Rao

The values are from 11:00pm on 03/11/2017 and return is 12:30 am 03/12/2017 .. 90 mins this trip spans midnight from 11:00pm to 12:20 am

Military time 11:00pm = 23:00:00 03/11/2017
Miltary Time 12:30am = 12:30:00 03/12/2017


Rick Lipkin

Correct value is not 90 minutes From night 11:00 pm to next day 12:30 pm is 13 hours and 30 minutes but not 90 minutes The formula's answer is correct. 90 minutes after 11 pm is 00:30 on the next day If you try with tEnd as "00:30:00" then you get 90 minutes. Please note that "12:30:0...

Rao I am having problems ... these are the values I am using : dEnd   := ctod("03/12/2017" )dStart := ctod("03/11/2017" )tStart := "23:00:00"  // 11:00 pmtEnd   := "12:30:00"  // 12:30 amnMin := ( CTOT( DTOC( dEnd ) + ...

Rao

Thank you ... I saw a similar solution on the internet but did not know how to change the value of the character to valtype "T" .. I agree with you on getting away from Date and Time ... DateTime is one of the foundation blocks in any Sql RDMS.

Thank you Again!
Rick Lipkin

The answer is 90 minutes ... I have no problem calculating time on the same date .. but spanning days ( midnight ) is giving me a headache .. any advice or algorithm would be appreciated. If you want the answer in minutes: ? ( CTOT( DTOC( dEnd ) + " " + tEnd ) - CTOT&#...

Uwe I found your post ... thank you, the only problem I have is when you span a day I do not get the correct calculation .. here is the code I used with your function : //  dStart = 03/11/2017//  dEnd   = 03/12/2017    ... notice dENd was incremented below     tSta...

Rick, You can try /* *     Return elapsed time between two days and/or times *  SYNTAX *     ELAPSED([ <dStart> ], [ <dEnd> ], ; *                <cTimeStart>, ...

To All Does anyone have a solution to find the elapsed time between two dates in minutes .. Date out             Time Out           military time06\18\2017         11:00pm ...