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jerem488

Add time To anothe time in an interval

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Hi everybody,

I want to do a script to calculate the date/time in an interval.

The interval is 9h-12h and 14h-19h. So When I want to add 3 hours of "2011/01/05 17:00:00" it must be "2011/02/05 10:00:00" and not "2011/01/05 20:00:00"

I think to use _DateDiff() and _DateAdd(), but don't say how proceed to do this.

Thanks in advance


Qui ose gagneWho Dares Win[left]CyberExploit[/left]

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What's the maximum hours you need to add?


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What's the maximum hours you need to add?

The maximum of time is 2 hours and 20 minutes.


Qui ose gagneWho Dares Win[left]CyberExploit[/left]

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#4 ·  Posted (edited)

9-12 and 14-19

if you added 3 to 17 you would be at 9 not at 10, no?

Edited by iamtheky

,-. .--. ________ .-. .-. ,---. ,-. .-. .-. .-.
|(| / /\ \ |\ /| |__ __||| | | || .-' | |/ / \ \_/ )/
(_) / /__\ \ |(\ / | )| | | `-' | | `-. | | / __ \ (_)
| | | __ | (_)\/ | (_) | | .-. | | .-' | | \ |__| ) (
| | | | |)| | \ / | | | | | |)| | `--. | |) \ | |
`-' |_| (_) | |\/| | `-' /( (_)/( __.' |((_)-' /(_|
'-' '-' (__) (__) (_) (__)

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9-12 and 14-19

if you added 3 to 17 you would be at 9 not at 10, no?

Not 9 but 10 because 17 to 19 => 2 hours and 09h to 10h do 3hours


Qui ose gagneWho Dares Win[left]CyberExploit[/left]

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Something like this should work:

#include <date.au3>

$sStartDateTime = "2011/01/05 11:00:00"
$iMinutesToAdd = 120 ; 2 hours
$sStartRange1 = "09:00:00"
$sEndRange1   = "12:00:00"
$sStartRange2 = "14:00:00"
$sEndRange2   = "17:00:00"

$sEndDateTime = _DateAdd("n", $iMinutesToAdd, $sStartDateTime)
If StringMid($sStartDateTime,12) >= $sStartRange1 And StringMid($sStartDateTime,12) <= $sEndRange1 Then   ; StartDateTime is in Range 1
    If StringMid($sEndDateTime,12) > $sEndRange1 Then $sEndDateTime = _DateAdd("h", 2, $sEndDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
Else
    If StringMid($sEndDateTime,12) > $sEndRange2 Then $sEndDateTime = _DateAdd("h", 16, $sEndDateTime)  ; EndDateTime exceeds Range 2 => add 16 hours
EndIf
ConsoleWrite($sEndDateTime & @CRLF)

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#7 ·  Posted (edited)

Something like this should work:

#include <date.au3>

$sStartDateTime = "2011/01/05 11:00:00"
$iMinutesToAdd = 120 ; 2 hours
$sStartRange1 = "09:00:00"
$sEndRange1   = "12:00:00"
$sStartRange2 = "14:00:00"
$sEndRange2   = "17:00:00"

$sEndDateTime = _DateAdd("n", $iMinutesToAdd, $sStartDateTime)
If StringMid($sStartDateTime,12) >= $sStartRange1 And StringMid($sStartDateTime,12) <= $sEndRange1 Then   ; StartDateTime is in Range 1
    If StringMid($sEndDateTime,12) > $sEndRange1 Then $sEndDateTime = _DateAdd("h", 2, $sEndDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
Else
    If StringMid($sEndDateTime,12) > $sEndRange2 Then $sEndDateTime = _DateAdd("h", 16, $sEndDateTime)  ; EndDateTime exceeds Range 2 => add 16 hours
EndIf
ConsoleWrite($sEndDateTime & @CRLF)

When I put :

$sStartDateTime = "2011/01/05 18:30:00"

with 2 hours :

$iMinutesToAdd = 120 ; 2 hours

the return value is 2011/01/06 12:30:00 !

and the result should be 2011/01/06 10:00:00

Edited by jerem488

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#8 ·  Posted (edited)

Not 9 but 10 because 17 to 19 => 2 hours and 09h to 10h do 3hours

if 17 + 1 = 18 and 17 + 2 = 19 and 17 + 3 = 10 Then there is no such animal as 19:30 (because 19 and 9 are essentially the same number) so:

18:30:00

with 2 hours :

$iMinutesToAdd = 120 ; 2 hours

and the result should be 2011/01/06 09:30:00

18:30 + 1 = 9:30 and 18:30 + 2 = 10:30

Once again I question the math and you may want to consider ending at 18:59:59 to avoid such confusion, which admittedly may only be mine.

Edited by iamtheky

,-. .--. ________ .-. .-. ,---. ,-. .-. .-. .-.
|(| / /\ \ |\ /| |__ __||| | | || .-' | |/ / \ \_/ )/
(_) / /__\ \ |(\ / | )| | | `-' | | `-. | | / __ \ (_)
| | | __ | (_)\/ | (_) | | .-. | | .-' | | \ |__| ) (
| | | | |)| | \ / | | | | | |)| | `--. | |) \ | |
`-' |_| (_) | |\/| | `-' /( (_)/( __.' |((_)-' /(_|
'-' '-' (__) (__) (_) (__)

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if 17 + 1 = 18 and 17 + 2 = 19 and 17 + 3 = 10 Then there is no such animal as 19:30 (because 19 and 9 are essentially the same number) so:

18:30 + 1 = 9:30 and 18:30 + 2 = 10:30

Once again I question the math and you may want to consider ending at 18:59:59 to avoid such confusion, which admittedly may only be mine.

Yes that's it ! So there has an error in the script. And I try to understand the error, but I don't see this error


Qui ose gagneWho Dares Win[left]CyberExploit[/left]

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#10 ·  Posted (edited)

This may be closer, just need to check the return from the _DateAdd (and decide what you want to do with those), as times like 23:30 would return 13:30 which is still out of range.

#include <date.au3>

$sStartDateTime = "2011/01/05 19:30:00"
$sStartRange1 = "09:00:00"
$sEndRange1   = "11:59:59"
$sStartRange2 = "14:00:00"
$sEndRange2   = "18:59:59"



    If StringMid($sStartDateTime,12) > $sEndRange1 And StringMid($sStartDateTime,12) < $sStartRange2 Then
    $sEndDateTime = _DateAdd("h", 2, $sStartDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
ElseIf StringMid($sStartDateTime,12) > $sEndRange2 Then
    $sEndDateTime = _DateAdd("h", 14, $sStartDateTime)  ; EndDateTime exceeds Range 2 => add 16 hours
Else
    $sEndDateTime = $sStartDateTime
EndIf

ConsoleWrite($sEndDateTime & @CRLF)
Edited by iamtheky

,-. .--. ________ .-. .-. ,---. ,-. .-. .-. .-.
|(| / /\ \ |\ /| |__ __||| | | || .-' | |/ / \ \_/ )/
(_) / /__\ \ |(\ / | )| | | `-' | | `-. | | / __ \ (_)
| | | __ | (_)\/ | (_) | | .-. | | .-' | | \ |__| ) (
| | | | |)| | \ / | | | | | |)| | `--. | |) \ | |
`-' |_| (_) | |\/| | `-' /( (_)/( __.' |((_)-' /(_|
'-' '-' (__) (__) (_) (__)

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#11 ·  Posted (edited)

This may be closer, just need to check the return from the _DateAdd (and decide what you want to do with those), as times like 23:30 would return 13:30 which is still out of range.

#include <date.au3>

$sStartDateTime = "2011/01/05 19:30:00"
$sStartRange1 = "09:00:00"
$sEndRange1   = "11:59:59"
$sStartRange2 = "14:00:00"
$sEndRange2   = "18:59:59"



    If StringMid($sStartDateTime,12) > $sEndRange1 And StringMid($sStartDateTime,12) < $sStartRange2 Then
    $sEndDateTime = _DateAdd("h", 2, $sStartDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
ElseIf StringMid($sStartDateTime,12) > $sEndRange2 Then
    $sEndDateTime = _DateAdd("h", 14, $sStartDateTime)  ; EndDateTime exceeds Range 2 => add 16 hours
Else
    $sEndDateTime = $sStartDateTime
EndIf

ConsoleWrite($sEndDateTime & @CRLF)

This don't work, and we can't specify time to add.

Like this it work :

#include <date.au3>

$sStartDateTime = "2011/01/05 16:30:00"
$iMinutesToAdd = 35
$sStartRange1 = "09:00:00"
$sEndRange1   = "12:00:00"
$sStartRange2 = "14:00:00"
$sEndRange2   = "17:00:00"

$sEndDateTime = _DateAdd("n", $iMinutesToAdd, $sStartDateTime)
If StringMid($sStartDateTime, 12) >= $sStartRange1 And StringMid($sStartDateTime, 12) <= $sEndRange1 Then   ; StartDateTime is in Range 1
    If StringMid($sEndDateTime,12) > $sEndRange1 Then $sEndDateTime = _DateAdd("h", 2, $sEndDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
    ConsoleWrite($sEndDateTime & " rr" & @CRLF)
ElseIf StringMid($sStartDateTime, 12) >= $sStartRange2 And StringMid($sStartDateTime, 12) <= $sEndRange2 Then   ; StartDateTime is in Range 1
    If StringMid($sEndDateTime,12) > $sEndRange2 Then $sEndDateTime = _DateAdd("h", 16, $sEndDateTime)   ; EndDateTime exceeds Range 1 => add 2 hours
    ConsoleWrite($sEndDateTime & " rr" & @CRLF)
EndIf
Edited by jerem488

Qui ose gagneWho Dares Win[left]CyberExploit[/left]

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Was just trying to simply show the math and how the _DateAdd return could be out of the established range. I would certainly use water's function if you need to set the hours dynamically, and you would still need to handle returns from the _DateAdd unless you find a cooler way to do it.

all apologies if I gave the impression I was solving all the issues at once.


,-. .--. ________ .-. .-. ,---. ,-. .-. .-. .-.
|(| / /\ \ |\ /| |__ __||| | | || .-' | |/ / \ \_/ )/
(_) / /__\ \ |(\ / | )| | | `-' | | `-. | | / __ \ (_)
| | | __ | (_)\/ | (_) | | .-. | | .-' | | \ |__| ) (
| | | | |)| | \ / | | | | | |)| | `--. | |) \ | |
`-' |_| (_) | |\/| | `-' /( (_)/( __.' |((_)-' /(_|
'-' '-' (__) (__) (_) (__)

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This code should handle the situation that $sStartDateTime is not within one of the ranges.

#include <date.au3>

$sStartDateTime = "2011/01/05 07:00:00"
$iMinutesToAdd = 120 ; 2 hours
$sStartRange1 = "09:00:00"
$sEndRange1 = "12:00:00"
$sStartRange2 = "14:00:00"
$sEndRange2 = "17:00:00"

; If $sStartDateTime is not within a range move $sStartDateTime to the start of the next range
$sStartTime = StringMid($sStartDateTime, 12)
If $sStartTime > $sEndRange1 And $sStartTime < $sStartRange2 Then $sStartTime = $sStartRange2  ; Time between Range 1 and Range 2 => move to begin of Range 2
If $sStartTime > $sEndRange2 Or $sStartTime < $sStartRange1 Then $sStartTime = $sStartRange1   ; Time between Range 2 and Range 1 => move to begin of Range 1
$sStartDateTime = StringLeft($sStartDateTime, 11) & $sStartTime
$sEndDateTime = _DateAdd("n", $iMinutesToAdd, $sStartDateTime)
If StringMid($sStartDateTime, 12) >= $sStartRange1 And StringMid($sStartDateTime, 12) <= $sEndRange1 Then ; StartDateTime is in Range 1
    If StringMid($sEndDateTime, 12) > $sEndRange1 Then $sEndDateTime = _DateAdd("h", 2, $sEndDateTime) ; EndDateTime exceeds Range 1 => add 2 hours
Else
    If StringMid($sEndDateTime, 12) > $sEndRange2 Then $sEndDateTime = _DateAdd("h", 16, $sEndDateTime) ; EndDateTime exceeds Range 2 => add 16 hours
EndIf
ConsoleWrite($sEndDateTime & @CRLF)

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Hi Guys,

Pardon my intrusion but am too stuborn to leave this alone without understanding it!

How does this work "if 17 + 1 = 18 and 17 + 2 = 19 and 17 + 3 = 10"

Obviously I have failed to get the point of this thread, please enlighten (then maybe headache will go away).

kylomas


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This is how I understand this thread:

The OP has two intervals (morning and afternoon). He has a point in time that can be between 00:00 and 24:00. He want's to add a number of minutes and the calculated time has to be within one of the intervals.

Example: If you have an helpdesk and a user reports an incident the helpdesk has to act within 2 hours during business time. So he wants to calculate when the SLA is violated.


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Tutorials:
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#16 ·  Posted (edited)

if you have two ranges (9, 10 , 11 , 12) (14, 15 , 16 , 17 , 18 , 19) and you have 17 + 3 you would be back at 9.

However the OP intended the clock to end at 18:59, such that there really is no 19 in the second range as the clock would roll to 9. which is why he replied

Not 9 but 10 because 17 to 19 => 2 hours and 09h to 10h do 3hours

To my first math question.

Hope that helps.

Edited by iamtheky

,-. .--. ________ .-. .-. ,---. ,-. .-. .-. .-.
|(| / /\ \ |\ /| |__ __||| | | || .-' | |/ / \ \_/ )/
(_) / /__\ \ |(\ / | )| | | `-' | | `-. | | / __ \ (_)
| | | __ | (_)\/ | (_) | | .-. | | .-' | | \ |__| ) (
| | | | |)| | \ / | | | | | |)| | `--. | |) \ | |
`-' |_| (_) | |\/| | `-' /( (_)/( __.' |((_)-' /(_|
'-' '-' (__) (__) (_) (__)

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water,

of course!!! thanks, headache much better now, goodnight

kylomas


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- Sir Winston Churchill

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