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Bad Juju
Joined: 19 Feb 2009
Posts: 11
Location: Southern California
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 Posted: Fri Apr 03, 2009 4:41 pm Post subject: Remove digits from a number |
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I have a number that is 12 digits long (eg. 800800022348). I need AHK to remove the first three digits so that I am left with a 9 digit number (eg. 800022348). I suppose I can just subtract 800000000000 but is there another way to accomplish this?
Thanks,
Shaun
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aCkRiTe
Joined: 21 Jul 2006
Posts: 555
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| Posted: Fri Apr 03, 2009 4:44 pm Post subject: |
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| Code: |
Num = 800800022348
StringTrimLeft, Num, Num, 3
MsgBox, %Num%
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sinkfaze
Joined: 18 Mar 2008
Posts: 5043
Location: the tunnel(?=light)
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| Posted: Fri Apr 03, 2009 4:52 pm Post subject: |
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Not as readable but shorter and faster:
| Code: | Num = 800800022348
Num:=SubStr(Num,4)
MsgBox, %Num% |
Changed the 3 to a 4 in the SubStr function.
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Last edited by sinkfaze on Fri Apr 03, 2009 5:27 pm; edited 1 time in total |
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aCkRiTe
Joined: 21 Jul 2006
Posts: 555
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| Posted: Fri Apr 03, 2009 5:06 pm Post subject: |
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| sinkfaze wrote: | Not as readable but shorter and faster:
| Code: | Num = 800800022348
Num:=SubStr(Num,3)
MsgBox, %Num% |
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Needs to be:
| Code: | Num = 800800022348
Num:=SubStr(Num,4)
MsgBox, %Num% |
and its not any faster...
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sinkfaze
Joined: 18 Mar 2008
Posts: 5043
Location: the tunnel(?=light)
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| Posted: Fri Apr 03, 2009 5:25 pm Post subject: |
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You are correct about my mistake, but SubStr does perform faster than StringTrimLeft, even if the increase in speed doesn't make a significant difference in this instance.
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aCkRiTe
Joined: 21 Jul 2006
Posts: 555
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| Posted: Fri Apr 03, 2009 5:41 pm Post subject: |
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| sinkfaze wrote: | | SubStr does perform faster than StringTrimLeft, even if the increase in speed doesn't make a significant difference in this instance. |
That I am not sure about, but I dont doubt that you are correct. When I said it wasnt faster, I was referring to this instance.
| Code: | StartTime := A_TickCount
Num = 800800022348
StringTrimLeft, Num, Num, 3
ElapsedTime := (A_TickCount - StartTime)/1000
MsgBox, %Num% - %ElapsedTime%
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and
| Code: |
StartTime := A_TickCount
Num = 800800022348
Num:=SubStr(Num,4)
ElapsedTime := (A_TickCount - StartTime)/1000
MsgBox, %Num% - %ElapsedTime%
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I get an elapsed time of 0 in both cases.
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Bad Juju
Joined: 19 Feb 2009
Posts: 11
Location: Southern California
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| Posted: Fri Apr 03, 2009 5:43 pm Post subject: |
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Outstanding! That worked wonderfully. Thank you very much!
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