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best way to determine if a string contains only the digits 0- 9

Posted on 2003-12-09
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Last Modified: 2010-05-03
What is the best way to determine if a string contains only the digits 0 to 9.

IsNumeric() doesn't seem to be sufficent, and I'd like to avoid looping through each character, but if that's the only way I guess I'll have to .

(would be easy in perl, but that doesn't help me either ;)

joe
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Question by:josephfluckiger
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17 Comments
 
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Expert Comment

by:bobbit31
ID: 9909097
Dim s as string

if Asc(s) >= 48 and Asc(s) <= 57 then
  msgbox "it is between 0 and 9"
end if
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Expert Comment

by:mccainz2
ID: 9909219
you could use a regular expression object to test for your pattern,,,
Will write a little function here and post in a bit....
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Assisted Solution

by:Dabas
Dabas earned 300 total points
ID: 9909241
Hi josephfluckiger,
Agree with mccainz2.
Regular expressions suits this very well, specially since you have perl experience

See http://oldlook.experts-exchange.com/Programming/Programming_Languages/Visual_Basic/Q_20528014.html
for a link to MS site where you can download regex library

Dabas
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Accepted Solution

by:
Dabas earned 300 total points
ID: 9909248
josephfluckiger,
Another way to do it is using Val

If trim(Val(sText)) = Trim(sText) then...

Dabas
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Assisted Solution

by:mccainz2
mccainz2 earned 200 total points
ID: 9909325
Dim objRXP As New RegExp
    objRXP.Pattern = "^\d[\d]*\d$"
    If objRXP.Test("1444444444") Then
        MsgBox "match"
    End If
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Expert Comment

by:mccainz2
ID: 9909331
Definitely check my Pattern , I am EXTREMELY rusty with patterns ... tested on just a few so be wary...
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Expert Comment

by:mccainz2
ID: 9909346
BTW:
To use the object simply add a reference to Microsoft VBscript Regualr Expressions
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Expert Comment

by:Dabas
ID: 9909351
Why not plainly "\d+" ?
\d = digit
+ = one or more

Dabas
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Author Comment

by:josephfluckiger
ID: 9909356
mccainz2, you mean I can have my cake and eat it too! that's fantastic, thanks so much for showing me how to do regx in VB.

Daba's answer does the trick.
CInt(Val(sText))
ended up doing this trick, since I don't even want 0.3, for example, to be a valid entry.

thanks so much all, all your responses were VERY helpful!
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Expert Comment

by:Dabas
ID: 9909361
mccainz2,
> add a reference to Microsoft VBscript Regualr Expressions
So that is where they are!

I was looking for Microsoft Regular Expressions, and could not find it!

Dabas
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Expert Comment

by:mccainz2
ID: 9909379
Why not plainly "\d+" ?

That pattern will match any string as long as it has one or more numbers in it ... example "sssssss1a" will return as a match
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Author Comment

by:josephfluckiger
ID: 9909393
yea, I've worked with Perl reg expressions for three years, and VB for one. I've always wanted reg expressions in VB, this great guys, thanks.
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Expert Comment

by:Dabas
ID: 9909413
What about "^\d+$" ?
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Expert Comment

by:Dabas
ID: 9909419
P.S
I only "discovered" regex about a week ago!

Dabas
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Expert Comment

by:mccainz2
ID: 9909425
I know what you mean!
I havent used in em about a year and its practically like I never used them before!!!
I'm sure theres a better expression for the task but at this time of night it's beyond me :)
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Expert Comment

by:mccainz2
ID: 9909432
"^\d+$"  looks like that one works!
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Expert Comment

by:Dabas
ID: 9909454
Sometimes I am a fast learner!

Thanks!
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