Check if string contains a number

Re: Check if string contains a number

How is the user entering the string?

Re: Check if string contains a number

Hi Grays

Try this Function;
Extract Numbers from Text Strings

Re: Check if string contains a number

Here's where a little bit of Linq can make life easier. This will return a boolean telling if there's at least one numeric character in the string.

(from items in inputString.ToCharArray Where Char.IsDigit(items)).Count <> 0


Dave

Re: Check if string contains a number

Quote from grays;192686

Thanks Dave but i found something i could use

Code

If HasNumber(fgncurrency) = True Then
    Err = 1
End If

then

Code

Function HasNumber(strData As String) As Boolean
    Dim iCnt As Integer
     
    For iCnt = 1 To Len(strData)
        If IsNumeric(Mid(strData, iCnt, 1)) Then
            HasNumber = True
            Exit Function
        End If
    Next iCnt
     
End Function

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Thanks grays - this is super helpful.

However, is there a way to modify it to be true if it contains any number except 0?

Quote from dentedduck

Re: Check if string contains a number

Here's where a little bit of Linq can make life easier. This will return a boolean telling if there's at least one numeric character in the string.

(from items in inputString.ToCharArray Where Char.IsDigit(items)).Count <> 0


Dave

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Hi Dave, I hope you get this... 7 years later

Could you elaborate on how your code is supposed to work? I can't seem to make it work?

from items in inputString.ToCharArray WhereChar.IsDigit(items)).Count <> 0

how do i write this?

ie: x = from items in inputString.ToCharArray WhereChar.IsDigit(items)).Count <> 0 ??

LINQ doesn't work in VBA as far as I know. Seems overkill here anyway when a simple Like operator will work:

Function HasNumber(sIn As String) As Boolean
    HasNumber = sIn Like "*#*"
End Function
Function HasNumberOtherThanZero(sIn As String) As Boolean
    HasNumberOtherThanZero = sIn Like "*[1-9]*"
End Function

Quote from rory

LINQ doesn't work in VBA as far as I know. Seems overkill here anyway when a simple Like operator will work:

Code

Function HasNumber(sIn As String) As Boolean
HasNumber = sIn Like "*#*"
End Function
Function HasNumberOtherThanZero(sIn As String) As Boolean
HasNumberOtherThanZero = sIn Like "*[1-9]*"
End Function

You're absolutely right ! Works like a charm. So simple yet so efficient.

Thanks rory !