Comments on: looking for isnumeric in vb.net?

By: Carter

Carter — Thu, 08 May 2008 14:11:10 +0000

Why reinvent the wheel? IsNumeric is part of .NET. Just use it.

By: Farren

Farren — Sun, 09 Sep 2007 00:44:28 +0000

This seems to work well for me.

Public Shared Function IsNumeric(ByVal sText As String) As Boolean

If Double.TryParse(sText, Globalization.NumberStyles.AllowDecimalPoint) Then
Return True
Else
Return False
End If

End Function

By: JimK

JimK — Thu, 23 Aug 2007 17:01:51 +0000

ok..that didnt paste well…removed some of my code..trying again

Public Shared Function IsNumeric(ByVal sText As String) As Boolean
Dim b As Boolean = True
Dim x As Integer = 0

For x = 0 To sText.Length - 1
Dim sChar As String = sText.Substring(x, 1)

If Asc(sChar) < 48 Or Asc(sChar) > Then
b = False
Exit For
End If
Next

Return b
End Function

By: JimK

JimK — Thu, 23 Aug 2007 17:00:10 +0000

I get nothing but false positives with the functions listed above

I created my own and is perfect for detecting ‘all numeric’ values…great for detecting table indexes (just make sure you trim the text first):

Public Shared Function IsNumeric(ByVal sText As String) As Boolean
Dim b As Boolean = True
Dim x As Integer = 0

For x = 0 To sText.Length - 1
Dim sChar As String = sText.Substring(x, 1)

If Asc(sChar) 57 Then
b = False
Exit For
End If
Next

Return b
End Function

By: DocXango

DocXango — Sun, 24 Jun 2007 00:22:48 +0000

I’ve had the best luck with “\b\d+(\.\d+)?\b” as my regular expression pattern since it will match doubles and floats as well.

My variation on Kevin’s function would be:
Public Shared Function isNumeric(ByVal textToCheck as String) as Boolean
return Regex.IsMatch(textToCheck, “\b\d+(\.\d+)?\b”)
End Function

Just a bit cleaner in my humble opinion.

By: Chris

Chris — Tue, 06 Mar 2007 16:10:14 +0000

The proper regex expression should be: ^\d+$
This will verify that the start of the value is numeric, not just search for any number in the supplied value

By: Troy

Troy — Wed, 29 Nov 2006 18:25:01 +0000

Unfortunately that regex expression doesn’t work in every case. I am parsing a file and I am getting some non-standard characters at the end of a number because of some bad data that was sent as the power failed. This expression still returns true and my program fails as it tries to convert the number with the bad characters….

By: Kevin

Kevin — Wed, 29 Nov 2006 10:20:08 +0000

oops, make that:
Public Shared Function IsNumeric(ByVal str As String) as Boolean
Return New Regex(”\d+”).Match(str).Success
End Function

By: Kevin

Kevin — Wed, 29 Nov 2006 10:19:00 +0000

Or even simpler:

Imports System.Text.RegularExpressions

Public Shared Function IsNumeric(ByVal str As String)
Return New Regex(”\d+”).Match(str).Success
End Function

By: janine

janine — Sun, 12 Feb 2006 12:27:47 +0000

thanks for the contribution peter - regular expressions to the rescue once again