asked on

Regular Expression Validator Validation Control Expression

I need a RegularExpressionValidator to validate a textbox that has only one character. It can have only the letters A-Z, or the numbers 0-9, or a dash, or an apostrophe, or a space. Any ideas what the ValidationExpression is?

Ravi Vaddadi

Try this

([A-Z]{1,1})|([0-9]{1,1})|(-)|(')|(\S)

Mark Franz

Try this;

SubjectString = "121234567890-=\][poiuytrewqq';l/., " 
Try
	If Regex.IsMatch(SubjectString, "[\w\x20\x27\x2D]") Then
		' Successful match
	Else
		' Match attempt failed
	End If
Catch ex As ArgumentException
	'Syntax error in the regular expression
End Try

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us1975mc

ASKER

SriVaddadi,

I am still able to enter other characters. Below is my code.

<asp:RegularExpressionValidator ID="Regex1" 
                runat="server" 
                ErrorMessage="RegularExpressionValidator" 
                ControlToValidate="Let1"
                ValidationExpression="([A-Z]{1,1})|([0-9]{1,1})|(-)|(')|(\S)" > 
            </asp:RegularExpressionValidator>

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us1975mc

ASKER

mgfranz,

I am really lost on this. Do I still need a RegularExpressionValidator? Where do I place the code? In the Let1_TextChanged?

Protected Sub Let1_TextChanged(ByVal sender As Object, ByVal e As System.EventArgs) Handles Let1.TextChanged
        Dim Subjectstring As String
        SubjectString = "121234567890-=\][poiuytrewqq';l/., "
        Try
            If Regex.IsMatch(SubjectString, "[\w\x20\x27\x2D]") Then
                Let2.Focus()
            Else
                Let1.Focus()
                MsgBox("Try Again")
            End If
        Catch ex As ArgumentException
            'Syntax error in the regular expression
        End Try
    End Sub

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Mark Franz

If you would rather not use ASCII parameters, you can use this;

SubjectString = "`~1!2@3#4$5%6^7&8*9(0)-_=+qwertyuiop[{]}\|asdfghjkl;:'zxcvbnm,<.>/? "
Try
	Dim RegexObj As New Regex("[A-Za-z0-9'\- ]")
	If RegexObj.IsMatch(SubjectString) Then
		' Successful match
	Else
		' Match attempt failed
	End If
Catch ex As ArgumentException
	'Syntax error in the regular expression
End Try

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Mark Franz

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us1975mc

ASKER

This is a solution for what I have ask but I guess I ask the wrong question. I will post another question tomorrow dealing with a different twist.

Thank you for your time and effort.