Get only numbers from string

Question

i want to get only numbers from string.

Lest say that this is my string :

324ghgj123

i want to get:

324123

Answer 1

You can actually combine some of these individual answers to create a single line solution that will either return an integer with the value 0, or an integer that is the concatenation of all of the numbers in the string. Not sure how useful that is, however -- this began as a method to create a string of numbers only....

    Dim TestMe = CInt(Val(New Text.StringBuilder((From ch In "123abc123".ToCharArray Where IsNumeric(ch)).ToArray).ToString))

Answer 2

str="something1234text456"   

Set RegEx = CreateObject("vbscript.regexp")
RegEx.Pattern = "[^\d+]"
RegEx.IgnoreCase = True
RegEx.Global = True
numbers=RegEx.Replace(str, "")(0)
msgbox numbers

Answer 3

resultString = Regex.Match(subjectString, @"\d+").Value;

will give you that number as a string. Int32.Parse(resultString) will then give you the number.

Answer 4

try this:

Dim mytext As String = "123a123"
Dim myChars() As Char = mytext.ToCharArray()
For Each ch As Char In myChars
     If Char.IsDigit(ch) Then
          MessageBox.Show(ch)
     End If
Next

or:

Private Shared Function Num(ByVal value As String) As Integer
    Dim returnVal As String = String.Empty
    Dim collection As MatchCollection = Regex.Matches(value, "\d+")
    For Each m As Match In collection
        returnVal += m.ToString()
    Next
    Return Convert.ToInt32(returnVal)
End Function

Answer 5

Or you can use the fact that a String is an Array of Chars.

Public Function getNumeric(value As String) As String
    Dim output As StringBuilder = New StringBuilder
    For i = 0 To value.Length - 1
        If IsNumeric(value(i)) Then
            output.Append(value(i))
        End If
    Next
    Return output.ToString()
End Function

Answer 6

you can use Regex for this

Imports System.Text.RegularExpressions

then on some part of your code

Dim x As String = "123a123&*^*&^*&^*&^   a sdsdfsdf"
MsgBox(Integer.Parse(Regex.Replace(x, "[^\d]", "")))