13,551,917 members
See more:
Hi...
I have One string.

00AA015949

How to Get After Chartres Value.
For Eg:

00AA015949
1)00AA
2)015949
Posted 22-Jan-13 20:28pm
SAKryukov 23-Jan-13 2:33am

Banned by Google? How about Bing? Both of them? Sorry, be more careful next time, do not search too much. :-)
—SA

## Solution 1

You need to see exactly two Microsoft documentation pages:
http://msdn.microsoft.com/en-us/library/system.string.substring.aspx[^],
Microsoft Q209354.

—SA
Abhinav-S 23-Jan-13 2:34am

5!
SAKryukov 23-Jan-13 2:37am

Thank you, Abhinav.
—SA

## Solution 2

`Regex split` functions[^] can give you an opportunity to filter out strings based on type.
SAKryukov 23-Jan-13 2:37am

Of course, depending on the formulation of the problem, which is actually totally missing in this "question". My 5.
—SA
Abhinav-S 23-Jan-13 2:47am

Thank you SA.

## Solution 3

```Dim strText as string="00AA015949"
Dim strText1 as string
Dim strText2 as string

strText1 = strText.SubString(0,4) ' 00AA
strText2 = strText.SubString(4) ' 015949```
Achal-Oza 29-Jan-13 6:59am

But Not Fix 00AA015949 this number.
0AA546545 ... How to get..???

## Solution 5

Is this what you are looking for?

```Imports System.Text.RegularExpressions
Module Module1

Sub Main
Dim txt As String ="00AA015949" ' Put value you want to test here!

Dim re1 As String="(\d+)"	'Integer Number 1
Dim re2 As String="((?:[a-z][a-z]+))"	'Word 1
Dim re3 As String="(\d+)"	'Integer Number 2

Dim r As Regex = new Regex(re1+re2+re3,RegexOptions.IgnoreCase Or _
RegexOptions.Singleline)
Dim m As Match = r.Match(txt)
If (m.Success) Then
Dim int1=m.Groups(1)
Dim word1=m.Groups(2)
Dim int2=m.Groups(3)
Console.WriteLine("(" & int1.ToString() & _
")" & "(" & word1.ToString() & ")" & _
"(" & int2.ToString() & ")")
End If
End Sub
End Module```

I generated this at: txt2re - regular expression generator[^]
v2

## Solution 6

```private sub getstringandnumber()

Dim lstring As String = "00AA015949"
Dim lstrChar As String
Dim lstrNum As String
Dim lblnChar As Boolean
Dim i As Integer

For i = 1 To Len(lstring)

If IsNumeric(Mid(lstring, i, 1)) Then
If lblnChar Then
lstrNum = lstrNum & Mid(lstring, i, 1)
Else
lstrChar = lstrChar & Mid(lstring, i, 1)
End If
Else
lstrChar = lstrChar & Mid(lstring, i, 1)
lblnChar = True
End If
Next

MsgBox(lstrChar & "   " & lstrNum)

End Sub```
v2

Top Experts
Last 24hrsThis month
 Maciej Los 210 OriginalGriff 210 ppolymorphe 165 Richard Deeming 160 Richard MacCutchan 144
 OriginalGriff 5,070 ppolymorphe 2,272 Richard MacCutchan 2,150 Wendelius 1,865 Maciej Los 1,695