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Hi...
I have One string.

00AA015949

How to Get After Chartres Value.
For Eg:

00AA015949
1)00AA
2)015949
Posted 22-Jan-13 21:28pm
Abhai Oza1.6K
Comments

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

## 5 solutions

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## 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
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## 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
Console.ReadLine()
End Sub
End Module```

I generated this at: txt2re - regular expression generator[^]
v2
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## 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```
Comments
Achal Oza 29-Jan-13 6:59am

But Not Fix 00AA015949 this number.
0AA546545 ... How to get..???
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## Solution 2

`Regex split` functions[^] can give you an opportunity to filter out strings based on type.
Comments

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.
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## 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
Comments
Abhinav S 23-Jan-13 2:34am

5!

Thank you, Abhinav.
—SA

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