13,201,452 members (69,384 online)
Rate this:
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
Abhai Oza1.7K

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

Rate this:

## 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!

Thank you, Abhinav.
—SA
Rate this:

## Solution 2

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

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.
Rate this:

## 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..???
Rate this:

## 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
Rate this:

## 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
 OriginalGriff 330 Richard MacCutchan 125 Dave Kreskowiak 110 GKP1992 75 Karthik Bangalore 70
 OriginalGriff 8,125 ppolymorphe 2,024 Karthik Bangalore 1,724 CPallini 1,208 Jochen Arndt 1,075