String parsing in VB

Question

For example, mystring is = "53 54 41 52 54 00 00 01 53 54 41 52 54 00 01 02 53 54 41 52 54 01 02 03"

What I want to do is every time I found "53 54 41 52 54" it will capture the remaining 3 bytes. So, based on the example of my string above, I want the result as below

When it first found 53 54 41 52 54, it will capture 00 00 01 then, when it found the 2nd 53 54 41 52 54, it will capture 00 01 02 then, finally when it found the 3rd 53 54 41 52 54, it will capture 01 02 03.

What is the best way to do this in VB?

Thank you in advance.

Appreciate your guidance.

You refer to the character pairs as bytes - is there a possibility of getting something like A2 or 162? — Andrew Morton
– Andrew Morton, Commented Feb 12, 2020 at 9:34

Caius Jard · Accepted Answer · 2020-02-12 09:10:45Z

2

You ca use a regular expression:

Dim r as New Regex("53 54 41 52 54 (?<b>\d\d \d\d \d\d)")

ForEach m as Match in r.Matches(mystring)
  MsgBox(m.Groups("b").Value)
Next m

Regular expression will seek your desired digits then take the subsequent 3 pairs of \digits and capture them into a group named b which you can retrieve as shown in the loop

answered Feb 12, 2020 at 9:10

Caius Jard

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Jon Roberts · Accepted Answer · 2020-02-12 09:22:43Z

0

An alternative (less neat, but possibly easier to read if unfamiliar with Regular Expressions) option would be:

    Const myString As String = "53 54 41 52 54 00 00 01 53 54 41 52 54 00 01 02 53 54 41 52 54 01 02 03"
    Const myMatch As String = "53 54 41 52 54"
    Dim Results As New List(Of String)
    Dim i As Integer = 0
    Do While i >= 0 AndAlso i < myString.Length
        i = myString.IndexOf(myMatch, i)
        If i >= 0 Then
            Results.Add(myString.Substring(i + 15, 8))
            i += 9
        End If
    Loop

answered Feb 12, 2020 at 9:22

Jon Roberts

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