[RESOLVED] Small help needed with this regex

[akhileshbc]

Hi...

I want to extract mark1=90 from the string mark1=90 For English3 & English2. And I tried the following code. But it doesn't seems to be working because it is returning blank value
BTW, I'm not good in RegEX.

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
        Dim strTemp As String
 
        strTemp = "mark1=90 For English"
 
        Dim aa As New System.Text.RegularExpressions.Regex("^[0-9]+$")
 
        MessageBox.Show(aa.Match(strTemp).Value)
 
    End Sub

Any help would be great... Thanks...

[aashish_9601]

Unless you are trying to learn Regex, these can easily achieved by just using the SubString Function

[akhileshbc]

Unless you are trying to learn Regex, these can easily achieved by just using the SubString Function

Yeah... I'm trying to learn RegEx...

[akhileshbc]

After some small modifications of the code based on another tutorial found on the internet:

Dim input As String
        Dim pattern As String = ".=[0-9]+"
        Dim rgx As New Regex(pattern, RegexOptions.IgnoreCase)
 
        input = "mark1=90 For English"
 
        Dim matches As MatchCollection = rgx.Matches(input)
        If matches.Count > 0 Then
            For Each match As Match In matches
                MsgBox("   " + match.Value)
            Next
        End If

This will give 1=90 as the result. But when I tried this pattern: ^.=[0-9]+, it's not showing anything

Why didn't it start from the beginning ? Because what I want to get is mark1=90

[Zeljko]

This will give you mark1=90 as result:

'regex with 1 capture group: (.*)
   Dim srgx As New System.Text.RegularExpressions.Regex("(.*)\sFor English3 & English2")
'text you need to parse
   Dim tmpstring As String = "mark1=90 For English3 & English2"
'Find Capture Group in:
   Dim matchG As System.Text.RegularExpressions.MatchCollection = srgx.Matches(tmpstring)
   MsgBox(matchG(0).Groups(1).ToString())

If you had multiple different entry strings, let me know to make a little change to above code, but then I will need a few more samples...

[akhileshbc]

Thanks Zeljko...

I'll provide some samples:

mark1=90 For English3 & English2
mark5=110, good work
mark11=2&he failed for malayalam

So,

• it is single lined
• contains the format some_characters=[numbers]some_other_characters
• i want to grab the string from the begining to =[numbers]

[Zeljko]

With this samples the new regex is:

Dim srgx As New System.Text.RegularExpressions.Regex("(mark\d*=\d*)")
MsgBox(matchG(0).Groups(1).ToString()) 
MsgBox(matchG(1).Groups(1).ToString()) 
MsgBox(matchG(2).Groups(1).ToString())

[Zeljko]

or if you need to get something like this with 3 rule:
"i want to grab the string from the begining to =[numbers]"

mark1=90 For English3 & English2
mark=110, good work
mar1=2&he failed for malayalam
 
The regex would be:
Dim srgx As New System.Text.RegularExpressions.Regex("(.*=\d*)")

[Zach_VB6]

Why not this:

"^mark(\d+)=(\d+)"

[minitech]

Why not ^\D+\d+ in multiline mode

[akhileshbc]

or if you need to get something like this with 3 rule:
"i want to grab the string from the begining to =[numbers]"

mark1=90 For English3 & English2
mark=110, good work
mar1=2&he failed for malayalam
 
The regex would be:
Dim srgx As New System.Text.RegularExpressions.Regex("(.*=\d*)")

Yeah.. That worked...

If don't you mind, can you give me a brief explanation on the pattern. Because I'm trying to learn.

Why not ^\D+\d+ in multiline mode

That doesn't display anything

Why not this:

"^mark(\d+)=(\d+)"

It will display only 1

[Zeljko]

If don't you mind, can you give me a brief explanation on the pattern. Because I'm trying to learn.

Here it is:

regex: (.*=\d*)
(.*=\d*) => anything inside parentheses is inside a 'capture group' - Capture group will return data to you ($1). If we have this regex: .*=(\d*) then it will return to us only last numbers (without any char before '=' sign and without equal sign)
(.*=\d*) => Matches any character, except for line breaks if dotall (global switch 's') is false. Dot will accept any text but not new line (that is what we need here)
(.*=\d*) => Matches 0 or more of the preceding token. Asterix tell us that we can have minimum 0 or preferably more of previous and in our case previous is '.' which stands for any char.
(.*=\d*) => Matches '='. Normal string.
(.*=\d*) => Matches any digit character (0-9). But it will match only one digit char!
(.*=\d*) => Matches 0 or more of the preceding token. Asterix tell us that we can have minimum 0 or preferably more of previous and in our case previous is '\d.' which stands for one digit char - we have single or two or three digit chars so we need this.

Just to let know while you are in learning phase:
With regex always exists multiple ways of doing one thing. Some ways are shorter, some longer, some better, some bulletproof...
One way is shown above...

[akhileshbc]

Thanks a lot...

added rep

I have even gone through some sites that provides information about RegEx. But most of them make it confusing and complicating rather than telling it in simple ways.

[Zeljko]

Glad to help but no reps received

[akhileshbc]

Glad to help but no reps received

Extremely sorry. I forgot to do that after marking the thread as RESOLVED. Because I was chatting with my friend at that time.

Please check it now... And thanks for reminding me.

[Zeljko]

Thanks reps are not necessary but they are only showing a sign of gracefulness...

[minitech]

My code probably doesn't display anything because it's grouped - but it will match what you're trying to capture.

[akhileshbc]

My code probably doesn't display anything because it's grouped - but it will match what you're trying to capture.

Can you explain ? I'm a bit confused

[minitech]

It doesn't have parenthesis, but the match will be correct. Check the .Value property of the Match objects returned.

[akhileshbc]

It doesn't have parenthesis, but the match will be correct. Check the .Value property of the Match objects returned.

        Dim srgx As New System.Text.RegularExpressions.Regex("^\D+\d+")
        'text you need to parse
        Dim tmpstring As String = "mark1=90 For English3 & English2"

        Dim matchG As System.Text.RegularExpressions.MatchCollection = srgx.Matches(tmpstring)
        MsgBox(matchG(0).Value.ToString)

Output: mark1

Dim tmpstring As String = "mark11=2&he failed for malayalam"

Output: mark11

Dim tmpstring As String = "mark5=110, good work"

Output: mark110

[minitech]

Ah, sorry:

^[^\=]+\=\d+

[akhileshbc]

Ah, sorry:

^[^\=]+\d+

Mini, that gives the same result as in my previous post

[minitech]

I forgot the = sign again, sorry. Post edited.

[akhileshbc]

That worked...

Could you please explain it like Zeljko did in post#12 ?

Thanks...

[minitech]

^, by itself, means the beginning of a line, and since the text you need is always at the beginning of the line, I use that.
\= is an = sign, escaped because it's used in some expressions.
[^(stuff)] means to match anything except for (stuff), so I match anything except for an = sign...
\= Then, I match an = sign...
\d+ and I finally match a number. \d means one digit and + means 1 or more.