[RESOLVED] Small help needed with this regex

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.

vb Code:

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...:wave:

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

Quote:

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Originally Posted by aashish_9601

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

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

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

vb Code:

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

This will give you mark1=90 as result:

VB.Net Code:

'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...

Thanks Zeljko...:wave:

I'll provide some samples:

Code:

---

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]

:wave:

With this samples the new regex is:

vb Code:

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())

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

VB.Net Code:

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*)")

Why not this:

Code:

---

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

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

Quote:

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Originally Posted by Zeljko

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

VB.Net Code:

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...:thumb:

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

Quote:

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Originally Posted by minitech

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

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That doesn't display anything :(

Quote:

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Originally Posted by Zach_VB6

Why not this:

Code:

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

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It will display only 1 :(

Quote:

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Originally Posted by akhileshbc

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

---

Here it is:

Code:

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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.

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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...

Thanks a lot...:wave:

added rep :thumb:

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. :)

Glad to help but no reps received :rolleyes:

Quote:

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Originally Posted by Zeljko

Glad to help but no reps received :rolleyes:

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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...:thumb: And thanks for reminding me.

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

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

Quote:

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Originally Posted by minitech

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

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Can you explain ? I'm a bit confused :confused:

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

Quote:

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Originally Posted by minitech

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

---

Code:

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        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

Code:

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

---

Output: mark11

Code:

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

---

Output: mark110

Ah, sorry:

Code:

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

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Quote:

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Originally Posted by minitech

Ah, sorry:

Code:

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^[^\=]+\d+

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

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

That worked...:thumb:

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

Thanks...:wave:

^, 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.