Avoiding Greedy matches and capturing Multiple lines using Regular Expressions (VB)

Posted on 2004-08-24
Last Modified: 2010-08-05
i need to extract some lines of strings between a specified "start" and "end". I made use of Microsoft Vbscript Regular Expressions 5.5, here is VB my code

Dim objReg As New RegExp
Dim objMatchCol As MatchCollection
With objReg
    .Global = True
    .IgnoreCase = True
    .Pattern ="start(.*?)end"
    sSourceString="start : Welcome to Experts-Exchnge : end"
    Set objMatchCol = .Execute(sSourceString)
End With
 It works smoothly but it is unable to catch multiple lines between start and end. See the following text

 : Welcome
How can i get these lines betwen 'start' and 'end' as a single match?
Please help

Question by:Shiju Sasidharan
  • 2
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  • 2
  • +1
LVL 84

Accepted Solution

ozo earned 80 total points
ID: 11887980
.Pattern ="start([\s\S]*?)end"

Assisted Solution

sgartner earned 20 total points
ID: 11889393
First, I'm not a VB expert, but the Microsoft web site indicated that the regular expression syntax for VB is the same as VBScript.  The period stands for any character in a RE, except for the carriage return.  So, for the example you have in VB you could simply use "^start(.*)end$", but this did not match the longer sample because the period *will not*match a line end character.

To match all of the text between start and end including multiple lines: "^start((.|\n)*)end$"

This assumes, like both samples, that the entire string begins with "start" and ends with "end".  If you expect other text to be in the middle you would remove the carat and dollar sign.

What this says is that the matched string would begin with "start" then be followed by either: any character (".") or a carriage return ("\n") zero or more times "*".  The outer parens will create the match block for the MatchCollection, the inner ones will be ignored as a block and only serve to group the two items.

One last thing that confuses me:  In VBScript (and Perl, etc.) the construct ".*?" is not legal, so you may need to adjust my RE syntax to match that of VB.

Sorry for submitting this twice, the first time I submitted early, and found that I can't edit my submissions on this site.  If a moderator reading this has the ability, please remove my previous post.

Expert Comment

ID: 11890485
Which OS are u using
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Expert Comment

ID: 11890708
Oh its Microsoft.....
No regular expressions match multiple lines. Your problem can be solved by writing a Perl script.
install the free software ActivePerl for Windows from

Copy the folowing to a .pl file say

$inSection = 0;
while (<STDIN>)
  if (m/end/)
    $inSection = 0;
  if ($inSection)
  if (m/start/)
    $inSection = 1;

Assume the source file is called temp.txt

use the following command
perl < temp.txt

You may need to provide the whole path if the perl script file or the temp.txt file are not in the same directory.
The required section will be printed to STDOUT which also be redirected to a file.

LVL 14

Author Comment

by:Shiju Sasidharan
ID: 11899816
Thank u all for posting comments...

hi sgartner
    I tried ur code  
but this doesnt seem to solve it. This code works only if the entire match is with in a single line.
Sample text i gave in question can be in the middle of a large string. I need to find all such matches from the entire string.

Hi ozo ,
    i tried ur code
                            .Pattern ="start([\s\S]*?)end"
 this also has the same effect, only takes matches with in a single line
Well, I am using VB6 , Os is Windows 2000

Hoping more comments....



Expert Comment

ID: 11899837

The ^ and $ force it to be by itself, which is how your samples were.  Just remove those two characters and it should work.

LVL 14

Author Comment

by:Shiju Sasidharan
ID: 11939350
Hi Ozo
Thank u for ur code, i am accepting ur answer
Code given by u is working good, I am really sorry that i posted a comment indicating that code given by u was not working, it was my mistake i gave an invalid string for verification.
it took only 20 milliseconds to execute my source string with the pattern u gave.

             .Pattern ="start([\s\S]*?)end"

Hi Scott
Thank u for ur posting, i accept ur answer as a supporting one
  after removing ^ and $ it looked greedy
       but putting a ? in ur code made successful matching
             .Pattern ="^start((.|\n)*?)end$"

  This pattern solved my problem, but it took 90 milliseconds to execute my source string



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