Perl print match


This should be simple but not getting what I want.

$line2 = '20110320222853  CN=Jones Test,OU=Users,OU=G01,OU=AAA,DC=well,DC=corp  CN=Test Jones  523999,ou=employee,o=abc,c=AN  CN=John Doe  878976,OU=abc,OU=Contacts,OU=Administrati
on,DC=well,DC=corp  cn=John Doe  878976,ou=employee,O=abc,C=AN  0  Modify  Success';

if ($line2 =~ /^(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})\s+(CN=.+DC=corp)\s+(CN=.*c=AN).*$/im) {
  print "$7\n";
}

From this $7 should print out:

CN=Jones Test,OU=Users,OU=G01,OU=AAA,DC=well,DC=corp

I want $7 to print out up to the space "\s+" after the first DC=corp

but I am getting everything up to the second DC=corp
such as:

CN=Jones Test,OU=Users,OU=G01,OU=AAA,DC=well,DC=corp  CN=Test Jones  523999,ou=employee,o=abc,c=AN  CN=John Doe  878976,OU=abc,OU=Contacts,OU=Administration,DC=well,DC=corp

Can someone tell me why I'm getting everything up to the second DC=corp in $7?  I would think the \s+ I have would separate that but sure is not.

Thanks,
bt707Asked:
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farzanjConnect With a Mentor Commented:
Try this


if ($line2 =~ /^(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})(\d{2})\s+(CN=.+?DC=corp)\s+(CN=.*c=AN).*$/im)
{
  print "$7\n";
}
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farzanjCommented:
Of course

Regular expressions are greedy.  So the .+ matches for as long as it can and would still be able to match the second instance of DC=corp.
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bt707Author Commented:
Yes that is for sure and just what I get, but need to figure out how to stop at the first DC=corp, any suggestions.
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farzanjCommented:
So the difference is .+?

The ? tells it not to be greedy.  Default is to go for the longest match.  This way, it would go for the shortest match.
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bt707Author Commented:
I actually already tried that figuring that's what I needed but messed up somewhere and thought it did not work.

Thanks for pointing that out, I missed is somewhere.

Thanks!!!!
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