in reply to Until there's nothing but spaces?

O.K. this is how I solved it, it seems to work pretty well. Tnx to all for their help! 
opendir (ID, "$dir");
        @dirlist = readdir(ID);
closedir(ID);

foreach $inode (@dirlist) {
        $i++;
        print "Files processed: $i\n";
        next if $inode =~ /^\./;
        my $sfile = $dir . $inode;

        open (IF, "<$sfile");
                @fcont = (<IF>);
        close (IF);

        open (OF, ">$sfile");
        foreach $line (@fcont) {
                if ($line =~ /themark/) {
                        print "starter string found\n";
                        $flag = 1;
                        next;
                }
                if ($flag == 1) {
                        if  (!($line =~ /\w+/)) {
                                $flag = 0;
                                next;
                        }
                        else {
                                print OF $line;
                        }
                }
        }
    close (OF);
}

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Re: Until there's nothing but spaces?
by mandog (Curate) on Oct 01, 2001 at 06:40 UTC 
            next if $inode =~ /^\./;

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    ...looks like it will match .bashrc .login .hidden_data and skip them.

    If you only want to skip '..' and '.' something you may want something (untested) like this: 

    m/^\.{1,2}$/

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    --mandog
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      Good thing you pointed out that this code is untested, because it is wrong! Someone posted this very same snippet a couple of weeks ago and merlyn pointed out a subtle way in which it can fail.

      But rather than elaborating a more and more complex regexp for such a simple operation, I like to spell things out more simply, but using something like:

        next if $inode eq '.' or $inode eq '..';

      This avoids pulling in the regexp machinery, and is much easier for the next maintainer (who may not be as good a rexexp wizard as yourself) to figure out what is going on.

      --
      g r i n d e r
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        Esteemed Sibling Grinder:

        Here you write "Good thing you pointed out that this code is untested, because it is wrong!" ....

        Thank you for your correction. At first I doubted you and the saint who's words you invoked.

        The correct regular expression is:
         /\A\.{1,2}\z/ My regular expression was: 

         /^\.{1,2}$/

<code> perldoc perlre

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        reads in part:
        ...The \A and \Z are just like ``^'' and ``$'', except that they won't match multiple times when the /m modifier is used, while ``^'' and ``$'' will match at every internal line boundary. ...

        Since there is no /m modifier in the picture I doubted you....

        However,  perldoc perlre continues:
        To match the actual end of the string and not ignore an optional trailing newline, use \z.

        As soon as I noticed that /z and /Z are as different as a and A, I realized you were correct.

        Of course you are also correct about using  eq

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