I can't make heads or tail of your version, bu this works for me.
#line.pl
#!/usr/local/bin/perl
open (FILE, $ARGV[0] ) or die "$!\n";
my $count = <FILE>;
chomp($count);
$line = 1;
while ( <FILE> ) {
print if $line++ > $count;
}
#line.txt
3
line1
line2
line3
line4
line5
line6
line7
[cgalpin@pooh perl]$ ./line.pl line.txt
line4
line5
line6
line7
hth
charles
On Mon, 22 Nov 1999, gnielson wrote:
> I am trying to write a perl script that will read a number from a file,
> use that as a counter, and only print out lines above that counter number.
> So if the number is 5 and the file is 10 lines, only the last 5 lines
> would be printed.
>
> The problem is when it prints out it misses one line, so instead of
> printing 5 lines it will print the last 4 lines.
>
> Any help much appreciated.
>
> # $text is the counter variable
>
> open(FD,"$file") || die "Can not open $file";
> while(<FD>) {
>
> my $lineNo = "0";
> while ($line = <FD>)
> {
> print "lineNo is $lineNo and text is $text\n";
>
> next if ($lineNo < $text);
> print $line;
> }
> }
> close FD;
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