how to get last value in an array in perl - Shell Programming and Scripting

meghana · #1 (**permalink**) 02-04-2008

Hi,

I have a set of file names say:

file1
file2
file3
..
filen

I may just do a ls -l or however it works to read this list into an array...

I need to pick the last file name that is "filen" from the list...

your help is greatly appreciated.....hope im clear on my question...

radoulov · #2 (**permalink**) 02-04-2008

Code:

zsh-4.3.4% ls
file1  file2  file3  file4  file5
zsh-4.3.4% perl -le'print $ARGV[-1]' file*
file5

V3l0 · #3 (**permalink**) 02-04-2008

Code:

ls file* | sort | tail -1

radoulov · #4 (**permalink**) 02-04-2008

Quote:

Originally Posted by V3l0

Code:

ls file* | sort | tail -1

Or

Code:

$ ls
file1  file2  file3  file4  file5
$ set -- file*;echo $_
file5

P.S. For some shells it should be:

Code:

set -- file*
echo $_

meghana · #5 (**permalink**) 02-04-2008

wow.. thanks for the really quick replies.. im doing this using a script tmp.pl ..

$ ls
file1 file2 file3 file4 file5
$ set -- file*;echo $_
file5

can u please tell me how to write this in a script??? thanks again

radoulov · #6 (**permalink**) 02-04-2008

If it's a perl script, use perl:

Code:

zsh-4.3.4% ls
file1  file2  file3  file4  file5  temp.pl
zsh-4.3.4% cat temp.pl 
#!/usr/bin/perl 
use warnings;

@files = <file*>;
print "$files[-1]\n"; 
zsh-4.3.4% ./temp.pl 
file5

meghana · #7 (**permalink**) 02-04-2008

i have got a solution ... its little different but it works.....

my @files = glob("file*");
my $last_file_name = pop(@files);

thanks for ur quick replies again........

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