Beginner wrote:
> Hi,
Hello,
> I have a number of jpegs I wanted to rename. I wrote a short script
> to do it but the new file name is not always generated correctly. The
> script should find the last letter in the filename (before the
> extension) and substitute it for '_a'.
>
> If you look at the results below you'll see that 'a' and 'b' fail but
> 'c' worked. I don't understand why.
>
> DSC00092a.jpg -> DSC00092a.jpg a
> DSC00093b.jpg -> DSC00093b.jpg b
> DSC00094c.jpg -> DSC00094_a.jpg c
> DSC00095d.jpg -> DSC00095d.jpg d
> DSC00096e.jpg -> DSC00096e.jpg e
> DSC00097f.jpg -> DSC00097f.jpg f
> DSC00098g.jpg -> DSC00098g.jpg g
> DSC00099h.jpg -> DSC00099h.jpg h
> DSC00100i.jpg -> DSC00100i.jpg i
> DSC00101j.jpg -> DSC00101_a.jpg j
> DSC00102k.jpg -> DSC00102_a.jpg k
> DSC00103l.jpg -> DSC00103l.jpg l
> ...snip
>
> Here the script, there isn't much to it. Can anyone explain why the
> substitute fails?
>
>
> #!/bin/perl
> # Active State 5.8.6.811
>
> use strict;
> use warnings;
> use File::Basename;
>
> my $dir = 'D:/Temp/jpegs/thumbs/';
> my @files = glob("${dir}*.jpg");
>
> foreach my $f (@files) {
> (my $l) = ($f =~ /([a-z]|[a-z][a-z])\.jpg/);
Perl's alternation always quits when the first alternative matches so in your
example ([a-z]) and ([a-z]|[a-z][a-z]) are equivalent. Perhaps you meant
([a-z]{1,2}) instead?
> (my $new = $f) =~ s/$l/_a/;
$f contains the complete path so you are probably modifying something other
than the file name.
You probably want something like:
( my $new = $f ) =~ s/([a-z]{1,2})(?=\.jpg\z)/_a/;
> my $basef = basename($f);
> my $basenew = basename($new);
> print "$basef -> $basenew $l\n";
> }
John
--
Perl isn't a toolbox, but a small machine shop where you can special-order
certain sorts of tools at low cost and in short order. -- Larry Wall
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