How to get first file in directory by alphabetical order? - perl

Question

I want to execute $filename = `ls *.gz | head -n 1`; through perl but I think the pipe is causing an error. Execution of -e aborted due to compilation errors.
This will be part of the perl script rather than run via -e.
What would be the correct way to do this?

Answer 1

How about
my $filename = (sort glob "*.gz")[0];
You state the alphabeticall order thus the sort, which by default uses the "standard string comparison order". Note that ls can be aliased, while its defaults also may depend on the system.
Going out to the shell would make sense only if you use some particular strengths of ls, that would take a lot of work to do in Perl. For mere sorting there is no reason to go out of your program. It is far less efficient and adds a whole list of new problems to solve.
A good point was raised by mob. Because of sort invocation, sort BLOCK|SUB LIST, one could wonder whether glob above could be taken in an unintended way, as a SUB. It's not, as the builtin certainly runs first. However, that's a little close and this is just clearer
my ($filename) = sort (glob "*.gz");

Answer 2

zdim's solution uses sort and glob, both of which do a lot of needless work.
The glob is something I explain in Wasting time thinking about wasted time, in which I look at some really bad benchmarks we had in Intermediate Perl.
The sort compares a bunch of files to each other, even if it already know that the two files it wants to compare cannot be the one you want. You only care about which one comes first. A linear scan does just fine (and we show an example of that in Learning Perl to find the maximum number in a list):
opendir my $dh, $dir or die "Could not open dir $dir";
my $first;
while( my $file = readdir $dh ) {
$first = $file if $file lt $first;
}
It's a bit more complicated as you probably want to filter out some files (the virtual dirs . and .., and maybe all the hidden files), but a grep handles that:
my $file = grep { ! /\A\.\.?\z/ } readdir $dh
Even better, this wouldn't even have to know how to get the next file. Some other sort of iterator would provide that without high_water knowing how it works. That's a more complicated example that I won't present here (and is one of those areas where Perl should look to Python, and that I demonstrate in Object::Iterate).
But, sometimes the easier, less pure thing is a good enough solution. Maybe not for this problem, but for some other problem. For example, if you have an external process (ls) generate all the input, all its resources can be freed once it does its job. A Perl pipe can do that:
$ perl -e 'open my $ph, q(-|), q(/bin/ls *.gz); print scalar <$ph>'
a.gz
Or, with head as well (notice the disappearing scalar) so I print all the lines instead of the first line (even though in this example there is only one line):
$ perl -e 'open my $ph, q(-|), q(/bin/ls *.gz | /usr/bin/head -n 1); print <$ph>'
a.gz
In your simple case, where you only get one line of input, the backticks can do the job:
$ perl -e 'print `/bin/ls *.gz | /usr/bin/head -n 1`'
a.gz
I don't know what you were having problems with your one liner, but that's something to consider: The shell things tend to be fragile and difficult to get right. Then, when you get it right, someone else messes it up. Or simple translations from Perl strings to the shell don't come out as expected.
I suspect that you had quoting issues, which is why I use generalized quoting, q(), inside my Perl one-liners. It gets even more fun on Windows where you can't use single ticks to quote the argument to -e, but if you use double ticks in a unix shell, you get interpolation.
Remember though, asking for an external process means you have to be really careful. I used /bin/ls to be sure I got what I wanted and not some other thing in my path (although you can also limit $ENV{PATH}). I write much more about that in Mastering Perl, although perlsec has some advice too.