print sprintf("file= %30s", $name), sprintf(" ext= %10s", $type),"\n";Personally, I'd use:
printf "file = %30s ext= %10s\n", $name, $type;
my($filename,$ext) = split('\..*',$file);While this is valid syntax, using a string as split's first argument might be confusing to beginners. Every string that is not a single space (\x20) is interpreted as a regex. Using slashes or another m// makes your intention clear.
my ($filename, $ext) = split /\..*/, $file;
Splitting on /\..*/ would return ('foo', undef) for 'foo.bar'.
Splitting on /\./ would probably fix this, but you don't want ('foo', 'bar', 'baz') or (using a limit) ('foo', 'bar.baz').
So using a regex without split would probably be best:
(The first .+ will grab as much as it can, because it is greedy. The /s was added just in case someone has a linefeed in his filename, the anchors are there just to clarify the code, they don't serve a real. I used .+ for dotfiles (filenames beginning with a dot are hidden files in *nix). The extention part is optional ( (?:)? ) because not all files have an extention.)my ($filename, $ext) = $file =~ /^(.+)(?:\.(.*))?$/s
In reply to Re: Re: Extract string from rear of string
by Juerd
in thread Extract string from rear of string
by Anonymous Monk
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