Read file into variable in Perl [duplicate]

问题

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Closed 9 years ago.

Possible Duplicate:
What is the best way to slurp a file into a string in Perl?

Is this code a good way to read the contents of a file into a variable in Perl? It works, but I'm curious if there is a better practice I should be using.

open INPUT, "input.txt";
undef $/;
$content = <INPUT>;
close INPUT;
$/ = "\n";

回答1:

I think common practice is something like this:

    my $content;
    open(my $fh, '<', $filename) or die "cannot open file $filename";
    {
        local $/;
        $content = <$fh>;
    }
    close($fh);

Using 3 argument open is safer. Using file handle as variable is how it should be used in modern Perl and using local $/ restores initial value of $/ on block end, instead of your hardcoded \n.

回答2:

use File::Slurp;
my $content = read_file( 'input.txt' ) ;

回答3:

Note that if you're in an environment where installing modules is possible, you may want to use IO::All:

use IO::All;
my $contents;
io('file.txt') > $contents;

Some of the possibilities get a bit crazy, but they can also be quite useful.

来源：https://stackoverflow.com/questions/4087743/read-file-into-variable-in-perl