How do I read the contents of a small text file into a scalar in Perl?

问题:

I have a small text file that I'd like to read into a scalar variable exactly as it is in the file (preserving line separators and other whitespace).

The equivalent in Python would be something like

buffer = ""  try:     file = open("fileName", 'rU')     try:         buffer += file.read()     finally:         file.close() except IOError:     buffer += "The file could not be opened." 

This is for simply redisplaying the contents of the file on a web page, which is why my error message is going into my file buffer.

回答1:

From the Perl Cookbook:

my $filename = 'file.txt'; open( FILE, '<', $filename ) or die 'Could not open file:  ' . $!;  undef $/; my $whole_file = <FILE>; 

I would localize the changes though:

my $whole_file = ''; {     local $/;     $whole_file = <FILE>; } 

回答2:

As an alternative to what Alex said, you can install the File::Slurp module (cpan -i File::Slurp from the command line) and use this:

use File::Slurp; # Read data into a variable my $buffer = read_file("fileName"); # or read data into an array my @buffer = read_file("fileName"); 

Note that this dies (well... croaks, but that's just the proper way to call die from a module) on errors, so you may need to run this in an eval block to catch any errors.

回答3:

If I don't have Slurp or Perl6::Slurp near by then I normally go with....

open my $fh, '<', 'file.txt' or die $!;  my $whole_file = do { local $/; <$fh> }; 

回答4:

There is a discussion of the various ways to read a file here.

回答5:

I don't have enough reputation to comment, so I apologize for making this another post.

@ Harold Bamford: $/ should not be an obscure variable to a Perl programmer. A beginner may not know it, but he or she should learn it. The join method is a poor choice for the reasons stated in the article linked by hackingwords above. Here's the relevant quotation from the article:

That needlessly splits the input file into lines (join provides a list context to ) and then joins up those lines again. The original coder of this idiom obviously never read perlvar and learned how to use $/ to allow scalar slurping.

回答6:

You could do something like:

$data_file="somefile.txt"; open(DAT, $data_file); @file_data = <DAT>; close(DAT); 

That'll give you the file contents in an array, that you can use for whatever you want, for example, if you wanted each individual line, you could do something like:

foreach $LINE (@file_data) {     dosomethingwithline($LINE); } 

For a full usage example:

my $result; $data_file = "somefile.txt"; my $opened = open(DAT, $data_file); if (!$opened) {     $result = "Error."; } else {     @lines = <DAT>;     foreach $LINE (@lines)     {         $result .= $LINE;     }     close(DAT); } 

Then you can use $result however you need. Note: This code is untested, but it should give you an idea.

回答7:

I'd tweak draegtun's answer like this, to make it do exactly what was being asked:

my $buffer; if ( open my $fh, '<', 'fileName' ) {     $buffer = do { local $/; <$fh> };     close $fh; } else {     $buffer = 'The file could not be opened.'; } 

回答8:

Just join all lines together into a string:

open(F, $file) or die $!; my $content = join("", <F>);  close F; 

(It was previously suggested to use join "\n" but that will add extra newlines. Each line already has a newline at its end when it's read.)

文章来源: How do I read the contents of a small text file into a scalar in Perl?