$input =~ s/\$Log:/\$History:/gi;
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"perl -p -i.bak -e 's/\$Log/\$History/gi' * " at the command line only creates filename.ext.bak, nothing else.

Boris, you're right, but I only showed a snippet of my code. I had something else being sent to the output file instead of $InputArray.

Below is the full script, which works on a single file. Actually, I don't want to create a new file, I just want to modify the existing file, but this is what I know to do so far. I'm thinking there's a way to use "readdir" to provide the list of files so that the first line would read something like " Open (InputFile, $FileArray)".

Thanks a LOT!!!



open( InputFile, "FirstFile.txt" );
open( OutputFile, ">SecondFile.txt" );
@InputArrays = <InputFile>;

foreach $InputArray ( @InputArrays)
{
$InputArray =~ s/\$Log/\$History/gi;
chomp ($InputArray);
@myColumns = split(/\\/, $InputArray);
print (OutputFile "\n$InputArray");
}

close( InputFile );
close( OutputFile );

perl -p -i.bak -e 's/\$Log/\$History/gi' *
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Creates backupfiles with the extension .ext and modifies the original files where all occurances of $Log are replaced with $History. This is done for every file in this case since you choice *.

 $InputArray is a scalar not an Array.
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To replace any occurance of $Log to $History: try this but on a backup please. the first argument is the dir where the files life.

#!/usr/bin/perl
use File::Find;
my @dirs = @ARGV;
@ARGV = ();
File::Find::find(
  {
    wanted => sub { push @ARGV, $File::Find::name if -f }
  },
  @dirs
);
local $^I;
while ( defined( $_ = <ARGV> ) ) {
  s/\$Log:/\$History:/gi;
  print;
}
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Boris