Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hello, I have a directory with multiple files. I'd like to create a new output file for each input file. I'm sure there is a good example somewhere here, but I haven't been able to find it through the search functions. Can someone point me to one? -William
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Re: How to create an output file for each input file
by apl (Monsignor) on Apr 02, 2008 at 20:17 UTC
    What exactly do you mean when you say you want to create a new output file for each input file? Will the output file contain selected lines from the input file? How will the output file be named, viz a viz the input file? Which search functions did you look at?

    And most importantly, what have you written to try to implement a solution to your question?

[reply]
      I'm trying read each file in a directory, change some of the text, then write it to a new file in another directory.

      Here's what I've written: 

      #get the names of all .tab files in inbox directory
opendir(DIR,$inbox);
@files = grep(/\.TAB$/,readdir(DIR));
closedir(DIR);


## loop through the TAB files###
foreach $file (@files) {
chomp;
        
$verifile = $outbox.$yr.$mn.$day.$hr.$min.$sec."_FIXED.TAB";
open(OUTPUT_FILE, ">>$verifile")||die "Can't open file: $!";
#string manipulation will go here
close OUTPUT_FILE;
        
}# end for each file

      [download]

      Edit: g0n - code tags

[reply]
[d/l]
        A few more questions:
        • Do the existing files end .TAB or .tab?
        • You will need to define $inbox and $outbox.
        • You will need to add something to the value of $verifile to reflect $file. (That is, /old/WARRANT.TAB should result in /new/WARRANT_080402_164500_FIXED.TAB which requires some separators as well)
[reply]

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