Mu... I have a quicker way, try this.
cat FileContainFileNames | xargs -n1 perl -pe "s/oldstring/newstring/g"
-i
Tor.
Richard Fernandez wrote:
>
> I just had a situation where I needed to replace one string with another
> string in 200 files.
> This is what I came up with, but I know there has to be a better way. Below
> is my code.
>
> "myfiles" contains a list of the files I need to scrub, one per line.
>
> -------8<-----------------8<-----------------------
> #!/usr/local/bin/perl -w
> use strict;
> $|++;
>
> my @files = `cat myfiles` or die;
> for (@files) {
>
> chomp;
> push @ARGV, $_;
> }
>
> $^I = ".bak"; # Got this from a previous message; thanks Peter!
> while (<>) {
>
> s#/u01/app/webMethodsFCS#/u02/app/webMethodsFCSclone#g;
> print;
>
> }
> ---------8<------------------8<-----------------------
>
> Seems to me there should be a way to provide the filenames on the command
> line
> w/o having to read the list into an array first, but I tried using xargs
> (this is unix) and a couple
> of other things but couldn't figure it out.
>
> Thanks for the help!
>
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