Hi Richard.
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;
>
> }
I don't think there's a better way, but it's a lot neater like this:
#!/usr/local/bin/perl -i.bak
use strict;
use warnings;
@ARGV = ( 'myfiles' );
my @files = <>;
chomp @files;
@ARGV = @files;
while (<>) {
s(/u01/app/webMethodsFCS)(/u02/app/webMethodsFCSclone)g;
print;
}
__END__
> ---------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.
@ARGV contains the (globbed, if you're on Unix) list of files on the
command line. You can alter @ARGV to pretend that something was
there.
Can't remember my Unix too well, but doesn't
script @myfiles
do what you want?
Cheers,
Rob
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