Replacing with a variable in while loop

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

Okay, I'm adding to a fairly large script that is already in use at my company. What I want to do is a "one liner", here.

while($var = <VAR>){
 chomp($_);
 #change token in url query string to $var
 $urlChop[1] =~ s/MATCH/$_/;
[download]

The problem I'm having (well, besides being new to Perl.) Is it only updates to the first instance of $var in the loop. Basically it is setting a POST parameter in the URL to something from a text file. ie:

http://someurl.com/some.php?id=1337&product=$var
[download]

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Re: Replacing with a variable in while loop by wind (Priest) on May 01, 2011 at 00:56 UTC
Do you just need global replacement? `$urlChop[1] =~ s/MATCH/$_/g;` [download]	[reply] [d/l]
Re: Replacing with a variable in while loop by cdarke (Prior) on May 01, 2011 at 06:41 UTC
You don't show any code where $var is used, other than when reading a record from the file. The chomp and the s are using $_, not $var. Is that correct? Maybe you could show the rest of the loop, or at least where you use $var.	[reply]
Re^2: Replacing with a variable in while loop by tekio (Novice) on May 02, 2011 at 00:43 UTC
I've tried using both $var and $_, as $_ should represent $var in this example. My question pertains to this example, guess I should have mentioned the first line in the example: `open(VAR, "<theinputfile.txt");` [download] The entire script is well over 500 lines, and the mentioned snippet is an example of where I'm having problems. As well as I could probably loose my job for posting company source code on the Interwebz. I was just wondering why $var (or $_ which is equal to $var in this scenario) is not updating with each iteration of the while loop. I've done similar things with both Pascal and PHP. It just bewilders me why, in the given example, the variable will not update with each iteration of the loop. Obviously, it is because it is being used as a substitution. I was really just trying to understand the logic behind why. I've figured out a solution to the problem, using another method. But I'd still like to know the logic behind, the why.	[reply] [d/l]
Re: Replacing with a variable in while loop by Mr. Muskrat (Canon) on May 02, 2011 at 20:38 UTC
As the others have already said, your code doesn't work because you aren't reading anything into `$_`. (You're reading into `$var` but using `$_` later.) Try this instead: `open(my $fh, '<', 'theinputfile.txt') or die "Unable to open theinputf +ile.txt, $!\n"; while(my $var = <$fh>) { # Reading each line one by one from theinputf +ile.txt into $var chomp($var); # chomp $var not $_ since $_ isn't used #change token in url query string to $var $urlChop[1] =~ s/MATCH/$var/; # replace with $var not $_ since $_ isn +'t used # ... whatever else you need to do }` [download]	[reply] [d/l]