providing filenames dynamically

manish.rathi has asked for the wisdom of the Perl Monks concerning the following question:

I want to enter file names to be worked with dynamically i.e I dont want to provide file names at the begnning of the program. I have written following code, but its not working properly.

print "file1 ="; 
$file1=<stdin>;
print "file2 =" ;
$file2=<stdin>;
@ARGV = {"$file1", "$file2"} ;
chomp($remove = <stdin>);
chomp($replace = <stdin>);
while(defined($line=<>)){
 $line =~ s/$remove/$replace/g;
 print "$line"; }
[download]

Pls let me know whats wrong.

Comment on providing filenames dynamically Download Code

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Re: providing filenames dynamically by ysth (Canon) on Feb 24, 2009 at 05:33 UTC
Change {} to () in the assignment to @ARGV. As is, you are assigning an anonymous hashref, which will stringify to something that is almost certainly not an existing file :) -- Online Fortune Cookie Search Office Space merchandise	[reply]
Re: providing filenames dynamically by balakrishnan (Monk) on Feb 24, 2009 at 06:01 UTC
You have to change your code like, `print "file1 ="; $file1=<stdin>; print "file2 =" ; $file2=<stdin>; @ARGV = ("$file1", "$file2") ; chomp($remove = <stdin>); chomp($replace = <stdin>); while(defined($line=<>)){ $line =~ s/$remove/$replace/g; print "$line"; }` [download] When your code seems not giving the expected result, you can use the perl debugger to find out issue. I debugged the code, after made the change from {} to () `main::(./dynamic_file.pl:22): print "file1 ="; DB<1> n main::(./dynamic_file.pl:23): $file1=<stdin>; DB<1> n file1 =file1 main::(./dynamic_file.pl:24): print "file2 =" ; DB<1> n main::(./dynamic_file.pl:25): $file2=<stdin>; DB<1> n file2 =file2 main::(./dynamic_file.pl:26): @ARGV = ("$file1", "$file2") ; DB<1> n main::(./dynamic_file.pl:27): chomp($remove = <stdin>); DB<1> p @ARGV file1 file2` [download]	[reply] [d/l] [select]
Re: providing filenames dynamically by PoorLuzer (Beadle) on Feb 24, 2009 at 05:30 UTC
chomp on filename input from stdin!	[reply]
Re: providing filenames dynamically by Anonymous Monk on Feb 24, 2009 at 08:49 UTC
Have you read perlintro?	[reply]
Re: providing filenames dynamically by apl (Monsignor) on Feb 24, 2009 at 13:23 UTC
I suggest you read Basic debugging checklist.	[reply]
Re: providing filenames dynamically by Marshall (Canon) on Feb 24, 2009 at 15:26 UTC
I think what you want is to process some file and replace some regex with another regex? If so, then consider this code. `die "Usage $0 remove_regex replace_regex" if (@ARGV !=2); my ($remove, $replace) = @ARGV; while(<stdin>) { s/$remove/$replace/g; print; }` [download] @ARGV should be considered a "read only" variable like in other languages. Here use like: replace.pl a b <infile >outfile If you want to pass filenames on command line, then same idea with getting values of @ARGV, but you will have to use a open() statement to read or write them.	[reply] [d/l]