Doesn't like a Scalar, as a folder path

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

I am trying to read the folder name from a text file and open the contents of the folder with the following-

my $lines; my @lines;

# read (1 line only). FOLDERZ.
open INPUT, "<Output1.txt";

@lines = <INPUT>;
close INPUT;

foreach $lines (@lines){
 print "Line: ", $lines, "\n";

        my $linesA;   
       #my @linesA = "c\:\\Other\\renamedir\\FOLDERZ";
        my @linesA = "c\:\\Other\\renamedir\\$lines";
        foreach $linesA (@linesA){
                my $directory = "$linesA";
                print ("Folder: ", $directory, "\n");
                opendir (DIR, $directory) or die $!;
                         while (my $fldr = readdir(DIR)) {
                         print "Files & Folders: ", $fldr, "\n";   }
                closedir(DIR);
          }
        }
[download]

Getting the following error-

Line: FOLDERZ

Folder: c:\Other\renamedir\FOLDERZ

Invalid argument at c:\temp\dir4E63.tmp\lineAF.pl line 22.
Press any key to continue . . .
[download]

Line 22: opendir (DIR, $directory) or die $!;

It works just fine if I replace $linesA with FOLDERZ. But doesn't like the scalar.
Please let me know.

Comment on Doesn't like a Scalar, as a folder path Select or Download Code

Replies are listed 'Best First'.
Re: Doesn't like a Scalar, as a folder path by McA (Priest) on Oct 27, 2014 at 18:47 UTC
Hi, you will probably have a CR/LF at the end of your filename. You have to do a `chomp $lines` to get rid of it. McA	[reply] [d/l]
Re^2: Doesn't like a Scalar, as a folder path by ArifS (Beadle) on Oct 27, 2014 at 18:55 UTC
good call. Totally forgot about that. Thanks.	[reply]
Re^3: Doesn't like a Scalar, as a folder path by Corion (Patriarch) on Oct 27, 2014 at 21:52 UTC
This is such a common error case that Perl even outputs a warning for it if you let it. Put `use warnings;` at the top of your program to tell Perl that you want it to tell you about common programming errors that it can recognize.	[reply] [d/l]
Re: Doesn't like a Scalar, as a folder path by Discipulus (Canon) on Oct 28, 2014 at 08:55 UTC
Why to you escape the colon in the path? it leads to a strange behaviour i cant explain: `perl -e "print qq(c\:\\other)" c:\other perl -e "print qq(c:\\other)" c:\other #Consider in Perl on win you can use the backslash too in path # perl -e "$path = qq(c:/windows); print qq(Exists and is a directory!\n +) if -e -d $path" Exists and is a directory!` [download] HtH L* UPDATE: as explained by Corion this is due to: Because \: interpolates to : in double quotes Basically, the leading backslash includes the next character as is (unless it's a special combination, like \n, \r , \a, \b , ... L* There are no rules, there are no thumbs.. Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.	[reply] [d/l]
Re^2: Doesn't like a Scalar, as a folder path by ikegami (Patriarch) on Oct 28, 2014 at 13:21 UTC
Basically, the leading backslash includes the next character as is (unless it's a special combination, like \n, \r , \a, \b , ... The leading backslash includes the next character as is if it's a non-word character.	[reply]


Your skill will accomplish what the force of many cannot
	PerlMonks