While is not working when using variable

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

I'm trying to run a while loop, after reading from a file. When I use variable, the while loop is not working. However when using absolute path, it is working fine. Please let me know of a way in which I can use the variable.

#    my $filename='$mydir\file_detail.txt';
#    open (FILE, '$filename');
    open (FILE, 'D:\SiebelAdmin\CSS\file_detail.txt');
    while (<FILE>)
        {}
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Re: While is not working when using variable by Anonymous Monk on Aug 31, 2015 at 08:32 UTC
Add `use autodie;` to the top of your program, it will generate a message explaining what happened	[reply]
Re^2: While is not working when using variable by parthodas (Acolyte) on Aug 31, 2015 at 08:47 UTC
On using autodie, it is showing the error message that no such file or directory. But when harcoding the same path, the script is running. Can't open('FILE', '$filename'): No such file or directory at sblmaint.pl	[reply]
Re^3: While is not working when using variable by Eily (Monsignor) on Aug 31, 2015 at 08:57 UTC
Try this: `print '$filename', "\n"; print "$filename", "\n";` [download] And for more information, you can read this tutorial on quotes in perl.	[reply] [d/l]
Re^4: While is not working when using variable by parthodas (Acolyte) on Aug 31, 2015 at 09:08 UTC
Re^3: While is not working when using variable by soonix (Chancellor) on Aug 31, 2015 at 14:46 UTC
In addition to Eilys hint, try `print $filename, "\n";` [download] Note that, if you have your complete filename in a variable, like in your OP (and as you should, as it eases debugging and error messages), quoting it in the open statement is superfluous. Besides, most Windows functions and many programs accept forward slashes, so you can write `my $filename="$mydir/file_detail.txt"; open (FILE, "<", $filename);` [download] Update: corrected quotes - d'oh! Thanks AnomalousMonk!	[reply] [d/l] [select]
Re: While is not working when using variable by locked_user sundialsvc4 (Abbot) on Aug 31, 2015 at 12:27 UTC
When you use single quotes to enclose a string, as you do in the now commented-out first line, symbols such as `$mydir` are not “interpolated.” They are not seen as variable-names, but simply literal characters. To get the behavior that you want, enclose the string in double quotes. (Yes, some other languages treat single vs. double quotes, with respect to interpolation, in precisely the opposite way.)