Use scalars to define a filename

akrrs7 has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Use scalars to define a filename by toolic (Bishop) on Oct 17, 2011 at 14:11 UTC
Perl thinks you have a variable named `$var1_`. my $fileName = "fileLocn/$var1_$var2.txt"; Change that to: `my $fileName = "fileLocn/${var1}_$var2.txt";` [download] Scalar value constructors	[reply] [d/l] [select]
Re^2: Use scalars to define a filename by akrrs7 (Acolyte) on Oct 17, 2011 at 14:16 UTC
Thanks Toolic...that fixed it.	[reply]
Re: Use scalars to define a filename by Ovid (Cardinal) on Oct 17, 2011 at 14:13 UTC
There are a couple of issues here. First, when you get the "string found or bareword..." error message, it will include a line number. That line number is likely not the line you're pointing to because there are no barewords or syntax errors in the code above. You'll want to post a larger code snippet. That being said, in Perl an identifier should start with a letter or underscore and is followed by zero or more word characters. A scalar variable in Perl is defined as a dollar sign ($) followed by an identifier. When you have a this line: `my $filename = "fileLocn/$var1_$var2.txt";` [download] When you fix your other issue, you'll probably get an error message similar to: `Global symbol "$var1_" requires explicit package name at temp.pl line 10.` (Note: If you're on a version of Perl less than 5.10, you won't see the variable name) Perl thinks that `$var1_` is a variable name. You need to tell Perl that the trailing underscore (_) is not part of the variable name. You can fix this one of two ways: `# escape the underscore my $fileName = "fileLocn/$var1\_$var2.txt"; # or wrap the identifier in curly braces my $fileName = "fileLocn/${var1}_$var2.txt";` [download] Also, I suspect that you wanted `fileLocn` to begin with a dollar sign? Cheers, Ovid The official Perl 6 wiki.	[reply] [d/l] [select]
Re: Use scalars to define a filename by ikegami (Patriarch) on Oct 17, 2011 at 14:15 UTC
The code you have does not result in that error. The code you have does result in a different error under `use strict`, though. `"fileLocn/$var1_$var2.txt"` [download] is the same as `"fileLocn/" . $var1_ . $var2 . ".txt"` [download] The problem being that there is no variable `$var1_`. Fix: `"fileLocn/${var1}_$var2.txt"` [download] [ Whoa! three people posted more or less the same while I was typing this! ]	[reply] [d/l] [select]
Re: Use scalars to define a filename by Utilitarian (Vicar) on Oct 17, 2011 at 14:13 UTC
The underscore is a legitimate character in a variable name,so you'll have to create to the file in one of the following ways `my $fileName = "fileLocn/".$var1."_$var2.txt"; my $fileName = "fileLocn/${var1}_$var2.txt"; my $fileName = sprintf("fileLocn/%s_%s.txt", $var1, $var2);` [download] Or one a a myriad others `print "Good ",qw(night morning afternoon evening)[(localtime)[2]/6]," fellow monks."`	[reply] [d/l] [select]
Re: Use scalars to define a filename by Anonymous Monk on Oct 17, 2011 at 14:06 UTC
See How do I post a question effectively? , Basic debugging checklist string found or bareword found where operator expected is a syntax error	[reply]


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