What does this mean?

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

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Re: What does this mean? by btrott (Parson) on Jul 18, 2000 at 22:00 UTC
That should be `$q->param('lang')` [download] Probably. The syntax you used makes Perl think that $q is a subroutine reference that you're trying to invoke with an argument of lang. $q is not a subroutine (CODE) reference. Hence the error.	[reply] [d/l]
(zdog) RE: (2) What does this mean? by zdog (Priest) on Jul 18, 2000 at 22:05 UTC
Heh! It is funny how obvious the mistake is, but you still can't put your finger on it, thanks. Where would I be without Perl Monks? Zenon Zabinski \| zdog \| zdog7@hotmail.com	[reply]
(jcwren) RE: What does this mean? by jcwren (Prior) on Jul 18, 2000 at 22:07 UTC
I agree with btrott, and I'll add another little tidbit that caused me to waste a few hours. If you want to change the value in the parameter you're accessing, it's `$p->param ('lang', "Perl spoken here");` [download] not `$p->param ('lang') = "Perl spoken here";` [download] I beat my head against a wall for over two hours trying to track down why my script wasn't doing what I thought it should be doing. It's there when you read the documentation, but it's easy to overlook, and it's a dumb mistake. --Chris e-mail jcwren	[reply] [d/l] [select]
(chromatic) RE: (jcwren) RE: What does this mean? by chromatic (Archbishop) on Jul 18, 2000 at 22:20 UTC
I've always preferred the pointy comma operator in this case. It makes it more obvious that you're doing some sort of assignation: `$q->param('lang' => 'Perl spoken here');`	[reply] [d/l]
Re: What does this mean? by cwest (Friar) on Jul 18, 2000 at 22:09 UTC
You are trying to use `$q` as a reference to an anonymous sub. I would expect some code above that line to look something like: `my $q = sub { my $var = shift; # 'lang' in this case return 1; }` [download] You would then call the above subroutine using `$q` as the reference. `$q->('lang'); # one way &{$q}('lang'); # another way` [download] Also, you can find out what kind of a reference `$q` is with the following (after `$q` is defined and assigned) `print ref($q);` [download] I don't know what this has to do with `CGI.pm` unless you assigned `$q` as a CGI class reference, and you're using it wrong, or re-assigned it or something... Perhaps you just forgot to put the sub's name in your statement... for instance: `$q->h1('lang'); # instead of $q->('lang')` [download] -- Casey	[reply] [d/l] [select]
Re: What does this mean? by ferrency (Deacon) on Jul 18, 2000 at 22:04 UTC
It would be much easier to fix your problem with a larger code snippet. `Not a CODE reference at ... line 53, <DATA> line 1. if ($q->('lang') eq '' \|\| $q->('lang') eq $bookinfo[1]) {` [download] This looks like you're dereferencing $q and calling it as a subroutine call. What does $q contain? If it's a hash reference, then to get the 'lang' key out, you should do this instead: `$q->{'lang'}` [download] Note the curly braces, instead of parentheses. Alan	[reply] [d/l] [select]
RE: Re: What does this mean? by ferrency (Deacon) on Jul 18, 2000 at 22:14 UTC
Oops- nevermind... I missed the part about "CGI.pm".	[reply]