Re: Re: Why is this param undefined? and Why isn't exit stopping the script from running?

Thanks for the reply. I realize that

<frame src="$script_name/top" scrolling="NO" name="topFrame">
<frame src="$script_name/main" name="mainFrame">
[download]

instantiates 2 copies of the script, but I thought the if (!$content) {exit;} would kill process #1 before it had child processes. This is not the case?? Why?
TIA
jg
_____________________________________________________
Think a race on a horse on a ball with a fish! TG

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Re: Re: Re: Why is this param undefined? and Why isn't exit stopping the script from running? by Kanji (Parson) on Mar 28, 2002 at 20:18 UTC
Ah, but the frames are second and third in the chain. The first link is the frameset builder, where `!$content` evaluates as false because `$content` is set if you're using the URL you mention in your original post. Propagating the query string so that it's set in the children frames is an easy fix, though ... `sub print_frameset { $script_name = $q->script_name; $query_string = $q->query_string; print <<EOF; <html><head><title>$TITLE</title></head> <frameset rows="147," frameborder="NO" border="0" framespacing="0" co +ls=""> <frame src="$script_name/top?$query_string" scrolling="NO" name="topFr +ame"> <frame src="$script_name/main?$query_string" name="mainFrame"> </frameset> EOF ; exit 0; }` [download] --k.	[reply] [d/l]
Re: Re: Re: Re: Why is this param undefined? and Why isn't exit stopping the script from running? by jerrygarciuh (Curate) on Mar 28, 2002 at 20:27 UTC
Thanks so much for your time!!!! That solved it and it might have cost me hours before I realized what was up. Thank you, jg _____________________________________________________ Think a race on a horse on a ball with a fish! TG	[reply] [d/l]