Re: Creating a javascript array from database data (Off-List)

Thanks for the clarification Sheridan - I think it must have been end of day madness on my part :) I see what's going on now with the 'print' line. Cheers, Neil Sheridan Saint-Michel wrote: > *************************************************************** > This message was virus checked with: SAVI 3.48 > last updated 14th August 2001 > *************************************************************** > > Think in terms of output... The JavaScript is literally going to > try and run the output of the PHP (It doesn't even know PHP > exists). PHP is outputting the source code for the HTML and > JavaScript. So what you are trying to do is get PHP to take > dog,cat,cow from MySQL and output: > > <Script Language="JavaScript"> > text = new Array("dog","cat","cow") > </Script> > > This block of code does exactly that, with the print"\"$text\""; > filling in dog cat and cow. I've added comments so you can > see how each line produces the above (I am using C-Style > comments as the // gets mucked up in e-mail) > > The parts in [] are comments within comments to show flow > control and are not output. > > else > { > /* <Script Language="JavaScript"> */ > echo "<Script Language=\"JavaScript\">\n"; > /* text=new Array( */ > echo "text = new Array("; > while ($i < $number) > > $text = mysql_result($result, $i, "name"); > /* "dog" [first time] */ > /* "cat" [second time] */ > /* "cow" [third time] */ > print "\"$text\""; > $i++; > if ($i < $number) > /* , [first time] */ > /* , [second time] */ > print ","; > else > /* ) [third time] */ > print ")\n"; > } > /* </Script> */ > echo "</Script>\n"; > } > > So if you can follow that (just walk all the way through > the loop for each of the three passes) you can see that > no part of the code (including the print "\"$text\"";) is > unnecessary > > Hope this helps > > Sheridan Saint-Michel > Website Administrator > FoxJet, an ITW Company > www.foxjet.com > > ----- Original Message ----- > From: Neil Freeman <neil.freeman@asa.co.uk> > To: Sheridan Saint-Michel <webmaster@foxjet.com> > Cc: php-general <php-general@lists.php.net> > Sent: Wednesday, August 15, 2001 12:09 PM > Subject: Re: [PHP] Creating a javascript array from database data > > > Thanks a lot for your help Sheridan, > > > > This now appears to work ok :) but if I remove the: > > > > print "\"$text\""; > > > > line (as it is not required), I receive errors complaining about a syntax > > error (related to what?) and that 'text' is undefined. As this line simply > > outputs the value to screen why should this cause a problem if it is > removed? > > Or could it be some other problem? > > > > Neil > > > > PS: Here is my updated code: > <Snip> -- -------------------------------- Email: neil.freeman@asa.co.uk neil_freeman@yahoo.com --------------------------------