Reply to Re: Parse error with php/java script

Your name:

Reply:


Posted by Justin Koivisto on 11/23/05 01:37

Ian Davies wrote:
> Thanks
> That helped
> But now get a syntax error on line 69
> which is
>
> echo "subselect.options[".$ctr."].value =
> \"".$Row2['TopicID']."\";\n";
>
> my debugger is saying
>
> if( mainselect.options[mainselect.selectedIndex].value == "12" ) {
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
> resource in e:\domains\i\mysite\user\htdocs\myfolder\Questions.php on line
> 67
> } }
>
> It would seem that there is a problem with
>
> mainselect.options[mainselect.selectedIndex].value

Or there's something wrong with the query. You blindly start a while
loop without first checking to see if the query has succeeded. In this
case, if $Result isn't a valid result resource, the query failed for
some reason... to help debug, use mysql_error() to see what the problem is.

--
Justin Koivisto, ZCE - justin@koivi.com
http://koivi.com

[Back to original message]


Удаленная работа для программистов  •  Как заработать на Google AdSense  •  England, UK  •  статьи на английском  •  PHP MySQL CMS Apache Oscommerce  •  Online Business Knowledge Base  •  DVD MP3 AVI MP4 players codecs conversion help
Home  •  Search  •  Site Map  •  Set as Homepage  •  Add to Favourites

Copyright © 2005-2006 Powered by Custom PHP Programming

Сайт изготовлен в Студии Валентина Петручека
изготовление и поддержка веб-сайтов, разработка программного обеспечения, поисковая оптимизация