Re: Works in html, not when echoed in PHP
| From: | David Robley | Date: | Wed, 13 Nov 2002 08:59:53 +0000 |
| Subject: | Re: Works in html, not when echoed in PHP | ||
| References: | 1 2 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-124184@lists.php.net to get a copy of this message | ||
In article <B9F73118.9D04%amerrick@mac.com>, amerrick@mac.com says...
> Folks,
>
> Can't see an answer anywhere in the archives, so here goes.
>
> This works fine in plain html:
>
> <body
> onload="start();initialize();onoff('mainmenu',section,'on')"
> onresize="window.location.reload(false)" topmargin="1"
> bottommargin="0"
> leftmargin="0" rightmargin="0">
>
> When I put it in PHP thus:
>
> echo "<body
> onload=\"start();initialize();onoff('mainmenu',section,'on')\"
> onresize=\"window.location.reload(false)\" topmargin=\"1\"
> bottommargin=\"0\" leftmargin=\"0\" rightmargin=\"0\">";
>
> When the page loads, I get an "Error: 'menuObj' is null or not an object"
>
> The onoff() function is what contains the menuObj, so I suspect the single
> quotes around the parameters mainmenu and on, but have tried everyway I can
> think of and can't get rid of the Error.
>
> The function is thus:
>
> function onoff (elemparent,elem,state) {
> if (loaded) {
> newstate = eval(elem+"_"+state);
> if (n4) {
> menuObj = eval (doc + elemparent + doc2 + elem);
> }
> else if (ie || n6) {
> menuObj = eval (doc + elem + doc2);
> }
>
> I would be grateful for any tips.
Not a JS expert, but: have you tried comparing the 'View source' of your
PHP script output with the expected code?
Cheers
--
David Robley
Temporary Kiwi!
Quod subigo farinam