In article <[EMAIL PROTECTED]>, [EMAIL PROTECTED] 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 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php