Re: Works in html, not when echoed in PHP

I'm not constructing it with, it is just included in the file. Does that clarify? Or did I misunderstand your point? On 11/13/02 1:47 AM, "Ernest E Vogelsinger" <ernest@vogelsinger.at> wrote: > You didn't show how you construct the JavaScript string in PHP - I'm sure > the problem is with quoting in there. > > Have a look at your output in the browser (do a "View Source" or whatever > your Browser calls it), and check the JS your PHP application constructed. > You'll spot the error... > > At 05:28 13.11.2002, Aaron Merrick said: > --------------------[snip]-------------------- >> 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. >> >> Thanks, >> Aaron >> >> >> -- >> PHP General Mailing List (http://www.php.net/) >> To unsubscribe, visit: http://www.php.net/unsub.php > --------------------[snip]--------------------