Re: Works in html, not when echoed in PHP

From: Date: Wed, 13 Nov 2002 07:47:56 +0000
Subject: Re: Works in html, not when echoed in PHP
References: 1  Groups: php.general 
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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]-------------------- -- >O Ernest E. Vogelsinger (\) ICQ #13394035 ^ http://www.vogelsinger.at/

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