- From: Larry Martell <larry.martell@gmail.com>
- Date: Wed, 8 Jan 2014 17:00:52 -0700
- To: www-dom@w3.org
I hope this is the correct forum for my question. If not I apologize and I ask that someone direct me to a more appropriate forum. (I have posted to the PHP list and to SO, but have not gotten a working solution from either.) I have a web page like this: <html> <head> <title>Main </title> </head> <frameset frameborder=1 border=1 rows="140,*"> <frame src="mainheader.php> <frameset frameborder=1 border=1 cols='75%,25%'> <frame src="main.php" name="main"> <frame src="selection.php" name="selection"> </frameset> </frameset> </html> >From the selection frame I send an ajax request. The response contains a complete new page and I want to replace the existing page with the response. If I do: document.open(); document.write(response); document.close(); or: var newDoc = document.open("text/html", "replace"); newDoc.write(response); newDoc.close(); The new content is rendered within the frame the request originated from. I then tried: top.document.body.innerHTML = response; and parent.document.body.innerHTML = response; In both case the entire doc was replaced, but only the top frame was rendered. If I look at the page source everything is there, and if I just hit return in the address bar all the frames are rendered. What would prevent it from rendering all the frames when I do the assignment? Thanks! -larry
Received on Thursday, 9 January 2014 00:01:20 UTC