- 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