I may be misunderstading something here but this does not look like a
cross-domain issue to me.

If you are making AJAX request on the same server that served the
page, then you should use absolute URI path without any protocol nor
domain name : $.get("/some/web/service"). This will work whatever
domain your user typed in his browser (www.example.com or
example.com).

If you are really making a cross-domain AJAX call, meaning that you
are calling another server (like yahoo.com or google.com) then
browsers will prevent you from doing that and you should rely on a
proxy webservice that you must create on your server.

Have I missed something obvious?


On Jul 10, 4:07 am, flycast <[EMAIL PROTECTED]> wrote:
> I have determined that the error was from a different script. Sorry
> about the confusion.
>
> After working on the document.domain I could not get it to work/help.
> The load statement would just not return any result.
>
> The solution that Jeffery offered above worked like a champ.
>
> Thanks for all your help.
>
> On Jul 9, 11:56 am, "Jeffrey Kretz" <[EMAIL PROTECTED]> wrote:
>
> > I had this same problem, which I solved by parsing the window.location.href
> > prior to making the ajax call, based on the current host used to access the
> > site.
>
> > var url = /(https?:\/\/[^\/]+)/.exec(window.location.href)[1] +
> > '/pathtomyresource';
>
> > JK
>
> > -----Original Message-----
> > From: jquery-en@googlegroups.com [mailto:[EMAIL PROTECTED] On
>
> > Behalf Of Erik Beeson
> > Sent: Wednesday, July 09, 2008 9:33 AM
> > To: jquery-en@googlegroups.com
> > Subject: [jQuery] Re: Cross domain problems
>
> > The protocol *must* be the same (if the page ishttps://.../thenthe
> > ajax request must also be tohttps://.../).
>
> > The port *must* be the same.
>
> > The host name *must* have the same SLD [1], and if the subdomains are
> > different, document.domain *must* be set to the SLD.
>
> > I'm certain that this can work as I do it all the time. If you're
> > still having trouble, could you create a page that demonstrates the
> > problem?
>
> > --Erik
>
> > [1]http://en.wikipedia.org/wiki/Second-level_domain
>
> > On 7/9/08,flycast<[EMAIL PROTECTED]> wrote:
>
> > >  I tried setting document.domain = 'site.com';
> > >  It works with a domain of site.com but notwww.site.com. I now get the
> > >  following message:
>
> > >  [Exception... "'Permission denied to call method XMLHttpRequest.open'
> > >  when calling method: [nsIDOMEventListener::handleEvent]" nsresult:
> > >  "0x8057001e (NS_ERROR_XPC_JS_THREW_STRING)" location: "<unknown>"
> > >  data: no]
>
> > >  This seems to be a different problem.
>
> > >  BTW...here is a Mozilla link to the issue of cross domain and the use
> > >  of 
> > > document.domain:http://www.mozilla.org/projects/security/components/same-origin.html
>
> > >  On Jul 9, 1:30 am, Alexsandro_xpt <[EMAIL PROTECTED]> wrote:
> > >  > Well, I thought this is security browser issue.
>
> > >  > I always solve this problem this way:
> > >  > Eg.:
>
> > >  > To ajax this:http://feedproxy.feedburner.com/undergoogle
>
> > > > I create
> > this:http://blog.alexsandro.com.br/application/load/feedproxy.feedburner.c..
> > .
>
> > >  > --
> > >  > Alexsandrowww.alexsandro.com.br
>
> > >  > On 9 jul, 00:18, "Erik Beeson" <[EMAIL PROTECTED]> wrote:
>
> > >  > > Add this somewhere in your javascript:
>
> > >  > > document.domain = 'site.com';
>
> > >  > > Google document domain
>
> > >  > > --Erik
>
> > >  > > On 7/8/08,flycast<[EMAIL PROTECTED]> wrote:
>
> > >  > > >  Simple problem (I think)...
>
> > >  > > >  I am new to JS and ajax.
>
> > >  > > >  I am building an ajax capability on a clients site. I am running
> > into
> > >  > > >  cross domain problems. If I get the page using the url
> > formhttp://www.site.com
> > >  > > >  but I do a load using the url form "http://site.com"; (www vs. no
> > www
> > >  > > >  in the url) I get nothing but a js error.
>
> > >  > > >  What is the best way to handle making sure that if the person is
> > at
> > >  > > >  the site with OR without the "www" in the url that the .load will
> > >  > > >  still work?

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