Replace this.url with the url of your form. this.url will actually only work if you are using a link.

Yes, the form should be submit without refreshing the page.

Let me know how it goes!

I’m trying to follow your example and I achieved to display the form via AJAX. But then the problem comes when hitting the “Submit” button. It jumps to the ‘url’ that I indicated in the ajax call.

In your code you have this.href ¿What exactly is that url?

What happens in your case when you hit “submit”.. stays in the same webpage (as Facebook comments, or jumps/refresh the page?)

I know that this post is 1 year ago and probably you don’t remember anymore about it, but it will be a great help. Thanks

Sorry I took a hiatus from my blog for a while. That block of code will return the content type as json for your jquery to read

It’ll return the var $html. Actually that might be a bit confusing since the $html is actually the $form.

If its after the return in your form function, it will not be hit, and the function will return normally.

Thanks for the post, I was wondering if you could explain this statement in more detail?

“Add this code just before the _page exits (but not after the exit(); !).”

I’m struggling with where I should where to place that code so only the json will be returned. Thanks!