hi oscar,
Thank you!!! that is awesome! i not even need to manipulate the form using
output data json!
thanks! but , $("xxx").live really is important...
Regards,
MH
On Fri, May 4, 2012 at 4:56 PM, Oscar Mederos <[email protected]> wrote:
> Hello Min,
>
> On Friday, May 4, 2012, 10:14:46 AM, you wrote:
>
> > hi oscar,
>
> > how do you make use of the particular method to be able to render
> > the form only? as you said make use of the
> > https://docs.djangoproject.com/en/1.4/ref/contrib/csrf/#ajax and
> > it able to render the form instead of page?
> > because i'm using kurtis method, if any validation error, i
> > retrieve the data from javascript and manipulate the
> > output in html only.
>
> What I usually do is the following:
> - Place the content of my form in a separate template file (eg.
> password-form.html)
> - Using jQuery, I do something like:
>
> //Override the behavior of the 'submit' event of the form.
> //Very important to use 'live' instead of 'click'. Otherwise,
> //if we change the HTML of the form, this function won't be triggered
> //next time we submit the form.
> $("#my-form").live('submit', function(e) {
> e.preventDefault();
> $.ajax({
> type: "post",
> //DRY. I suppose you already have the url where
> //you want to make the POST request in the "action"
> //tag of the <form>
> url: $("#my-form").attr("action"),
> //This automatically get all the values from the
> //inputs in the form (eg. a=1&b=2)
> data: $("#my-form").serialize(),
> dataType: "json",
> success: function(data) {
> //If there was an error...
> if (data.error == 1) {
> //All we have to do is replace the body of
> //the <form>..</form> with the new HTML
> //rendered value of the form returned from
> //the server
> $("#my-form").replaceWith(data.message);
> }
> else {
> //Do whatever you want here
> }
> }
> });
> });
>
> 'data' is what the view should return. What I usually do is the
> following:
> * If there was an error validating the form, then:
> - "data.error" will be 1
> - "data.message" will have the form rendered
> * If the form was validated without problems
> - "data.error" will be 0
> - "data.message" will have some success message (eg. "Your password
> was changed successfully).
>
> - Now, in the view... how do I return the rendered form as HTML?
>
> The "password-form.html" template should look like:
>
> <form id="my-form" action="/some/url">
> {% csrf_token %}
> ...
> </form>
>
> And the code of the view could be something like the following (of
> course, checking that the request method was POST, etc).
>
> def view(request):
> # Create the bounded form (as you usually do)
> f = MyForm(request.POST)
> if form.is_valid():
> form.save()
> d = {'error': 0, 'message': 'Some success message'}
> else:
> d = {'error': 1}
> # Here we render the entire HTML text of the form. You can pass
> # anything you want in the context...
> form_html = render_to_string('password-form.html',...,
> context_instance=RequestContext(request))
> d['message'] = form_html
> response = simplejson.dumps(d)
> return HttpResponse(response, mimetype='application/json')
>
> I wrote all of it in the editor of my email client, so if something
> does not work, just let me know.
>
> Again, very important to use the snippet provided in
> https://docs.djangoproject.com/en/1.4/ref/contrib/csrf/#ajax
>
>
> > Regards,
> > MH
>
>
> --
> Oscar Mederos
> [email protected]
>
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