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 omede...@gmail.com -- You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com. To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/django-users?hl=en.
