如何使用Django,Ajax,jQuery提交表单而不刷新页面?
我是django的新手。 我需要简单的例子。 如何使用Django,Ajax,jQuery提交表单(post)而不刷新页面?
这是我的表单,视图和模板:
views.py
from django.shortcuts import * from django.template import RequestContext from linki.forms import * def advert(request): if request.method == "POST": form = AdvertForm(request.POST) if(form.is_valid()): print(request.POST['title']) message = request.POST['title'] else: message = 'something wrong!' return render_to_response('contact/advert.html', {'message':message}, context_instance=RequestContext(request)) else: return render_to_response('contact/advert.html', {'form':AdvertForm()}, context_instance=RequestContext(request))
forms.py(使用“ModelForm”表单)
from django import forms from django.forms import ModelForm from linki.models import Advert class AdvertForm(ModelForm): class Meta: model = Advert
模板(formsHTML代码)
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{% if message %} {{ message }} {% endif %}
如果你打算用jquery使用ajax提交,你不应该从你的视图中返回html ..我建议你这样做:
HTML:
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js
$('#form').submit(function(e){ $.post('/url/', $(this).serialize(), function(data){ ... $('.message').html(data.message); // of course you can do something more fancy with your respone }); e.preventDefault(); });
views.py
import json from django.shortcuts import * from django.template import RequestContext from linki.forms import * def advert(request): if request.method == "POST": form = AdvertForm(request.POST) message = 'something wrong!' if(form.is_valid()): print(request.POST['title']) message = request.POST['title'] return HttpResponse(json.dumps({'message': message})) return render_to_response('contact/advert.html', {'form':AdvertForm()}, RequestContext(request))
这样你就可以将响应放在message
div中。 而不是返回普通的HTML,你应该返回json。
$('#form-id').submit(function(e){ $.post('your/url', $(this).serialize(), function(e){ ... }); e.preventDefault(); });