我自己喜歡的按鈕：Django + Ajax - 如何？

Question

所以我一直有麻煩將這一觀點爲Ajax調用：

def company_single(request, slug): 
    company = get_object_or_404(CompanyProfile, slug=slug) 
    company_list = CompanyProfile.objects.get(slug=slug) 

    try: 
     tcompany = CompanyLikes.objects.get(company=company_list) 
     total_likes = tcompany.likes 
     user_liked = CompanyLikes.objects.get(user=request.user) 
    except: 
     total_likes = 0 


    instance, created = CompanyLikes.objects.get_or_create(company=company_list) 
    likes_form = CompanyLikesForm(request.POST or None, instance=instance) 

    if likes_form.is_valid(): 
     this = likes_form.save(commit=False) 
     try:  
      if user_liked: 
       this.likes -=1 
       this.user.remove(request.user) 
     except: 
      this.user.add(request.user) 
      this.likes += 1 
     this.save() 

    return render_to_response('company.html', locals(), context_instance=RequestContext(request)) 

我想我需要jQuery和JSON，但我不知道如何實現它在這裏做我自己「喜歡按鈕「爲我的網站。任何想法/建議？

Answer 1

我會給你一個例子。你只是從中學習並相應地做出改變。

myapp.models.py（簡化公司模式）：

from django.db import models 
from django.contrib.auth.models import User 
from django.template.defaultfilters import slugify 


class Company(models.Model): 
    name = models.CharField(max_length=255) 
    slug = models.SlugField() 
    likes = models.ManyToManyField(User, related_name='likes') 

    @property 
    def total_likes(self): 
     """ 
     Likes for the company 
     :return: Integer: Likes for the company 
     """ 
     return self.likes.count() 

    def save(self, *args, **kwargs): 
     self.slug = slugify(self.name) 
     super(Company, self).save(*args, **kwargs) 

myapp.urls.py（URL爲一個視圖）：

url(r'^like/$', 'myapp.views.like', name='like'), 

myapp.views。 py（查看）：

from django.http import HttpResponse 
try: 
    from django.utils import simplejson as json 
except ImportError: 
    import json 
from django.shortcuts import get_object_or_404 
from django.contrib.auth.decorators import login_required 
from django.views.decorators.http import require_POST 

from myapp.models import Company 


@login_required 
@require_POST 
def like(request): 
    if request.method == 'POST': 
     user = request.user 
     slug = request.POST.get('slug', None) 
     company = get_object_or_404(Company, slug=slug) 

     if company.likes.filter(id=user.id).exists(): 
      # user has already liked this company 
      # remove like/user 
      company.likes.remove(user) 
      message = 'You disliked this' 
     else: 
      # add a new like for a company 
      company.likes.add(user) 
      message = 'You liked this' 

    ctx = {'likes_count': company.total_likes, 'message': message} 
    # use mimetype instead of content_type if django < 5 
    return HttpResponse(json.dumps(ctx), content_type='application/json') 

模板：

• 如果Django < 1.3使用url標籤不帶引號周圍的網址名稱這樣{% url like %}
• 如果Django > 1.3 and < 1.5就應該添加：

  <input type="button" id="like" name="{{ company_slug }}" value="Like" /> 
  
  <script> 
  $('#like').click(function(){ 
       $.ajax({ 
         type: "POST", 
         url: "{% url 'like' %}", 
         data: {'slug': $(this).attr('name'), 'csrfmiddlewaretoken': '{{ csrf_token }}'}, 
         dataType: "json", 
         success: function(response) { 
           alert(response.message); 
           alert('Company likes count is now ' + response.likes_count); 
         }, 
         error: function(rs, e) { 
           alert(rs.responseText); 
         } 
        }); 
      }) 
  </script> 
  

  使用url標籤模板一些指令{% load url from future %} at top level of your template and enclosed your URL name with quotes as I have done in my answer

• If Django >= 1.5 then simply remove {% load url from future %}和封閉的網址名稱以作爲{% load url from future %}被標記爲被棄用，並將在Django 1.9

Answer 2

寫這這裏，因爲我沒有足夠的信譽評論和編輯都必須至少6個字符被刪除引號。在Django的新版本中，您需要將路徑傳遞給視圖函數或將url的名稱作爲字符串傳遞給url模板標記。因此上面的模板第7行是：

url: "{% url 'like' %}", 

這裏是documentation部分實現此了。