django-dynamic-formset为每个formset创建两个添加/删除链接

出于某种原因，我对django-dynamic-formset的实现表现得有点滑稽。

它为我的模板中的每个动态formset创建了两个添加/删除链接。

我一直在试图弄清楚这一点，并且它得到了我的最好。

这是我正在谈论的错误的屏幕截图

我也可以提供登录，如果您想自己触发错误，这里是我的模板：

   $(function() { $('#ingredients tbody tr').formset({ prefix: '{{ ingredients_formset.prefix }}', formCssClass: 'dynamic-ingredients' }); $('#steps tbody tr').formset({ prefix: '{{ steps_formset.prefix }}', formCssClass: 'dynamic-steps' }); })    
 {% csrf_token %}  Ingredients:  
 
 {% csrf_token %}  Steps:  
  

这是forms.py

 class RecipeForm(forms.ModelForm): reset_recipe = forms.CharField(widget=forms.HiddenInput(), required = False) class Meta: model = Recipe widgets = {'original_cookbook':forms.HiddenInput(), 'pub_date':forms.HiddenInput(), 'author':forms.HiddenInput()} fields =("name", "picture","ingredients","steps","prep_time","type",'reset_recipe') class CookbookForm(forms.ModelForm): class Meta: model = Cookbook class StepsForm(forms.Form): Step = forms.CharField() StepsFormSet = formset_factory(RecipeForm, StepsForm, can_order=True, can_delete=True, extra = 0) class IngredientsForm(forms.Form): Ingredient = forms.CharField() IngredientsFormSet = formset_factory(RecipeForm, IngredientsForm, can_order=True, can_delete=True, extra = 0) 

和观点：

 def createrecipe(request): RecipeForm = RecipeForm(request.POST) IngredientsFormSet = formset_factory(IngredientsForm) StepsFormSet = formset_factory(StepsForm) if request.method == 'POST': form = RecipeForm(request.POST) ingredients_formset = IngredientsFormSet(request.POST, request.FILES, prefix='ifs') steps_formset = StepsFormSet(request.POST, request.FILES, prefix='sfs') if form.is_valid() and ingredients_formset.is_valid() and steps_formset.is_valid(): print "form is valid" new_recipe = form.save(commit=False) new_recipe.original_cookbook = request.user.cookbooks.all()[0] new_recipe.author = request.user.cookbooks.all()[0].name new_recipe.steps = steps_formset new_recipe.ingredients = ingredients_formset new_recipe.save() cookbooks = request.user.cookbooks.all() cookbook = cookbooks[0] cookbook.recipes.add(new_recipe) form.save_m2m() t = loader.get_template('cookbook/create_form.html') c = RequestContext(request, { 'form': new_recipe, }) data = { 'replace': True, 'form': t.render(c), 'success': True, } json = simplejson.dumps(data) return HttpResponse(json, mimetype='text/plain') else: print "form is invalid" form = RecipeForm(request.POST) ingredients_formset = IngredientsFormSet(request.POST, request.FILES, prefix='ifs') steps_formset = StepsFormSet(request.POST, request.FILES, prefix='sfs') t = loader.get_template('cookbook/create_form.html') c = RequestContext(request, { 'form':form, }) data ={ 'form': t.render(c), 'success': False, } json = simplejson.dumps(data) return HttpResponse(json, mimetype='text/plain') 

Answer 1

您可以使用ELO80KA提供的令人惊叹的django-dynamic-formset jQuery插件

我现在正在项目中使用它，让我告诉你我现在是怎么做的，让我们来看看标记：

 
 {% for form in p_i_c_formset.forms %} 
 {{ form.id }} 
year
 {{ form.year }} 
product_category
 {{ form.product_category }} 
insurance_company
 {{ form.insurance_company }} 
value
 {{ form.value|attr:"placeholder:insurance_claims" }} 
result
 {{ form.result }} 
 {% endfor %} {{ p_i_c_formset.management_form }}  

然后，在javascript中你需要做正确的选择器（你的formset的字段所在的位置），这是插件将获取代码以使用正确的id和名称克隆它的地方，不要忘记将其设置为“ formset的前缀“：

 $(function(){ $('form#my-personal-info #id_p_i_c_formset_div .p_i_c_child_formset').formset({ prefix: '{{ p_i_c_formset.prefix }}' }); }); 

该插件将自动创建删除和添加按钮。