1π
It is necessary to return HttpResponseRedirect after dealing with POST data, the tip is not specific to Django, itβs a good web practice in general.
Also, try to maintain both GET and POST request separately, so try below view:
def recipe_create_view(request):
context = {}
form="" # for the error of variable refrenced before assignment.
IngredientFormset=""
formset=""
if request.method == 'POST':
form = RecipeForm(request.POST)
IngredientFormset = formset_factory(IngredientForm)
formset = IngredientFormset(request.POST)
if form.is_valid():
if formset.is_valid():
form.save()
print("made a recipe")
for form in formset:
child = form.save(commit=False)
child.recipe = parent
child.save()
print("made a Ingredient")
return redirect('some_success_path_name')
else:
print("formset is not valid")
else:
print("form is not valid")
else: # GET method
print("request method is GET")
form = RecipeForm()
IngredientFormset = formset_factory(IngredientForm)
formset = IngredientFormset()
context['form'] = form
context['formset'] = formset
return render(request, 'recipes/create_recipe.html', context)
π€Sunderam Dubey
0π
add action
in your HTML form and POST
in small case.
<form action="/your_backend_url_to_view/" method="post">
{% csrf_token %}
<label>recipe</label>
<p>{{form}}</p>
<label>ingredients</label>
{% for form in formset %}
<ul>
<label>name</label>
<li>{{ form.name }}</li>
<label>quantity</label>
<li>{{ form.quantity }}</li>
</ul>
{% endfor %}
<div>
<input type="submit" value="submit" class="button-33" role="button">
</div>
</form>
π€shivankgtm
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