python - 如何在 django 中更新现有记录而不是创建新记录？

Question

所以我在views.py中有以下函数:

def recipe_edit(request, pk):
    recipe = get_object_or_404(Recipe, pk=pk)
    if request.method == "POST":
        initial = {'title': recipe.title, 'description': recipe.description}
        form = RecipeForm(request.POST, initial=initial)
        if form.is_valid():
            current_user = request.user
            data = form.cleaned_data
            recipe_data=Recipe.objects.create(user=current_user, title=data['title'], description=data['description'])
            recipe_data.save( force_insert=False, force_update=False, using=None)
            return HttpResponseRedirect('recipe_detail', pk=recipe.pk)
    else:
        initial = {'title': recipe.title, 'description': recipe.description}
        form = RecipeForm(initial=initial)
    return render(request, 'recipe_edit.html', {'form': form, 'recipe':recipe})

但是当我提交表单时，它实际上创建了一条新记录，而不是编辑旧记录。有什么建议如何更新旧记录而不是创建新记录吗？

Answer 1

您应该很明显，您专门在 is_valid block 中调用 create ，因此您自然会创建一条记录。然而，与始终创建一样，通过这样做，您将绕过模型表单为您提供的所有帮助。

您应该传递实例，而不是传递initial；然后在 is_valid block 中您应该调用 form.save。

def recipe_edit(request, pk):
    recipe = get_object_or_404(Recipe, pk=pk)
    if request.method == "POST":
        form = RecipeForm(request.POST, instance=recipe)
        if form.is_valid():
            recipe = form.save(commit=False)
            recipe.user = request.user
            recipe.save()
            return redirect('recipe_detail', pk=recipe.pk)
    else:
        form = RecipeForm(instance=recipe)
    return render(request, 'recipe_edit.html', {'form': form, 'recipe':recipe})