Re: Absolute beginner question -- recipes

You could do something like this:

Instead of 

recipe['steps'] = Steps.objects.filter(recipe_id=pk)

Do this:

steps = Steps.objects.filter(recipe_id=pk)
context = {
        'recipe': recipe,

    }

On Monday, February 29, 2016 at 1:00:25 AM UTC+5:30, Simon Gunacker wrote:

Dear Django Community,

as a first django project I tried to write a site to manage cooking recipes. My model is quite simple:

* 1 recipe has n steps
* 1 step has a description and n ingredients, each of a certain amount defined by a certain unit

all in all, I have 5 tables (Recipe, Step, Ingredient, Unit, StepIngredient).

After reading through some documentation and trying around a little bit, I got my model and my admin site (I had to install django-nested-inline for the admin site to fit my expectations).

Now I am struggeling with my views. There is a simple index view wich basically lists all recipes:

class IndexView(generic.ListView):
    context_object_name = 'recipes_list'

    def get_queryset(self):
        return Recipe.objects.all()

The hard part is setting up the details view. It is supposed to show the recipes name as a title and then list all the required ingredients and all the required steps. It tried something like:

def detail(request, pk):
    recipe = Recipe.objects.get(id=pk)
    recipe['steps'] = Steps.objects.filter(recipe_id=pk)
    template = loader.get_template('recipe/recipe_detail.html')
    context = {
        'recipe': recipe,
    }
    return HttpResponse(template.render(context, request))

but it seems I am not allowed to modify the recipes object. Any ideas how to pass all the data which belongs to recipe to the template?

Thank you,

Simon

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