Re: Absolute beginner question -- recipes

I might be wrong here, but this is what I understand:

Let's say you have something like this:

class Recipe(models.Model):
    name = models.CharField(max_length=255)
    description = models.TextField()


class Step(models.Model):
    step_number = models.IntegerField()
    description = models.TextField()
    recipe = models.ForeignKey(Recipe)

now, if I do the following in the shell:

recipe = Recipe.objects.get(id=1)
recipe.steps

wouldn't this give an error?

To get the step, I could do something like

recipe.step_set

but I don't think I'll be able to do recipe.steps



On Monday, February 29, 2016 at 8:08:35 AM UTC+5:30, James Schneider wrote:

On Sun, Feb 28, 2016 at 11:14 AM, Simon Gunacker <simon.g...@gmail.com> 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()

You can actually get rid of the get_queryset() override entirely here by properly setting the default view model to Recipe.

 

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?

Whoa...why are you manually setting recipe['steps']? Isn't there already a relationship between Recipe (the model) and Step (Step should have a FK back to Recipe)? You should not need an explicit second call to Step.objects. If that call is necessary, then I would be highly suspicious that your models are not defined properly (sanely).

I would remove that second line where you set recipe['steps'] entirely. If that breaks something, your models are probably incorrect.

Assuming you run just the first line (recipe = Recipe.objects.get(id=pk)), you should be able to do the following in the Django shell:

recipe.steps # return a list of all of the steps associated with that recipe

recipe.steps[2].ingredients # return a list of ingredients for what is presumably step 3 of the recipe

There shouldn't be any need for manually querying for the steps or their ingredients after the original Recipe.objects.get() call. 

-James

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