CODEWITHSUNDEEP
pythondjango
Tikue Anazodo
Tikue Anazodo

Reputation: 329

How to show whether user already liked object instance in Django

I have the following table in my Django model, Dishes and Likes. On my home page I am showing a list of all the dishes in the database, and I have a like button on each dish. For dishes that the user has liked I want to indicate that they already liked it, so that they can unlike it and vice versa. I have been trying different approaches for the last few days but can't seem to figure anything out. Here is the code for my latest unsuccessful attempt. 

#dishes table
class Dishes(models.Model):
    name = models.CharField(max_length=40, unique=True)

    def liked(dish, user):
        try:
            user_upvoted = Likes.objects.get(dish=dish, user=user)
        except:
            user_upvoted = None
        if user_upvoted:
            return True
        else:
            return False


#upvotes
class Likes(models.Model):
    dish = models.ForeignKey(Dishes)
    user = models.ForeignKey(User)
    date_added = models.DateTimeField(auto_now_add=True)

def home(request):
    this_user = auth.models.User.objects.get(id=1)

    dishes = models.Dishes.objects.all()
    for dish in dishes:
         models.Dishes.voted(dish, this_user)

    `enter code here`return render_to_response('frontend/home.html', { 'dishes': dishes, })

Upvotes: 4

Views: 1082

Answers (2)

Burhan Khalid
Burhan Khalid

Reputation: 174622

Adding a ManyToMany makes this an easy problem to solve:

class Dishes(models.Model):
    name = models.CharField(max_length=40, unique=True)
    likes = models.ManyToManyField(Likes)

class Likes(models.Model):
    dish = models.ForeignKey(Dishes)
    user = models.ForeignKey(User)
    date_added = models.DateTimeField(auto_now_add=True)

Adjust your view like this:

from django.contrib.auth.decorators import login_required
from django.shortcuts import render

@login_required
def home(request):
    dishes = Dishes.objects.all()
    return_list = []
    for dish in dishes:
        return_list.append((dish, dish.likes_set.filter(user=request.user)))
    return render(request, 'dish_list.html', {'dishes': return_list})

Your template is where you do the "toggle":

 {% for dish, liked in dishes %}
     {{ dish.name }}
     {% if liked %}
          You already like this dish.
     {% else %}
          Like this dish now, its yummy!
     {% endif %}
  {% endfor %}

Or, if you can't change your model, adjust your view code like this:

@login_required
def home(request):
    dishes = Dishes.objects.all()
    return_list = []
    for dish in dishes:
        return_list.append((dish,
                            Likes.objects.filter(user=request.user, dish=dish)))
    return render(request, 'dish_list.html', {'dishes': return_list})

The idea is that the list of objects you return to the template is already flagged for the user that is logged in.

The login_required decorator makes sure that the view is only called when a user is logged in. Otherwise, it will redirect the user to the login page.

The render shortcut will make sure that RequestContext is always passed from your views.

Upvotes: 2

PepperoniPizza
PepperoniPizza

Reputation: 9112

You need a new model and assuming you are using django User model, you could have something like this:

models.py

from django.contrib.auth.models import User

class DishLiked(models.Model):

    dish = models.ForeignKey(Dish,
        related_name='liked_by'
    )

    like = models.ForeignKey(Like,
        related_name='like_by'
    )

    user = models.ForeignKey(User,
        related_name='liked_dishes'
    )

views.py

Note: also note they way of getting user from request:

def home(request):
    user = request.user
    liked = DishLiked.objects.filter(
        user__id=user.id
    )
    dish_id_list = []
    for liked_dish on liked:
        dish_id_list.append(liked_dish.dish.pk)

    dishes_liked_by_user = Dish.objects.filter(id__in=dish_id_list)

Upvotes: 0

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