CODEWITHSUNDEEP
pythondjangodjango-rest-framework
D_P
D_P

Reputation: 862

How to return queryset instead of list?

models

class Permission(models.Model):
    name = models.CharField(max_length=250, unique=True)  

class PermissionDetail():
    perm = models.ForeignKey(
        Permission, on_delete=models.CASCADE, related_name='perm_details')
    group = models.ForeignKey(
        'Group', blank=True, null=True, related_name='group_perms', on_delete=models.CASCADE)

class Group():
    name = models.CharField(max_length=250)
    users = models.ManyToManyField(User, blank=True, related_name='groups')
    permissions = models.ManyToManyField(Permission, blank=True, related_name='permissions')

what I did:

 user_groups = user.groups.all()
 perm_details = []
 for group in user_groups:
     perm_details.append(group.group_perms.filter(perm=perm))
 return perm_details

This returns the correct list of data but I want queryset instead of list.

How can I return the queryset instead of making list here ?

I tried like this but Is's not right.

PermissionDetail.objects.filter(perm=perm, groups__pk__in=user_groups)

Upvotes: 0

Views: 407

Answers (2)

Astros
Astros

Reputation: 177

The list you made perm_details contains the query that you appended; which means you could access anything inside the model Group if you iterate over that list.

But you don't need to create a list, you could simply add the user to your filter

Upvotes: 0

Iain Shelvington
Iain Shelvington

Reputation: 32294

You should be able to do this in a single query

PermissionDetail.objects.filter(
    perm=perm,
    group__users=user
)

You may need to add .distinct() if you get duplicate results

Upvotes: 1

Related Questions