Django related_name import Objects instead of foreign_keys

Question

When I importing my CSV file in db.sqlite3, I don't know how to import foreign_key instead of "Object".

This is what I've tried.

# import_csv.py (manage.my custom command)

import csv
from django.core.management.base import BaseCommand
from models import Site, Association

class Command(BaseCommand):
    help = "Import Command"

    def handle(self, *args, **options):
        with open(_file, newline='') as csvfile:
            reader = csv.DictReader(csvfile, delimiter=";")
            for row in reader:
                localsite, created = Site.objects.get_or_create(name=row["locSite"])
                distantsite, created = Site.objects.get_or_create(name=row["disSite"])
                csv_line, created = Association.objects.get_or_create(
                    localName=row["Name"],
                    locSite=localsite.id,
                    disSite=distantsite.id,
                    ...
                )


# models.py

class Site(models.Model):
    name = models.CharField(max_length=5, unique=True, help_text="Site Local")
    objects = models.Manager()


class Association(models.Model):
    localName = models.CharField(max_length=50, help_text="Nom Local")
    locSite = models.ForeignKey(Site, null=True, on_delete=models.SET_NULL, related_name='local_site_set')
    disSite = models.ForeignKey(Site, null=True, on_delete=models.SET_NULL, related_name='distant_site_set')

Django Admin panel : add record

Thx for help

score 1 · Answer 1 · edited May 06 '22 at 12:22

Are you looking for this: foreignkey_id to set this field

import csv
from django.core.management.base import BaseCommand
from models import Site, Association

class Command(BaseCommand):
    help = "Import Command"

    def handle(self, *args, **options):
        with open(_file, newline='') as csvfile:
            reader = csv.DictReader(csvfile, delimiter=";")
            for row in reader:
                localsite, created = Site.objects.get_or_create(name=row["locSite"])
                distantsite, created = Site.objects.get_or_create(name=row["disSite"])
                csv_line, created = Association.objects.get_or_create(
                    localName=row["Name"],
                    locSite_id=localsite.id,
                    disSite_id=distantsite.id  # This is the way to add foreign key if you know or if you want to create
                    ...
                )


# models.py

class Site(models.Model):
    name = models.CharField(max_length=5, unique=True, help_text="Site Local")
    objects = models.Manager()


class Association(models.Model):
    localName = models.CharField(max_length=50, help_text="Nom Local")
    locNomSite = models.ForeignKey(Site, null=True, on_delete=models.SET_NULL, related_name='local_site_set')

To display different name in django admin panel you need to resgister your model in admin.py as follows

class CustomAssociationAdmin(admin.ModelAdmin):
    form = MyInvoiceAdminForm


class CustomAssociationAdminForm(forms.ModelForm):
    person = YourModelChoiceField(queryset=Site.objects.all()) 
    class Meta:
          model = Invoice
      
class YourModelChoiceField(forms.ModelChoiceField):
     def label_from_instance(self, obj):
         return "%s"% (obj.name)

admin.site.register(CustomAssociationAdmin, Association)

Same. It is diplaying "Site object (1)" .... in the select field in back office. I would like to display Site names. — MrSo, May 06 '22 at 12:00
@MrSo I could suggest you this method but you can read this SO https://stackoverflow.com/questions/6836740/how-to-change-foreignkey-display-text-in-the-django-admin — Deepak Tripathi, May 06 '22 at 12:14

score 1 · Accepted Answer · answered May 06 '22 at 12:11

1

class Site(Model):
    name = models.CharField(max_length=5, unique=True, help_text="Site Local")
    objects = models.Manager()

    def __str__(self):
        return self.name

If you want to display the Site name in the admin panel instead of object then you need to add str method to the model class

answered May 06 '22 at 12:11

kjaw

515
2
9

2

This will change the name everywhere whereever model objects are accessed. – Deepak Tripathi May 06 '22 at 12:13
I don't think that's a problem in this case – kjaw May 06 '22 at 12:15
1

Obviously wherever site will be accessed now name will be displayed, but its OP choice. – Deepak Tripathi May 06 '22 at 12:16
1

I prefer this method. Even if, as Deepak said, It's better to register a new model – MrSo May 06 '22 at 12:16

Django related_name import Objects instead of foreign_keys

2 Answers2