How can i filter a models field in django?

Question

im quite new in backend development so i got a basic question.

I've three different models one named Campaigns;

class Campaigns(models.Model):
   
   channel = models.ForeignKey(Channels, blank=False, verbose_name="Channel Name", on_delete=models.CASCADE)
   campaign_name = models.CharField(max_length=255, blank=False, verbose_name="Campaign Name")

Second one is;

class CampaignDetails(models.Model):
   channel = models.ForeignKey(Channels, blank=False, null=True, verbose_name="Channel Name", on_delete=models.CASCADE)
   name = models.ForeignKey(Campaigns, blank=False, null=True, verbose_name="Campaign", on_delete=models.CASCADE)

And the last one is;

Class Channels(models.Model):
   name = models.CharField(max_length=50, null=False, blank=False, verbose_name="Tv Channel")

I want to filter name in CampaignDetails by channel. Like if i choose channel 1 i want to filter name by campaign names that under that channel. How can i manage that?

Any help is appreciated, thank you.

Answer 1

I don't fully understand your question. so I can give you some decision.For example You can filter it :

 channel=Channel.objects.get(pk=1) channel.campaigndetails_set.all() #or channel.campaigndetails_set.filter(name='your name')

I'm not sure about the answer, but i think it will help you

Answer 2

champaigns = Campaigns.objects.filter(chanel=Channels.objects.get(name="Chanel name"))
for i in champaigns:
     print(i.channel)

same goes for the other models you can do that by other methods too however you find it easy

in template

{%for i in champaigns%}
{{i.channel}}
{%endfor%}

Answer 3

I don't know why you have 2 channels. But if you want to filter it by channel

campaign_details = CampaignDetails.objects.filter(channel__name="insert_channel_name_here")

Well if you want to clean up. Remove channel from campaign details, and change the name of the foreign key "name" to campaign

campaign_details = CampaignDetails.objects.filter(campaign__channel__name="insert_channel_name_here")