django-filter over models

René L. Hechavarría

unread,

Sep 13, 2018, 3:26:20 AM9/13/18

to django...@googlegroups.com

Hello,

I have a model, Jobs and Persons, Jobs are listed in ListView, i need to add a filter (django-filter), if i filter over field in Jobs all work good, but i need to filter also for Persons. this is and example

@property
def get_persons(self):
    return self.persons_all.all()

That allow to add persons to Jobs model in de html. How i filter over persons in Jobs view?

Thanks in advance.

Mateusz

unread,

Sep 18, 2018, 11:15:02 PM9/18/18

to Django users

In that answer I assume Jobs are connected anyhow with Persons with a ForeignKey field like so:

File /appname/models.py:

from django.db import models


class Job(models.Model):
    name = models.CharField(max_length=100)

    def __str__(self):
        return self.name


class Person(models.Model):
    name = models.CharField(max_length=100)
    surname = models.CharField(max_length=100)

    job = models.ForeignKey(Job, on_delete=models.SET_NULL, null=True, default=None)

    def __str__(self):
        return self.name + ' ' + self.surname

File /appname/views.py:

from django.shortcuts import render
from django.views.generic.list import ListView

from appname.models import Job, Person


class JobsWithPersonsView(ListView):
    queryset = Job.objects.all()
    template_name = 'jobs_list.html'

File /appname/templates/jobs_list.html:

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Jobs</title>
</head>
<body>
    {% for job in object_list %}
        {{ job.name }}
        {% for person in job.person_set.all %}
            {{ person.name }} {{ person.surname }}
        {% empty %}
            Nobody works there.
        {% endfor %}
    {% empty %}
        No jobs.
    {% endfor %}
</body>
</html>

And others you probably know how to deal with...
File /urls.py:

from django.contrib import admin
from django.urls import path, include

urlpatterns = [
    path('admin/', admin.site.urls),
    path('appname/', include('appname.urls'))
]

File /appname/urls.py:

from django.urls import path
from appname import views

urlpatterns = [
    path('jobslist/', views.JobsWithPersonsView.as_view(), name='jobslist')
]

BTW: Your model names should be in singular form, if you want to be able to use plural forms in some cases, use verbose_name_plural (docs).

Mateusz

unread,

Sep 18, 2018, 11:47:52 PM9/18/18

to Django users

Sorry, my previous answer was senseless, I realized that just after adding it.

1) You can filter jobs in appname/views.py:

queryset = Job.objects.filter(name__contains="job")

2) and people in template /appname/templates/jobs_list.html:

{% for job in object_list %}

    <h1>{{ job.name }}</h1>

    {% for person in job.person_set.all %}

        {% if person.name|length <= 4 %}
            {{ person.name }} {{ person.surname }}
        {% endif %}


    {% empty %}
        Nobody works there.
    {% endfor %}

I still find that 2) inefficient, but could do anything else. Read something about registering filters that I do not understand that well, sorry.

brajesh kumar

unread,

Sep 19, 2018, 10:44:20 PM9/19/18

to Django users

I think this answers my question , trying to implement it.

Reply all

Reply to author

Forward