Django FileField: How to return filename only (in template)

Question

I've got a field in my model of type FileField. This gives me an object of type File, which has the following method:

File.name: The name of the file including the relative path from MEDIA_ROOT.

What I want is something like ".filename" that will only give me the filename and not the path as well, something like:

{% for download in downloads %}
  <div class="download">
    <div class="title">{{download.file.filename}}</div>
  </div>
{% endfor %}

Which would give something like myfile.jpg

Manuelmanuela

Answer 1

In your model definition:

import os

class File(models.Model):
    file = models.FileField()
    ...

    def filename(self):
        return os.path.basename(self.file.name)

Maas

worked great but needed the param to be passed into basename as self.file.name. I think it needs this as self.file is a file object rather than the string to the file. – Manuelmanuela

adding the @property decorator avoids calling file.filename() and simply file.filename can be used – Mountie

Surprisingly this solution even works with S3 storage backend. – Panacea

It works for me. Could you please some information to explain how it works? – Ohl

@user3327344 Which is the part that you have trouble with? – Maas

Answer 2

You can do this by creating a template filter:

In myapp/templatetags/filename.py:

import os

from django import template


register = template.Library()

@register.filter
def filename(value):
    return os.path.basename(value.file.name)

And then in your template:

{% load filename %}

{# ... #}

{% for download in downloads %}
  <div class="download">
      <div class="title">{{download.file|filename}}</div>
  </div>
{% endfor %}

Cathleencathlene

This is also a nice approach. Perhaps a little more portable? – Pencil

As you wrote the filter it should be {{download|filename}} – Regularly

don’t forget the __init__.py file to ensure the templatetags directory is treated as a Python package. – Galileo

Answer 3

You could also use the cut filter in your template

{% for download in downloads %}
  <div class="download">
    <div class="title">{{download.file.filename|cut:'remove/trailing/dirs/'}}</div>
  </div>
{% endfor %}

Gunning

Answer 4

A modern solution using Pathlib:

from pathlib import Path

class File(models.Model):
    file = models.FileField()
    ...

    def filename(self):
        return Path(self.file.name).name

Galantine

Answer 5

You can access the filename from the file field object with the name property.

class CsvJob(Models.model):

    file = models.FileField()

then you can get the particular objects filename using.

obj = CsvJob.objects.get()
obj.file.name property

Villeneuve

Accessing obj.file.name returns the entire path in the media dir, e.g. it returns tasks/132/foo.jpg rather than foo.jpg. – Dybbuk