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

Ludwik Trammer · Accepted Answer · 2014-11-19T08:49:17.017

224

In your model definition:

import os

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

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

edited Nov 19 '14 at 08:49

answered Apr 21 '10 at 14:31

Ludwik Trammer

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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. – John Apr 21 '10 at 14:49
19

adding the @property decorator avoids calling file.filename() and simply file.filename can be used – spedy May 09 '17 at 11:59
8

Surprisingly this solution even works with `S3` storage backend. – Stan Reduta Feb 27 '20 at 08:59
It works for me. Could you please some information to explain how it works? – xie186 Aug 13 '20 at 19:37
@user3327344 Which is the part that you have trouble with? – Ludwik Trammer Aug 20 '20 at 20:42

score 60 · Answer 2 · answered Oct 28 '10 at 19:29

60

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 %}

answered Oct 28 '10 at 19:29

rz.

19,861
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This is also a nice approach. Perhaps a little more portable? – Ben Keating Aug 18 '11 at 06:35
3

As you wrote the filter it should be {{download|filename}} – Emil Stenström Oct 15 '14 at 14:41
6

don’t forget the `__init__.py` file to ensure the templatetags directory is treated as a Python package. – shuckc Apr 20 '16 at 07:08

score 3 · Answer 3 · answered Dec 09 '21 at 18:50

3

You could also use 'cut' in your template

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

answered Dec 09 '21 at 18:50

Olaleye Abiola

81
3

score -2 · Answer 4 · answered Sep 16 '16 at 09:00

-2

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

answered Sep 16 '16 at 09:00

DeadDjangoDjoker

534
2
8
20

Accessing `obj.file.name` returns the entire path in the media dir, e.g. it returns `tasks/132/foo.jpg` rather than `foo.jpg`. – shacker Apr 07 '19 at 06:08

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

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