Access to file uploaded with FileField Django

Question

I'm trying to create a Django form with a filefield to upload a pdf based on a model.

#models.py
class ProductionRequest(models.Model):
     ...
     contract_file = models.FileField('Contract file (PDF)', upload_to='contracts/')
     ...

I can upload the file and save it in my object, but when I try to show the file in my template with this code

 {{ prod_req.contract_file }}

it only show me the path to the file "contracts/file_uploaded.pdf". How can I make a link to download (or open in a new tab ) the file ?

Plus : when I try to open the file from Django admin, with this link

Contract file (PDF) : Currently: contracts/file_uploaded.pdf

I don't show me the file, i got this :

Page not found (404) Request URL: http://127.0.0.1:8000/admin/appname/productionrequest/1/change/contracts/file_uploaded.pdf/change/

How can I access to my file ?

https://stackoverflow.com/questions/9462999/how-to-download-a-filefield-file-in-django-view might help — Tom Dalton, Nov 06 '17 at 17:30

score 1 · Accepted Answer · answered Nov 06 '17 at 16:55

It works just like a Python file. So, the way you have just refers to a file object.

To get the text, you need to use f.read(). So, in your template, instead of

{{ prod_req.contract_file }}

use

{{ prod_req.contract_file.read }}

BUT, note that calling read moves the cursor to the end of the file. The first time you call read on a file, it returns all the text. The second time, it returns nothing. This is only a problem if you need the text twice, but in that case, store the text directly on the context object.

Let me know if anything was unclear.

Thank you now I am able to get my file ! – dxtr69 Nov 13 '17 at 15:47 — dxtr69, Nov 13 '17 at 15:47

Access to file uploaded with FileField Django

1 Answers1