Question

I have the following django model:

RESOURCE_DIR = os.path.join(settings.MEDIA_ROOT, 'resources')

class Resource(models.Model):
    title = models.CharField(max_length=255)
    file_name = models.FilePathField(path=RESOURCE_DIR, recursive=True)

and I want to give the URL to the file in a template so that the user can view it or download it.

If I use {{ resource.file_name }} in the template it outputs the full path of the file on the server, e.g. if RESOURCE_DIR='/home/foo/site_media/media' it outputs '/home/foo/site_media/media/pdf/file1.pdf' whereas what I want is 'pdf/file1.pdf'. In the admin or in a modelform the option is displayed as '/pdf/file1.pdf' in the select widget. So obviously it is possible to do what I asking. Of course the extra slash is not important. If I were setting recursive=False then I could just remove the part of the path before the last slash.

How can I get the same result as the modelform or admin?

Answer 1

I figured out that you can retrieve the path argument to the FilePathField using resource._meta.get_field(‘file_name’).path It seems best to do this in the model. So the model becomes:

RESOURCE_DIR = os.path.join(settings.MEDIA_ROOT, 'resources')

class Resource(models.Model):
    title = models.CharField(max_length=255)
    file_name = models.FilePathField(path=RESOURCE_DIR, recursive=True)

    def url(self):
        path = self._meta.get_field('file_name').path
        return self.file_name.replace(path, '', 1)

then in the template you can put:
{{ MEDIA_URL }}resources{{ resource.url }}