Friday, June 27, 2014

dynamically upload file to Django model FileField

This post was inspired by this SO Q&A which is for an ImageField but which I adapted for an FileField.

Does your app generate some content that is too large or not appropriate for a database? You can store it as a FileField which points to a file in your MEDIAROOT folder. How do you upload the generated content to the FileField? Creation is a bit similar to ManyToManyField if you are familiar with using that.
  1. Add the FileField to your model and set blank=True, null=True so that you can instantiate it without setting the FileField.
  2. Create the object leaving off the FileField. Save the instance.
  3. When you retrieve the FileField from your model you get a FieldFile (note the word order swap) which allows you to interact with the File object (a Django version of a Python file object). You could save the content to disk then call the FieldFile.save() method, but you can skip this unnecessary step. Let's assume the content can be serialized as a JSON object. The following code will upload the content to Django. Also note that FieldFile.save() content must be a django.core.files.File object not a Python file object.
    from StringIO import StringIO
    import json
    from django.core.files import File
    f = StringIO(json.dumps(my_content, indent=2, sort_keys=True))
    try:
        my_model_object.my_file_field.save('my_file_name.json', File(f))
        # `FieldFile.save()` saves the instance of the model by default
    finally:
        f.close()  # `StringIO` has no `__exit__` method to use `with`
Setting the FileField to null=True and blank=True is only necessary if you want to upload a file object, otherwise you can pass file name as the FileField when you construct the database object. EG:
my_model_object(my_char_field='some other model fields', my_file_field='my_file_field.json')
This will upload 'my_file_field.json' from disk if it is a valid path.

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