Django FileField (or ImageField) open() method returns None for valid file?

Question

let me put it like this:

model.py:

class Task(models.Model):     ...     seq_file = models.FileField(upload_to='files/', blank=True, null=True)     ... 

ajax.py (I'm using dajaxice but it doesn't matter):

... def startTask(request, name):     task = Task.objects.get(task_name=name)     data = task.seq_file.open()     filename = os.path.join(settings.MEDIA_ROOT ,task.seq_file.name)     if not os.path.isfile(filename):         raise Exception, "file " + filename + " not found."     sequences = parser.parse(data.read())     ... 

this returns:

File "/home/mnowotka/Dokumenty/MgrFuncAdnot/app/django-gui/src/gui/ajax.py", line 43, in startTask sequences = parser.parse(data.read())  AttributeError: 'NoneType' object has no attribute 'read' 

but:

... def startTask(request, name):     task = Task.objects.get(task_name=name)     filename = os.path.join(settings.MEDIA_ROOT ,task.seq_file.name)     if not os.path.isfile(filename):         raise Exception, "file " + filename + " not found."     data = open(filename)       sequences = parser.parse(data.read())     ... 

works perfectly! Why?

(I'm using django 1.3)

Answer 1

because open method of models.FileField doesn't return anything

you can just use:

task.seq_file.read() 

and you don't need calculate path of file for checking if file exist. you can use task.seq_file.path:

if not os.path.isfile(task.seq_file.path):     .... 

Answer 2

A FileField will give you a file-like object and there is no need to call open() on it. In your example, just call task.seq_file.file.

Why is that? There are many storage backends for FileField, and many of them are not backed by a file in disk (think of S3 storage, for example). I guess that is why the documentation says it returns a file-like object, not a file. For some kinds of storage the "open" method makes no sense.