[Fixed]-Django models.FileField(upload_to=function1, default=function2)

You need to use instance instead of self. self does not exist in function1 and function2 scope:

def function1(instance, filename):
    return os.path.join(instance.uid.username, str(instance.shortname), '.jpg')

def function2(instance):
    return os.path.join(MEDIA_ROOT, str(instance.username), 'tmp.jpg')

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2)

You can found docs for FileField.upload_to here

You get function2() takes exactly 1 argument, (0 given) because function2 is expecting for instance param, and if you try to pass uid like this:

class MyModel(models.Model):
    uid = models.ForeignKey(User)
    shortname = models.CharField()
    qr = models.FileField(upload_to=function1, default=function2(uid))

will raise an error because at this point uid is a models.ForeignKey instance. You could try to do this in a pre_save signal:

def function2(sender, instance, *args, **kwargs):

    if not instance.qr:
        instance.qr = os.path.join(MEDIA_ROOT, username, 'tmp.jpg')

pre_save.connect(function2, sender=MyModel)

You need to be using instance instead of self. Also, function2 will not be being passed an instance.