python - Django FileField(或ImageField)open()方法为有效文件返回None？

Question

让我这样说:

模型.py:

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

ajax.py(我用的是dajaxice但没关系):

...
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())
    ...

返回:

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'

但是:

...
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())
    ...

完美运行!为什么？

(我使用的是 django 1.3)

Answer 1

因为 models.FileField 的 open 方法没有返回任何内容

你可以使用:

task.seq_file.read()

而且你不需要计算文件的路径来检查文件是否存在。你可以使用 task.seq_file.path:

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