How to update a model but return unmodified model in Django?
I'm using django-piston to write a RESTful Web Service and have a problem.
in models.py:
class Status(models.Model):
user = models.ForeignKey(User)
content = models.TextField(max_length=140)
class StatusReply(models.Model):
user = models.ForeignKey(User)
reply_to = models.ForeignKey(Status, related_name='replies')
content = models.TextField(max_length=140)
has_read = models.BooleanField(default=False, help_text="has the publisher of the status read the reply")
in handlers.py:
class StatusHandler(BaseHandler):
allowed_methods = ('GET', 'POST', 'DELETE' )
model = Status
fields = ('id',
('user', ('id', 'username', 'name')),
'content',
('replies', ('id',
('user', ('id', 'username', 'name')),
'content',
'has_read'),
),
)
@need_login
def read(self, request, id, current_user): # the current_user arg is an instance of user created in @need_login
try:
status = Status.objects.get(pk=id)
开发者_如何学Go except ObjectDoesNotExist:
return rc.NOT_FOUND
else:
if status.user == current_user: #if current_user is the publisher of the status, set all replies read
status.replies.all().update(has_read=True)
return status
In the handler, it returned a specific status by id. Now I want to return the status before status.replies.all().update(has_read=True)
but also do the update operation in database. How to do it? Thanks in advance.
Not sure if I understand what you need. As I understand your code, status.replies.all().update(has_read=True)
doesn't change status
but only changes the replies. If that's true, the code should do what you want. If it isn't, you could make a copy of status
and return the copy:
if status.user == current_user:
old_status = status.make_copy()
status.replies.all().update(has_read=True)
return old_status
return status
Or do you just want the method to return early and do the database update asynchronously? Then you should have a look at celery and maybe this nice explanation.
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