我有django项目与房间,我想创建实时聊天的房间与频道库,这个聊天工作与数据从url类example.com/房间名称/人名,但我有房间有静态的url(示例如下),如何调整我的项目。我想在url中显示request.user.username而不是person_name,并且我想使用room_detail url而不是每次都应该输入的room_name
views.py
class RoomDetail(DetailView):
model = Room
template_name = 'rooms/room_detail.html'
context_object_name = 'room_detail'
slug_field = 'invite_url'
slug_url_kwarg = 'url'
consumers.py
class Consumer(WebsocketConsumer):
def connect(self):
self.room_name=self.scope['url_route']['kwargs']['room_name']
self.room_group_name='chat_%s' % self.room_name
self.person_name=self.scope['url_route']['kwargs']['person_name']
async_to_sync(self.channel_layer.group_add)(
self.room_group_name,
self.channel_name
)
async_to_sync(self.channel_layer.group_send)(
self.room_group_name,
{
"type":"chat_message",
"message":self.person_name+" Joined Chat"
}
)
self.accept()
def disconnect(self, code):
async_to_sync(self.channel_layer.group_send)(
self.room_group_name,
{
"type":"chat_message",
"message":self.person_name+" Left Chat"
}
)
async_to_sync(self.channel_layer.group_discard)(
self.room_group_name,
self.channel_name
)
def receive(self, text_data=None, bytes_data=None):
text_data_json=json.loads(text_data)
message=text_data_json['message']
async_to_sync(self.channel_layer.group_send)(
self.room_group_name,
{
'type':'chat_message',
'message':self.user+" : "+message
}
)
def chat_message(self,event):
message=event['message']
self.send(text_data=json.dumps({
'message':message
}))
urls.py
path('rooms/<url>/', RoomDetail.as_view(), name='room_detail'),发布于 2020-04-05 07:48:45
我不确定,我知道你会的。但是如果你需要获取用户名,你可以简单地使用
self.scope['user'].username而不是从urls模式中获取它。
https://stackoverflow.com/questions/61034863
复制相似问题