17

我正在使用 Django 信号，但它们似乎被接收了两次，即使发出一次。这是我正在使用的代码（这是一个将 Uploadify 与 Django 一起使用的简单包装器）...

# Signal-emitting code... emits whenever a file upload is received
# ----------------------------------------------------------------
upload_recieved = django.dispatch.Signal(providing_args=['data'])

def upload(request, *args, **kwargs):
    if request.method == 'POST':
        if request.FILES:
            print 'sending signal'
            upload_recieved.send(sender='uploadify', data=request.FILES['Filedata'])
    return HttpResponse('True')

# Signal-receiving code...
# ----------------------------------------------------------------    
def upload_received_handler(sender, data, **kwargs):
    print 'upload received handler'

print 'connecting signal'
upload_recieved.connect(upload_received_handler)

（我刚刚注意到我的信号拼写错误）

我敢肯定你注意到了那里的打印语句。在控制台上，这就是它所显示的内容：

(server starts)
connecting signal

...

sending signal
upload received handler
upload received handler     # << == where is this 2nd one coming from?
127.0.0.1 - - [25/Sep/2009 07:28:22] "POST /uploadify/upload/ HTTP/1.1" 200 -

（同样奇怪的是为什么 Django 会在信号触发后报告页面 POST？）

4

2 回答 2

22

这发生在我之前，这是由于您连接信号的模块被导入两次。为了确保信号没有连接两次，您可以设置 dispatch_uid：

upload_recieved.connect(upload_received_handler, dispatch_uid="some.unique.string.id")

更新 它实际上记录在这里：http ://code.djangoproject.com/wiki/Signals#Helppost_saveseemstobeemittedtwiceforeachsave

于 2009-09-26T01:20:11.520 回答
0

您可以检查函数中的“created”参数，您正在使用分别返回 True 和 False 的信号连接。仅当创建新对象时，Created 才会为 True。

def task_feedback_status_handler(sender, instance, created, **kwargs):
    if created:
        do something
post_save.connect(task_feedback_status_handler, sender=Feedback)
于 2020-08-20T06:55:33.843 回答