python - 装饰器在其装饰的函数调用之前运行？

举个例子:

def get_booking(f=None):
    print "Calling get_booking Decorator"
    def wrapper(request, **kwargs):
        booking = _get_booking_from_session(request)
        if booking == None:
            # we don't have a booking in our session.
            return HttpRedirect('/')
        else:
            return f(request=request, booking=booking, **kwargs)
    return wrapper

@get_booking
def do_stuff(request, booking):
    # do stuff here

我遇到的问题是，甚至在我调用要修饰的函数之前，就已经调用了@get_booking decorator。
开始输出:

Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
Calling get_booking Decorator
[26/Oct/2008 19:54:04] "GET /onlinebooking/?id=1,2 HTTP/1.1" 302 0
[26/Oct/2008 19:54:05] "GET /onlinebooking/ HTTP/1.1" 200 2300
[26/Oct/2008 19:54:05] "GET /site-media/css/style.css HTTP/1.1" 200 800
[26/Oct/2008 19:54:05] "GET /site-media/css/jquery-ui-themeroller.css HTTP/1.1" 200 25492

在这一点上，我什至没有调用过一个装饰过的函数。
我刚刚开始使用装饰器，所以也许我缺少了一些东西。

最佳答案

我相信python装饰器只是语法糖。

@foo
def bar ():
    pass

和...一样

def bar ():
    pass
bar = foo(bar)

如您所见，即使没有调用bar也将调用foo。这就是为什么您看到装饰器函数的输出的原因。对于您将装饰器应用到的每个函数，您的输出应只包含一行。