python - 我的装饰器中需要该请求，但我不知道如何将其放入其中

Question

我的除雾器功能...

def validate_captcha(view):
    '''Decorator to validate a captcha based on settings'''

    def failure():
        return HttpResponse('You need to complete the captcha, please refresh and try again')

    if request.method == 'POST':

        url = "https://www.google.com/recaptcha/api/siteverify"
        values = {
            'secret': settings.GOOGLE_RECAPTCHA_SECRET_KEY,
            'response': request.POST.get(u'g-recaptcha-response', None),
            'remoteip': request.META.get("REMOTE_ADDR", None),
        }

        data = urllib.urlencode(values)
        req = urllib2.Request(url, data)
        response = urllib2.urlopen(req)
        result = json.loads(response.read())

        # result["success"] will be True on a success
        if result["success"]:
            return view
        else:
            return fail
    return fail

然后是我的观点...

@validate_captcha
def sendemail(request):
    ...

我可以将请求放入装饰器参数中，但是当我将其放入 View 参数中时，它是未定义的。我尝试用其他几种方式调用它，但没有成功，你会如何把它放在那里？

Answer 1

您需要有一个包装函数:

def validate_captcha(view):
    def wrap(request, *args, **kwargs):
        if request.method == 'POST':

            url = "https://www.google.com/recaptcha/api/siteverify"
            values = {
                'secret': settings.GOOGLE_RECAPTCHA_SECRET_KEY,
                'response': request.POST.get(u'g-recaptcha-response', None),
                'remoteip': request.META.get("REMOTE_ADDR", None),
            }

            data = urllib.urlencode(values)
            req = urllib2.Request(url, data)
            response = urllib2.urlopen(req)
            result = json.loads(response.read())

            # result["success"] will be True on a success
            if result["success"]:
                return view
            else:
                return fail
        return fail
    return wrap

一定要研究一下关于 Python 中装饰器的这个很棒且非常详细的概述(我个人认为，这是有史以来最好的答案之一):

• https://stackoverflow.com/a/1594484/771848