Question

我有网址和相应的视图如下。

url(r'^(?P<token>.*?)/ack$', views.api_ACK, name='device-api_ack')

def api_ACK(request, token):
    """
    Process the ACK request comming from the device
    """
    logger.info('-> api_ACK', extra={'request': request, 'token' : token, 'url': request.get_full_path()})
    logger.debug(request)
    if request.method == 'GET':
        # verify the request
        action, err_msg = api_verify_request(token=token, action_code=Action.AC_ACKNOWLEDGE)
        return api_send_answer(action, err_msg)

我想从另一个视图api_send_answer

以请求方法调用api_ACK函数作为GET

我正在/device/LEAB86JFOZ6R7W4F69CBIMVBYB9SFZVC/ack视图中api_send_answer创建一个网址，如下所示。

def api_send_answer(action, err_msg, provisional_answer=None):        

    last_action = create_action(session,action=Action.AC_ACKNOWLEDGE,token=last_action.next_token,timer=500)
    url = ''.join (['/device/',last_action.next_token ,'/',Action.AC_ACKNOWLEDGE])
    logger.debug('Request Url')
    logger.debug(url)
    response = api_ACK(request=url,token=last_action.next_token) # This is wrong

现在从api_send_answer开始重定向到api_ACK视图，但如何使用请求方法调用api_ACK作为GET？

请帮助..任何建议对我有帮助

Answer 1

这一行 response = api_ACK(request=url,token=last_action.next_token)是错误的，因为视图需要HttpRequest个对象而你会给他提供网址。

如果您需要向用户返回查看响应，可以使用重定向：

def api_send_answer(action, err_msg, provisional_answer=None):        

    last_action = create_action(session,action=Action.AC_ACKNOWLEDGE,token=last_action.next_token,timer=500)
    url = ''.join (['/device/',last_action.next_token ,'/',Action.AC_ACKNOWLEDGE])
    logger.debug('Request Url')
    logger.debug(url)
    return HttpResponseRedirect(url)

如果您需要对视图响应执行其他操作，则必须使用HttpRequest object而不是url作为参数。

Django：使用get方法调用URL

1 个答案: