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django - 在视图中创建函数。“分配前参考”

问题描述

我正在尝试创建一个简单的函数来接收从先前视图传递的请求数据。我想使用数据来创建示例中的条件（注释代码）。那么我创建：

def get_request(field_name):
    data = request.GET.get('%s'%(field_name), False)
    return data

在我的views.py中我使用：

from .helper import get_request

def search_simple(request):
    #brand=data from previous view
    get_request(brand) #this raises an exception
    request.GET.get('brand', False) #it work good
    # and them i wont using somethin like this
    #if get_request(brand):
    #    do something
    return render(request, 'search/search_results_brand.html')

这会引发以下异常：

local variable 'brand' referenced before assignment

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/wyniki-wyszukiwania/?brand=2&model=&province=&type_car=&transmission=&price_st=&price_end=&year_production_st=&year_production_end=&car_mileage=60do100&fuel_type=BenzynaLPG&engine_power=120-150&condition=Uszkodzony&automatic_air_conditioning=on&speed_limit=on&parking_heating=on&light_xenon=on&automatic_air_conditioning=on&speed_limit=on&parking_heating=on&light_xenon=on&search=

Django Version: 2.2.4
Python Version: 3.7.0
Installed Applications:
['django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django.contrib.humanize',
 'app',
 'app_profile',
 'django.contrib.sites',
 'allauth',
 'allauth.account',
 'allauth.socialaccount',
 'allauth.socialaccount.providers.google',
 'allauth.socialaccount.providers.facebook',
 'crispy_forms',
 'django_select2',
 'widget_tweaks']
Installed Middleware:
['django.middleware.security.SecurityMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware']



Traceback:

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\exception.py" in inner
  34.             response = get_response(request)

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\base.py" in _get_response
  115.                 response = self.process_exception_by_middleware(e, request)

File "C:\Users\tymot\AppData\Local\Programs\Python\Python37-32\lib\site-packages\django\core\handlers\base.py" in _get_response
  113.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "C:\Users\tymot\Desktop\car-app\app_rama\app\views.py" in search_simple
  36.     get_request(brand)

Exception Type: UnboundLocalError at /wyniki-wyszukiwania/
Exception Value: local variable 'brand' referenced before assignment

如何使用我的功能正确获得品牌。品牌并不总是传递下去。

标签： djangoview

解决方案

问题是当你应该传递一个字符串时，Python 认为你正在将一个变量传递给你的函数。尝试替换get_request(brand)为get_request('brand').

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