我在urls.py文件一堆網址是有login_required裝飾Django的測試,模擬的有效網址請求
# Index Page
url(r'^$', login_required(views.IndexPage.as_view()), name='index'),
# Schedule urls
url(r'^schedules/$', login_required(views.ScheduleListView.as_view()),
name='schedule-list'),
url(r'^schedule/(?P<pk>[\d]+)/$',
login_required(views.ScheduleDetailView.as_view()),
name='schedule-detail'),
url(r'^schedule-freeze/(?P<pk>[\d]+)/$',
login_required(views.freezeSchedule),
name='schedule-freeze'),
url(r'^schedule-create/$', login_required(views.ScheduleCreate.as_view()),
name='schedule-create'),
url(r'^schedule-delete/(?P<pk>[\d]+)$',
login_required(views.ScheduleDelete.as_view()),
name='schedule-delete'),
url(r'^schedule-update/(?P<pk>[\d]+)/$',
login_required(views.ScheduleUpdate.as_view()),
name='schedule-update'),
url(r'^schedule-generate/(?P<pk>[\d]+)/$',
login_required(views.scheduleGenerate), name='schedule-generate'),
# Client urls
url(r'^clients/$', login_required(views.ClientList.as_view()),
name='client-list'),
url(r'^client/(?P<slug>[\w-]+)/$',
login_required(views.ClientDetail.as_view()), name='client-detail'),
url(r'^client-create/$', login_required(views.ClientCreate.as_view()),
name='client-create'),
url(r'^client-delete/(?P<slug>[\w-]+)/$',
login_required(views.ClientDelete.as_view()), name='client-delete'),
url(r'^client-update/(?P<slug>[\w-]+)/$',
login_required(views.ClientUpdate.as_view()), name='client-update'),
# And so on ....
對於每一個管窺我試圖寫一個測試確保未經授權的用戶在嘗試訪問視圖時被重定向到登錄頁面。如果可能的話,我希望能夠在單個代碼塊中實現這一點,而不是爲每個URL編寫單個測試。
我已經試過類似如下:
list_urls = [e for e in get_resolver(urls).reverse_dict.keys() if isinstance(e, str)]
for url in list_urls:
# Fetches the urlpath e.g. 'client-list'
namedspaced_url = 'reports:' + url
path = reverse(namedspaced_url)
response = self.client.get(path)
self.assertEqual(response.status_code, 302)
self.assertRedirects(response, reverse('login') + '?next=' + path)
list_urls
返回我的urls.py文件中所有命名的網址即['schedule-create', 'server-detail', 'schedule-list', 'schedule-update', 'index', ....]
名單的問題
這一段代碼:reverse(namedspaced_url)
where thi s導致的問題是,每個網址有不同的正則表達式模式,即一些採取slu some一些採取pk的
所以行path = reverse(namedspaced_url)
將適用於簡單的URL,如那些指向ListViews但將失敗的更復雜的URL,如那些指向DetailViews需要蛞蝓的/ PK的,即path = reverse(namedspaces_url, args=[1945])
是否有可能暫時拒絕/忽略Django的模式匹配/路由強制要求去通過(無論通過參數的個數)
還是我必須使用有效的kwargs/args手動爲每個URL編寫一個測試來滿足regex?
是否有另一種完全不同的方法可以爲我的所有login_required()視圖編寫測試?
更新 使用自省,我想出了以下的怪物,以解決我的問題
def test_page_redirects_for_unauthorised_users(self):
url_dict = get_resolver(urls).reverse_dict
url_list = [e for e in get_resolver(urls).reverse_dict.keys() if
isinstance(e, str)]
for url in url_list:
patterns = url_dict[url][0][0][1]
matches = [1 if e == 'pk' else "slug" if e == 'slug' else None for
e in patterns]
path = reverse('reports:' + url, args=matches)
response = self.client.get(path)
self.assertEqual(response.status_code, 302)
self.assertRedirects(response, reverse('login') + '?next=' + path)
您應該能夠內省url解析器中的項目以獲取命名參數。然後,您可以以合適的值調用反向,例如, '1945'代表'pk',或'my-slug'代表'slug'。 – Alasdair
你又來了!我玩弄了這個想法,並被勸告反對它,因爲這意味着我不得不拿出樣本值來匹配我的網址中的每個正則表達式模式(這很快就會變得混亂)。我想知道是否有一個更清潔的解決方案,因爲我肯定有人必須在這個問題之前解決一些問題。但是,謝謝你的建議!(如果這是解決此問題的唯一已知方法,那麼我想我只需要這樣做) –