google app engine - How to refactor this Python code? -

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  वर्ग मेनपेज (webapp.RequestHandler): डीईएफ (स्वयं) प्राप्त करें: user = users.get_current_user () tasks_query = टास्क.ऑल () कार्यों = कार्य_क्वैरी.फेट (1000) यदि उपयोगकर्ता: url = users.create_logout_url (स्वयं। अनुरोध) अन्य: url = users.create_login_url (self.request.uri) template_values ​​= {'कार्य': कार्य , 'Url': url} पथ = os.path.join (os.path.dirname (__ फ़ाइल__), 'index.html') self.response.out.write (टेम्पलेट। रेंडर (पथ, टेम्पलेट_विभिन्न)) श्रेणी गैजेट ( Webapp.RequestHandler): def (स्वयं) प्राप्त करें: user = users.get_current_user () tasks_query = Task.all () कार्यों = tasks_query.fetch (1000) अगर उपयोगकर्ता: url = users.create_logout_url (self.request.uri) अन्य: Url = users.create_login_url (self.request.uri) template_values ​​= {'कार्य': कार्य, 'url': url} path = os.path.join (os.path.dirname (__ फ़ाइल__), 'गैजेट। Xml') self.response.out.wr आईटीई (टेम्पलेट। रेंडर (पथ, टेम्पलेट_मूल्य))

वास्तव में यह आपकी क्या निर्भर करता है भविष्य में दो वर्गों के बीच सामान्य होने की उम्मीद है रिफैक्टरिंग का उद्देश्य आम अबाधियों की पहचान करना है, कोड की लाइनों की संख्या को कम करने के लिए नहीं।

उस ने कहा, मान लें कि दो अनुरोध केवल टेम्पलेट में ही भिन्न होंगे: 

  वर्ग टास्कलिस्ट पृष्ठ (webapp.RequestHandler): डीईएफ़ (स्वयं) प्राप्त करें: उपयोगकर्ता = users.get_current_user () tasks_query = Task.all () कार्यों = tasks_query.fetch (1000) यदि उपयोगकर्ता: url = users.create_logout_url (स्वयं। Request.uri) अन्य: url = users.create_login_url (self.request.uri) template_values ​​= {'कार्य': कार्य, 'url': url} path = os.path.join (os.path.dirname (__ फ़ाइल__), Self.template_name ()) self.response.out.write (template.render (path, template_values)) क्लास मेनपेज (टास्कलिस्ट पृष्ठ): def template_name (self): वापसी 'index.html' श्रेणी गैजेट (टास्कलिस्ट पृष्ठ): def template_name ( स्वयं): 'गैजेट.एक्सएमएल' वापस करें

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