Google App Engine：创建自定义错误页面

Question

The application that I have been working on takes a input from a user and accordingly outputs a bunch of results based on the keyword. I use web scraping to scrap the results.

Here's the python code for the same:

import os
import webapp2
import jinja2
from google.appengine.ext import db
import urllib2
import re

template_dir = os.path.join(os.path.dirname(__file__), 'templates')
jinja_env = jinja2.Environment(loader = jinja2.FileSystemLoader(template_dir), autoescape=True)

class Handler(webapp2.RequestHandler):
    def write(self, *a, **kw):
        self.response.out.write(*a, **kw)
    def render_str(self, template, **params):
        t = jinja_env.get_template(template)
        return t.render(params)
    def render(self, template, **kw):
        self.write(self.render_str(template, **kw))

class MainPage(Handler):
    def get(self):
        self.render("formrss.html")
    def post(self):
        x = self.request.get("rssquery")
        url = "http://www.quora.com/" + x + "/rss"
        content = urllib2.urlopen(url).read()
        content = content.decode('utf-8')
        allTitles =  re.compile('<title>(.*?)</title>')
        allLinks = re.compile('<link>(.*?)</link>')
        list = re.findall(allTitles,content)
        linklist = re.findall(allLinks,content)
        self.render("frontrss.html", list = list, linklist = linklist)



app = webapp2.WSGIApplication([('/', MainPage)], debug=True)

Now, since I obtain the final url by combining the keyword entered by the user, there's always a chance that the url obtained from the keyword is invalid. Hence, in that case, the user gets a 404 error.

Here's a live demo of the application:

http://quorable.appspot.com/

How can I avoid this error? Or rather, how can I display a custom message for the user in case the url obtained is not available? I don't want the user to think that the application is broken or malfunctioning, when instead the keyword entered by the user is invalid.

Answer 1

import...

def handleNone(request, response, exception):
    logging.exception(exception) # if you need some logging
    response.headers['Content-Type'] = 'text/html'
    response.write('No feed for this keywords. Go <a href="/">back</a>')
    response.set_status(404)

app = webapp2.WSGIApplication([('/', MainPage)], debug=True)

app.error_handlers[404] = handleNone

By the way, you should better do a redirect if the Quora server responds 404. Check urllib2.urlopen() reference docs :

req = urllib2.Request(yourURL)
try: urllib2.urlopen(req)
except URLError as e:
    if e.code == 404:
        redirect_to_a_no_results_page() 
else:
       fetch_the_results()