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Get URL when handling urllib2.URLError

开发者 https://www.devze.com 2023-03-16 11:36 出处:网络
This pertains to urllib2 specifically, but custom exception handling more generally.How do I pass additional information t开发者_如何学Co a calling function in another module via a raised exception?I\

This pertains to urllib2 specifically, but custom exception handling more generally. How do I pass additional information t开发者_如何学Co a calling function in another module via a raised exception? I'm assuming I would re-raise using a custom exception class, but I'm not sure of the technical details.

Rather than pollute the sample code with what I've tried and failed, I'll simply present it as a mostly blank slate. My end goal is for the last line in the sample to work.

#mymod.py
import urllib2

def openurl():
    req = urllib2.Request("http://duznotexist.com/")
    response = urllib2.urlopen(req)

#main.py
import urllib2
import mymod

try:
    mymod.openurl()
except urllib2.URLError as e:
    #how do I do this?
    print "Website (%s) could not be reached due to %s" % (e.url, e.reason)


You can add information to and then re-raise the exception.

#mymod.py
import urllib2

def openurl():
    req = urllib2.Request("http://duznotexist.com/")
    try:
        response = urllib2.urlopen(req)
    except urllib2.URLError as e:
        # add URL and reason to the exception object
        e.url = "http://duznotexist.com/"
        e.reason = "URL does not exist"
        raise e # re-raise the exception, so the calling function can catch it

#main.py
import urllib2
import mymod

try:
    mymod.openurl()
except urllib2.URLError as e:
    print "Website (%s) could not be reached due to %s" % (e.url, e.reason)


I don't think re-raising the exception is an appropriate way to solve this problem.

As @Jonathan Vanasco said,

if you're opening a.com , and it 301 redirects to b.com , urlopen will automatically follow that because an HTTPError with a redirect was raised. if b.com causes the URLError , the code above marks a.com as not existing

My solution is to overwrite redirect_request of urllib2.HTTPRedirectHandler

import urllib2

class NewHTTPRedirectHandler(urllib2.HTTPRedirectHandler):
    def redirect_request(self, req, fp, code, msg, headers, newurl):
        m = req.get_method()
        if (code in (301, 302, 303, 307) and m in ("GET", "HEAD")
            or code in (301, 302, 303) and m == "POST"):
            newurl = newurl.replace(' ', '%20')
            newheaders = dict((k,v) for k,v in req.headers.items()
                              if k.lower() not in ("content-length", "content-type")
                             )
            # reuse the req object
            # mind that req will be changed if redirection happends
            req.__init__(newurl,
                headers=newheaders,
                   origin_req_host=req.get_origin_req_host(),
                   unverifiable=True)
            return req
        else:
            raise HTTPError(req.get_full_url(), code, msg, headers, fp)

opener = urllib2.build_opener(NewHTTPRedirectHandler)
urllib2.install_opener(opener)
# mind that req will be changed if redirection happends
#req = urllib2.Request('http://127.0.0.1:5000')
req = urllib2.Request('http://www.google.com/')

try:
    response = urllib2.urlopen(req)
except urllib2.URLError as e:
    print 'error'
    print req.get_full_url()
else:
    print 'normal'
    print response.geturl()

let's try to redirect the url to an unknown url:

import os
from flask import Flask,redirect

app = Flask(__name__)

@app.route('/')
def hello():
    # return 'hello world'
    return redirect("http://a.com", code=302)

    if __name__ == '__main__':
    port = int(os.environ.get('PORT', 5000))
    app.run(host='0.0.0.0', port=port)

And the result is:

error
http://a.com/

normal
http://www.google.com/
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