Fetching url in python with google app engine

Question

I'm using this code to send Http request inside my app and then show the result:

def get(self):
      url = "http://www.google.com/"
      try:
          result = urllib2.urlopen(url)
          self.response.out.write(result)
      except urllib2.URLError, e:

I expect to get the html code of google.com page, but I get this sign ">", what the wrong with that ?

IanGSY · Accepted Answer · 2014-04-12 22:00:04Z

5

Try using the urlfetch service instead of urllib2:

Import urlfetch:

from google.appengine.api import urlfetch

And this in your request handler:

def get(self):
    try:
        url = "http://www.google.com/"
        result = urlfetch.fetch(url)
        if result.status_code == 200:
            self.response.out.write(result.content)
        else:
            self.response.out.write("Error: " + str(result.status_code))
    except urlfetch.InvalidURLError:
        self.response.out.write("URL is an empty string or obviously invalid")
    except urlfetch.DownloadError:
        self.response.out.write("Server cannot be contacted")

See this document for more detail.

answered Apr 12, 2014 at 22:00

IanGSY

3,7141 gold badge25 silver badges41 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

user3516596 Over a year ago

Thank you for your response, but what the difference ?

IanGSY Over a year ago

Not much, in fact when running on App Engine the urllib2 library performs HTTP requests using App Engine's URL fetch service, which runs on Google's scalable HTTP request infrastructure, so behind the scenes they do the same thing, see developers.google.com/appengine/docs/python/urlfetch/…

Gwyn Howell · Accepted Answer · 2014-04-12 21:59:36Z

4

You need to call the read() method to read the response. Also good practice to check the HTTP status, and close when your done.

Example:

url = "http://www.google.com/"
try:
    response = urllib2.urlopen(url)

    if response.code == 200:
        html = response.read()
        self.response.out.write(html)
    else:
        # handle

    response.close()

except urllib2.URLError, e:
    pass

answered Apr 12, 2014 at 21:59

Gwyn Howell

5,4142 gold badges34 silver badges48 bronze badges

Collectives™ on Stack Overflow

Fetching url in python with google app engine

2 Answers 2

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related