Asynchronous fetch request with Google App Engine

Question

I'm reading the documentation for asynchronous fetch requests in GAE. Python isn't my first language, so I'm having trouble finding out what would be best for my case. I don't really need or care about the response for the request, I just need it to send the request and forget about it and move on to other tasks.

So I tried code like in the documentation:

from google.appengine.api import urlfetch

rpc = urlfetch.create_rpc()
urlfetch.make_fetch_call(rpc, "http://www.google.com/")

# ... do other things ...

try:
    result = rpc.get_result()
    if result.status_code == 200:
        text = result.content
        # ...
except urlfetch.DownloadError:
    # Request timed out or failed.
    # ...

But this code doesn't work unless I include try: and except, which I really don't care for. Omitting that part makes the request not go through though.

What is the best option for creating fetch requests where I don't care for the response, so that it just begins the request, and moves on to whatever other tasks there are, and never looks back?

Answer 1

Just do your tasks where the

# ... do other things ...

comment is. When you are otherwise finished, then call rpc.wait(). Note that it's not the try/except that's making it work, it's the get_result() call. That can be replaced with wait().

So your code would look like this:

from google.appengine.api import urlfetch

rpc = urlfetch.create_rpc()
urlfetch.make_fetch_call(rpc, "http://www.google.com/")

# ... do other things ... << YOUR CODE HERE

rpc.wait()

Answer 2

If you don't care about the response, the response may take a while, and you don't want your handler to wait for it to complete before returning a response to the user, you may want to consider firing off a task queue task that makes the request rather than doing it within your user-facing handler.