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()