[Tutor] URL test function.

Steven D'Aprano steve at pearwood.info
Sat Oct 23 02:45:47 CEST 2010

On Sat, 23 Oct 2010 01:54:47 am Jeff Honey wrote:

> def is_ready():
>  with settings(
>   warn_only=True
>  ):
>   try:
>    urllib2.urlopen('http://'+env.host_string+'\:8080')
>   except urllib2.URLError:
>    time.sleep(10)
>    urllib2.urlopen('http://'+env.host_string+'\:8080')

That will only try twice, then give up with an exception.

> I am trying to get the 'host_string' environment variable to plug in
> here but not sure if the parent function will pass along that
> information nor if I can call it like this.

Have you tried it to see what happens?

> Additionally, I have 
> never used try/except so I don't know if this will just try twice and
> quit. As it is now, it raises URLError and bombs out.

What does the error say?

I would try something like this:

def urlopen_retry(url, maxretries=10, time_to_wait=10):
    """Call urlopen on a url. If the call fails, try again up to a
    maximum of maxretries attempts, or until it succeeds. Delay 
    between attempts by time_to_wait seconds, increasing each time."""
    if maxretries < 1:
        return None
    for i in range(maxretries):
            return urllib2.urlopen(url)
        except urllib2.URLError:
            # Delay by (e.g.) 10 seconds, then 20, then 30, then ... 
    raise  # re-raise the last exception

and then call it with:

def is_ready():
   url = 'http://' + env.host_string + '\:8080'
   with settings(warn_only=True):
       return urlopen_retry(url)

Hope this helps.

Steven D'Aprano

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