Try - Except in Python for given amount of time

Question

I am using Marathon (Java desktop application testing tool) to automate regression testing. Marathon uses Jython so I can use Java libraries and Python libraries. As my script goes through filling out certain fields, various fields appear (or don't appear) based on values that I entered in previous fields. I need to skip over the fields that aren't there for the obvious reason. When fields are disabled, but still there, this is fine because I can use

    if Component.isEnabled():
        #do something
    else:
        #do something

The problem is when the component doesn't exist. Is there any way, in Java, to test for the existence of a component? For example, Component.exists() would suit my needs, but there isn't such a method in the component class.

I would prefer to solve my problem by using an if Component.exists(): statement, but I am able to get around it using a try, except block. However, this leads to major execution time issues for the script. It tries to find the component for ~2 or 3 minutes before it throws an exception. The only way I could see around this issue is if there was some sort of statement like try for x seconds and move on if component is not found. Is there any way to limit the amount of time you try any given statement?

On the Java side you can use a `FutureTask` and an `Executor`. `FutureTask` implements `Future` which has a `.get()` with timeout (throwing `TimeoutException` if timeout) — fge, Mar 20 '14 at 15:20

score 2 · Answer 1 · edited May 23 '17 at 12:05

I found some code to throw a time-out exception in your code as an answer to another Stackoverflow-question: What should I do if socket.setdefaulttimeout() is not working?, it however only works on linux-machines as stated in the link.

Concretely:

import signal, time

class Timeout():
  """Timeout class using ALARM signal"""
  class Timeout(Exception): pass

  def __init__(self, sec):
    self.sec = sec

  def __enter__(self):
    signal.signal(signal.SIGALRM, self.raise_timeout)
    signal.alarm(self.sec)

  def __exit__(self, *args):
    signal.alarm(0) # disable alarm

  def raise_timeout(self, *args):
    raise Timeout.Timeout()

# Run block of code with timeouts
try:
  with Timeout(60):
    #do something

except Timeout.Timeout:
  #do something else

This will try "do something" for 60 seconds and moves on if the execution exceeds 60 seconds...

I appreciate the response, but unfortunately I don't have a Linux machine. — Nick L, Mar 21 '14 at 18:56
This example worked for what I was looking for. However, it seems I can not catch the error name. For instance, if you catch the exception like this `except Exception as e:` and do a `print(e)`, nothing is printed. — muammar, Jul 16 '20 at 18:31

score 1 · Answer 2 · answered Nov 03 '14 at 10:56

this is the solution to your problem:

import signal
def time_out(self):
    def signal_handler(signum, frame):
        raise Exception("Timed out!")

    signal.signal(signal.SIGALRM, signal_handler)
    signal.alarm(3)  
    all_games = []
    try:
        while True:
            #do what you want to do

    except Exception as msg:
        print msg

3 is the time for the function time out

best regards, Shlomy

Try - Except in Python for given amount of time

2 Answers2