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Home/ Questions/Q 442743
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T21:06:07+00:00 2026-05-12T21:06:07+00:00

I am using Marathon 2.0b4 to automate tests for an application. A shortcoming of

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I am using Marathon 2.0b4 to automate tests for an application.

A shortcoming of wait_p, one of the script elements provided by Marathon, is that its default timeout is hardcoded to be 60 seconds. I needed a larger timeout due to the long loading times in my application.
[I considered patching Marathon, but didn’t want to maintain parallel versions etc., so figured that a better solution would actually be a workaround at the test script level.]

def wait_p_long(times, compID_name, ppty_name, ppty_value, compID_cell=None):
    from marathon.playback import *
    """Wrapper around wait_p which takes exactly the same parameters as wait_p,
    except that an extra first parameter is used to specify the number of times
    wait_p is called"""
    for i in range(1, times):
        try:
            wait_p(compID_name, ppty_name, ppty_value, compID_cell)
        except:
            if (i < times):
                print "wait_p failed, trying again"
            else:
                raise

wait_p is short for "wait property", and it takes in 3 compulsory and one optional argument (the argument’s names are rather self-explanatory), and what it does is wait for a speicifed property of the specified component to be equals to the specified value.

What the above method (Jython) intends to do is take one extra parameter, times, which specifies the number of times to attempt wait_p, suppressing the exceptions up until the last try.

However, this method isn’t working for me, and I am afraid there might be some syntactical or logical error somewhere in there. Any comments from python / jython gurus out there?

Thanks!

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T21:06:08+00:00Added an answer on May 12, 2026 at 9:06 pm

    Two things:

    • range(1, times) should almost certainly be range(times); what you wrote is equivalent to for (int i=1; i < times; i++)
    • Because of what I just explained, if (i < times) will always be True in your except block

    If this doesn’t help with your problem, please describe how exactly your results are differing from what you expect.

    The results would look something like:

    def wait_p_long(times, compID_name, ppty_name, ppty_value, compID_cell=None):
        from marathon.playback import *
        """
        Wrapper around wait_p which takes exactly the same parameters as wait_p,
        except that an extra first parameter is used to specify the number of times
        wait_p is called.
        """
        for i in range(times):
            try:
                wait_p(compID_name, ppty_name, ppty_value, compID_cell)
            except:
                if i == times - 1:
                    raise
                else:
                    print "wait_p failed, trying again"
    
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