TimeOut Exception handling

Question

I have written the code as like below by providing delay (time.sleep()), but I wanted modify code by invoking try exceptions functionality. So to modify below code.

def min_power(self):
        power_min_list = ["Carrier", "@#$%", "-1", "101", "22"]
        time.sleep(6)
        for i in range(len(power_min_list)):
            if i == 0:
                pencil()
            power_min = web.find_element_by_id("powmin")
            power_min.clear()
            power_min.send_keys(power_min_list[i])
            submit()
            time.sleep(3)
            var1 = web.find_element_by_xpath("//div[@id='grp_powmin']//span[contains(text(),'Number must be between 0 and 100.')]").text
            var2 = "Number must be between 0 and 100."
            if var1 == var2:
                pass
            else:
                print ("Failed - invalid data written successfully")
            if i == 3:
                power_min = web.find_element_by_id("powmin")
                power_min.clear()
                power_min.send_keys(power_min_list[4])
                submit()
        time.sleep(6)
        var = power_min_list[4]
        var3 = web.find_element_by_xpath("//*[@id='modulation_table']/tbody/tr[1]/td[3]").text
        if var == var3:
            pass
        else:
            print ("Failed - Power Min wrong Data is written successfully")

here the question is how to modify this code by invoking try exception block ? and that will be generic for entire definition. — Ramesh, Jul 11 '17 at 05:11

score 0 · Answer 1 · answered Jul 11 '17 at 09:57

0

You can use wait for element instead time.sleep()

Selenium waitForElement

You cannot remove sleep in second line in method min_power?

answered Jul 11 '17 at 09:57

qchar90

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TimeOut Exception handling

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