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Consider a scenario where in QuickTest, the synchronization timeout is set to 10 seconds and the global timeout is set to 10 seconds. However, the actual time taken by the object to be visible is 25 seconds. How long will QuickTest wait for the synchronization to take place?
a) 20 seconds
b) 25 seconds
c) 10 seconds
d) 35 seconds
20 seconds.
QuickTest won't be able to identify the object, and the test will fail since it takes 5 more seconds for the object to appear.
Global Timeout + Sync. Timeout = Total Timeout
10 seconds
[QUOTE=jainbrijesh;25454]Consider a scenario where in QuickTest, the synchronization timeout is set to 10 seconds and the global timeout is set to 10 seconds. However, the actual time taken by the object to be visible is 25 seconds. How long will QuickTest wait for the synchronization to take place?
a) 20 seconds
b) 25 seconds
c) 10 seconds
d) 35 seconds[/QUOTE]
Hi Brijesh,
Really a nice objective type question asked in QTP!!!
The answer will be [B]20 seconds [/B]as explained by Anshoo.
Regards,
Ganesan
The question is, "How long will QuickTest wait for the synchronization to take place?". I believe that QTP initializes for 10 seconds (Global timeout) before QTP will begin the sync timeout. So QTP will wait 10 seconds for the object to appear after the global timeout. Make sense?
You are absolutely correct. As the script loads, global timeout kicks in and after that, sync step timeout initializes, which is 10 seconds. But, the total timeout is: global timeout + sync step timeout. In our scenario, as qtp initiates and waits for the object to load, it spends 10 seconds of global timeout and then, 10 seconds of sync step timeout. This is the total timeout instructed to qtp to wait for an object to load, therefore, when both timeouts “burn-out”, qtp fails the script as the object takes 25 seconds (5 seconds longer) to appear. Yes, what you said is correct. Qtp will wait for 10 seconds after the global timeout. But in all, it will wait for 20 seconds. This wait is also the amount of time qtp waits for synchronization to take place. Let me make use of the “insert done…” example from the flights application. If you insert a synchronization point of 10 seconds with a global timeout of 20 seconds, qtp will wait for a total of 30 seconds for “insert done…” to appear. However, it doesn’t take that long to load and appears before the total timeout (global+syncstep), so the script moves on. Now, try the same with a sync step timeout of 0. See how long qtp waits? it will wait if global timeout<>0. In other words, since sync step timeout is 0, qtp will only wait if global timeout>=1. If you use the default value of global timeout (20 seconds), with sync step timeout of 0 seconds, the total timeout will be 20 seconds. It will only wait for 20 seconds for any object to appear. In other words, qtp will only wait 20 seconds for synchronization to take place. I hope i didn’t go in circles. It’s a friday.
Please see the attachment. I don't know why the Editor won't format my stuff. :(
Thank you for the explanation. I stand corrected but you have to admit that the exercise brought out the thinking caps.
Thank you too, Fred. :)
[QUOTE=Anshoo_Arora;25591]You are absolutely correct. As the script loads, global timeout kicks in and after that, sync step timeout initializes, which is 10 seconds. But, the total timeout is: global timeout + sync step timeout. In our scenario, as qtp initiates and waits for the object to load, it spends 10 seconds of global timeout and then, 10 seconds of sync step timeout. This is the total timeout instructed to qtp to wait for an object to load, therefore, when both timeouts “burn-out”, qtp fails the script as the object takes 25 seconds (5 seconds longer) to appear. Yes, what you said is correct. Qtp will wait for 10 seconds after the global timeout. But in all, it will wait for 20 seconds. This wait is also the amount of time qtp waits for synchronization to take place. Let me make use of the “insert done…” example from the flights application. If you insert a synchronization point of 10 seconds with a global timeout of 20 seconds, qtp will wait for a total of 30 seconds for “insert done…” to appear. However, it doesn’t take that long to load and appears before the total timeout (global+syncstep), so the script moves on. Now, try the same with a sync step timeout of 0. See how long qtp waits? it will wait if global timeout<>0. In other words, since sync step timeout is 0, qtp will only wait if global timeout>=1. If you use the default value of global timeout (20 seconds), with sync step timeout of 0 seconds, the total timeout will be 20 seconds. It will only wait for 20 seconds for any object to appear. In other words, qtp will only wait 20 seconds for synchronization to take place. I hope i didn’t go in circles. It’s a friday.
Please see the attachment. I don't know why the Editor won't format my stuff. :([/QUOTE]
Very good explanation Anshoo....
I think if you face the problem with the alignment, then you can use shift enters, inserting HTML tags manually for example <b </b> for bold etc., it may solve you.
Regards,
Ganesan