How to handle the window twice

[shiv 1]

For $i = 1 To 10
        If WinExists("My window") Then
        ControlClick("My window","NO","[CLASS:Button; INSTANCE:1]")
        Sleep(2000)
        WinWaitActive("Check your window","test: yahoo.com",2)
        ControlClick("Check your window","test: yahoo.com","Button2")
ExitLoop
        EndIf
        Sleep(1000)
        Next

Some time "My window" is pop up 1 time and some time it is pop up twice.

The above code handle this window "My window" successfully for the 1st time its thrown but when it occurs 2nd time it unable to handle it.

Please provide some refactoring for the above code so that it can handle the pop up window "My window" 2 times

[JohnOne 1,589]

while 1
        If WinExists("My window") Then
        ControlClick("My window","NO","[CLASS:Button; INSTANCE:1]")
        Sleep(2000)
        WinWaitActive("Check your window","test: yahoo.com",2)
        ControlClick("Check your window","test: yahoo.com","Button2")
;ExitLoop
        EndIf
        Sleep(1000)
wend

[shiv 1]

Hi John thanks a lot,

But suppose if window  "My Window" is not pop up , not for once then for how long would it wait for the window to exist.

And if window not exist then would it get stuck there or will execute my code further.

Actually in my application this window occurs after 2 days. so i just want to write a code which checks the presence of this window. if the window exist then it can handle it .(here the probability of occurring the window is 2 times)

thanks

[NewPlaza 4]

The code above will wait 1 second.... Then loop again...

But if it does find 'MyWindow' then it will wait a maximum of 4 seconds for 'CheckYourWindow' to be active.

Edited March 14, 2014 by NewPlaza

[shiv 1]

Its not only about that how long would it wait. PLEASE go through again my query above

[NewPlaza 4]

How often do you run that bit of code?

Edited March 14, 2014 by NewPlaza

[JohnOne 1,589]

Most code has a main while loop.

In the absence of solid information about when and when not your target window appears twice or how long you're prepared to wait for a second window, logic suggests you check for it infinitely in that main while loop.