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lTz

open if and else

Question

hi guys, lets say i have this button x , when i press x i want to display msg fail if the item broke or (else)  msg succes if the item dont broke, but every time i get the same msg fail.

Does not matter if the object succeeds or not i get the fail msg.

How can i make this function to display succes when the item dont broke?

def test(self, PopupMessage):
            if PopupMessage:
                self.PopupMessage(localeInfo.BIO_FAIL) // msg say fail
            else:
                self.PopupMessage(localeInfo.BIO_SUC) // msg sa succes

 

Ty for help

 

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def __ClickSystemButton(self, func):
		try:
			if func == 1:
				self.Dosonthing()
			elif func == 2:
				self.PopupMessage(localeInfo.BIO_FAIL) // msg say fail
		except:
			import exception
			exception.Abort("SystemDialog.__LoadSystem.__ClickSystemButton")

 

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