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Re: [Question #121539]: openApp parameter support

 

Question #121539 on Sikuli changed:
https://answers.launchpad.net/sikuli/+question/121539

NMANOS posted a new comment:
Thanks RayMan,
Eventually that's the Function I use:

def launchWinApp(command, windowTitle=''):
	type("r", KEY_WIN)
	wait(1)
	App.focus('Run')
	paste(command.encode('string-escape'))
	type(Key.ENTER)
	wait(5)
	if exists(windowTitle):
		winApp = App(windowTitle)
	else:
		winApp = App.focusedWindow()
		windowTitle = winApp.text().splitlines()[0]
	print 'Returning Application with Title: %s' % windowTitle
	return winApp

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