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[Chetan <question698815@xxxxxxxxxxxxxxxxxxxxx>]

New question #698815 on SikuliX:
https://answers.launchpad.net/sikuli/+question/698815

Hello
i would like to launch application acrobat.exe and mention the file name to open
i am able to do via command line and also via subprocess.Popen.
however it does not work with openApp
tried searching forum but no lock
below is the code

command line #works
"C:\\Program Files\\Adobe\\Acrobat Trunk\\Acrobat\\Acrobat.exe" C:\\Test.pdf   

   path=r"C:\Program Files\Adobe\Acrobat Trunk\Acrobat\Acrobat.exe"
    #path="C:\\Program Files\\Adobe\\Acrobat Trunk\\Acrobat\\Acrobat.exe"
    file="C:\\test.pdf"

#works
    subprocess.Popen([path,file])

    
# Does NOT work 
  openApp(path+file)

any pointers would be appreciated

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