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Re: [Question #267140]: App.open() open exe file twice on windows 7. How to launch only one instance?

To: sikuli-driver@xxxxxxxxxxxxxxxxxxx
From: RaiMan <question267140@xxxxxxxxxxxxxxxxxxxxx>
Date: Wed, 20 May 2015 07:07:16 -0000
Reply-to: question267140@xxxxxxxxxxxxxxxxxxxxx
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Question #267140 on Sikuli changed:
https://answers.launchpad.net/sikuli/+question/267140

    Status: Open => Answered

RaiMan proposed the following answer:
this works on Win 7 for me:

    String pgmPath = "\"C:\\Program Files\\Microsoft Games\\Minesweeper\\MineSweeper.exe\"";
    App pgm = new App(pgmPath);
    pgm.open();
    RunTime.pause(2);
    pgm.focus();

Take care:
The program path has to obey the rules how it has to be specified on command line, which in this case means, that I have to add extra apostrophes before and after (need to be escaped of course), because the path contains blanks.

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