4

In Alias mode, this fails.

Here's what I'm running:

OSX 10.11.1 (El Capitan)

Python 3.5 via Anaconda (with the Jupyter 4.1.0 Notebook)

py2app is the newest version (0.9)

Here's what I'm doing:

Create script and save as APP_OSX.py

Enter the following commands in terminal:

    py2applet --make-setup APP_OSX.py

    rm -rf build dist

    python setup.py py2app -A

This is after I installed a regular version of Python 3.5.1 from python.org (because there was an issue with using py2app and my Anaconda Python 3.5.1 version).

Then I find the bundled app under the 'dist' folder and open it and a box pops up with the name of my app and "error".

From the console, I get this:

3/30/16 7:37:18.972 PM APP_OSX[5819]: 2016-03-30 19:37:18.971 APP_OSX[5819:746261] APP_OSX Error
3/30/16 7:37:21.511 PM sharedfilelistd[242]: SecTaskLoadEntitlements failed error=22
3/30/16 7:37:21.585 PM Console[5822]: Failed to connect (_consoleX) outlet from (NSApplication) to (ConsoleX): missing setter or instance variable
3/30/16 7:37:25.893 PM WindowServer[161]: send_datagram_available_ping: pid 349 failed to act on a ping it dequeued before timing out.
3/30/16 7:37:51.601 PM Console[5822]: Persistent UI failed to open file file:///Users/mi/Library/Saved%20Application%20State/com.apple.Console.savedState/window_3.data: Permission denied (13)

Not sure if this helps, but the script uses the following libraries:

Numpy

Pandas

Easygui

Thanks in advance!

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  • "UI failed to open file" seems like a permissions problem. On a fresh install of py2app try bundling a simple hello world python script (`print 'hellow world'`), and see what happens. That will help us figure out if your application has anything to do with the error. – James Apr 13 '16 at 01:04

0 Answers0