我宁愿最终用户可以访问我的 Python 代码，以便他们查看并改进代码，所以我对隐藏代码的选项不感兴趣。我发现我可以让 wx 应用程序（尚未尝试使用 Qt）使用不同的标准 Python 解释器以指定的应用程序名称运行，方法是在我安装它们时将它们包装成一个包（例如，作为 conda 包安装过程的一部分） ）。这是一个简短的脚本，它使用执行脚本的 Python 创建具有指定名称的 .app 包。使用创建的软链接（参见下面的 pyAlias）运行脚本，如下所示
./MyApplication.app/Contents/MacOS/MyApplication /path/MyApplication.py
做这项工作。
from __future__ import division, print_function
import os,sys
appName = 'MyApplication' # name of app
scriptdir = '.'
iconfile = os.path.join(scriptdir,'MyApplication.icns') # optional icon file
if __name__ == '__main__':
wrapApp = os.path.join(scriptdir,appName+'.app','Contents')
os.makedirs(wrapApp)
wrapPy = os.path.join(wrapApp,'MacOS')
os.makedirs(wrapPy)
fp = open(os.path.join(wrapApp,'PkgInfo'),'w')
fp.write('APPL????\n')
fp.close()
fp = open(os.path.join(wrapApp,'Info.plist'),'w')
fp.write('''<?xml version="1.0" encoding="UTF-8"?>\n<!DOCTYPE plist SYSTEM "file://localhost/System/Library/DTDs/PropertyList.dtd">\n<plist version="0.9">\n<dict>\n <key>CFBundleIconFile</key>\n <string></string>\n <key>CFBundlePackageType</key>\n <string>APPL</string>\n <key>CFBundleGetInfoString</key>\n <string>Created in makeApp.py (Brian Toby/QMake</string>\n <key>CFBundleSignature</key>\n <string>????</string>\n <key>CFBundleExecutable</key>\n <string>{:}</string>\n <key>CFBundleIdentifier</key>\n <string>com.continuum.python</string>\n <key>NSPrincipalClass</key>\n <string>NSApplication</string>\n</dict>\n</plist>\n'''.format(appName))
fp.close()
pyAlias = os.path.join(wrapPy,appName)
pythonpath = os.path.realpath(sys.executable)
os.symlink(pythonpath,pyAlias)
if os.path.exists(iconfile):
shutil.copyfile(iconfile,oldicon)
print('Use',pyAlias,'to run wxPython scripts')
以下是一个较旧的答案，我将保留任何历史价值。现在在我的应用程序中，我使用了它们之间的混合，在其中我创建了一个拖放式 AppleScript 包（如下所示），我在其中放置了一个指向 Python 的软链接（命名为与我的应用程序匹配），如上所示。
基于 Christopher Bruns 的回答，以及“如何通过 Python 为 Python 脚本创建 Mac 应用程序包”中的脚本，这里是一个 Python 脚本，它为用户 Python 脚本创建一个包（应用程序），它将显示应用程序名称而不是菜单中的“Python”。为此，它会尝试定位 Python 的捆绑版本，并使用应用程序名称符号链接到该版本。我用 wxpython 脚本对其进行了测试，但它也应该适用于 Qt。
用户脚本从其原始位置运行，而不是放置在应用程序中。如果您想将您的脚本（s）放入一个包（与 python 一起）以进行重新分发，请参阅py2app。
#!/usr/bin/env python
'''This creates an app to launch a python script. The app is
created in the directory where python is called. A version of Python
is created via a softlink, named to match the app, which means that
the name of the app rather than Python shows up as the name in the
menu bar, etc, but this requires locating an app version of Python
(expected name .../Resources/Python.app/Contents/MacOS/Python in
directory tree of calling python interpreter).
Run this script with one or two arguments:
<python script>
<project name>
The script path may be specified relative to the current path or given
an absolute path, but will be accessed via an absolute path. If the
project name is not specified, it will be taken from the root name of
the script.
'''
import sys, os, os.path, stat
def Usage():
print("\n\tUsage: python "+sys.argv[0]+" <python script> [<project name>]\n")
sys.exit()
version = "1.0.0"
bundleIdentifier = "org.test.test"
if not 2 <= len(sys.argv) <= 3:
Usage()
script = os.path.abspath(sys.argv[1])
if not os.path.exists(script):
print("\nFile "+script+" not found")
Usage()
if os.path.splitext(script)[1].lower() != '.py':
print("\nScript "+script+" does not have extension .py")
Usage()
if len(sys.argv) == 3:
project = sys.argv[2]
else:
project = os.path.splitext(os.path.split(script)[1])[0]
# find the python application; must be an OS X app
pythonpath,top = os.path.split(os.path.realpath(sys.executable))
while top:
if 'Resources' in pythonpath:
pass
elif os.path.exists(os.path.join(pythonpath,'Resources')):
break
pythonpath,top = os.path.split(pythonpath)
else:
print("\nSorry, failed to find a Resources directory associated with "+str(sys.executable))
sys.exit()
pythonapp = os.path.join(pythonpath,'Resources','Python.app','Contents','MacOS','Python')
if not os.path.exists(pythonapp):
print("\nSorry, failed to find a Python app in "+str(pythonapp))
sys.exit()
apppath = os.path.abspath(os.path.join('.',project+".app"))
newpython = os.path.join(apppath,"Contents","MacOS",project)
projectversion = project + " " + version
if os.path.exists(apppath):
print("\nSorry, an app named "+project+" exists in this location ("+str(apppath)+")")
sys.exit()
os.makedirs(os.path.join(apppath,"Contents","MacOS"))
f = open(os.path.join(apppath,"Contents","Info.plist"), "w")
f.write('''<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
<plist version="1.0">
<dict>
<key>CFBundleDevelopmentRegion</key>
<string>English</string>
<key>CFBundleExecutable</key>
<string>main.sh</string>
<key>CFBundleGetInfoString</key>
<string>{:}</string>
<key>CFBundleIconFile</key>
<string>app.icns</string>
<key>CFBundleIdentifier</key>
<string>{:}</string>
<key>CFBundleInfoDictionaryVersion</key>
<string>6.0</string>
<key>CFBundleName</key>
<string>{:}</string>
<key>CFBundlePackageType</key>
<string>APPL</string>
<key>CFBundleShortVersionString</key>
<string>{:}</string>
<key>CFBundleSignature</key>
<string>????</string>
<key>CFBundleVersion</key>
<string>{:}</string>
<key>NSAppleScriptEnabled</key>
<string>YES</string>
<key>NSMainNibFile</key>
<string>MainMenu</string>
<key>NSPrincipalClass</key>
<string>NSApplication</string>
</dict>
</plist>
'''.format(projectversion, bundleIdentifier, project, projectversion, version)
)
f.close()
# not sure what this file does
f = open(os.path.join(apppath,'Contents','PkgInfo'), "w")
f.write("APPL????")
f.close()
# create a link to the python app, but named to match the project
os.symlink(pythonapp,newpython)
# create a script that launches python with the requested app
shell = os.path.join(apppath,"Contents","MacOS","main.sh")
# create a short shell script
f = open(shell, "w")
f.write('#!/bin/sh\nexec "'+newpython+'" "'+script+'"\n')
f.close()
os.chmod(shell, os.stat(shell).st_mode | stat.S_IXUSR | stat.S_IXGRP | stat.S_IXOTH)