我正在研究Orange,我在OSX(10.6.5)中得到了这个名字,菜单名是'Python'而不是橙色。这是一个python / qt应用程序。我需要改变什么?
澄清:
我的info.plist:
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE plist PUBLIC "-//Apple Computer//DTD PLIST 1.0//EN" "http://www.apple.com/DTDs/PropertyList-1.0.dtd">
<plist version="1.0">
<dict>
<key>CFBundleDisplayName</key>
<string>YOORANGE</string>
<key>CFBundleExecutable</key>
<string>Orange</string>
<key>CFBundleIconFile</key>
<string>orange.icns</string>
<key>CFBundleIdentifier</key>
<string>si.ailab.Orange</string>
<key>CFBundleInfoDictionaryVersion</key>
<string>6.0</string>
<key>CFBundleName</key>
<string>Orange</string>
<key>CFBundleGetInfoString</key>
<string>Orange, component-based data mining software</string>
<key>CFBundlePackageType</key>
<string>APPL</string>
<key>CFBundleSignature</key>
<string>Orng</string>
<key>CFBundleShortVersionString</key>
<string>1.0.0</string>
<key>CFBundleVersion</key>
<string>1.0.0</string>
<key>CFBundleDocumentTypes</key>
<array>
<dict>
<key>CFBundleTypeExtensions</key>
<array>
<string>ows</string>
</array>
<key>CFBundleTypeName</key>
<string>Orange Canvas Schema</string>
<key>CFBundleTypeOSTypes</key>
<array>
<string>OWSf</string>
</array>
<key>CFBundleTypeIconFile</key>
<string>schema.icns</string>
<key>CFBundleTypeRole</key>
<string>Viewer</string>
<key>LSIsAppleDefaultForType</key>
<true/>
</dict>
</array>
</dict>
</plist>
这是Orange启动脚本,经过修改后尝试使用符号链接。显然,有些工作可行:)
#!/bin/bash
BUNDLE_DIR=`dirname $0`/../
BUNDLE_DIR=`perl -MCwd=realpath -e 'print realpath($ARGV[0])' $BUNDLE_DIR`/
FRAMEWORKS_DIR="$BUNDLE_DIR"Frameworks/
CANVAS_FILE="$FRAMEWORKS_DIR"Python.framework/Versions/2.6/lib/python2.6/site-packages/orange/OrangeCanvas/orngCanvas. pyw
cp "$FRAMEWORKS_DIR"Python.framework/Resources/Python.app/Contents/MacOS/{Python-32,AWESOME}
PYTHONEXECUTABLE="$FRAMEWORKS_DIR"Python.framework/Resources/Python.app/Contents/MacOS/AWESOME
PYTHONHOME="$FRAMEWORKS_DIR"Python.framework/Versions/2.6/
#DYLD_FRAMEWORK_PATH="$FRAMEWORKS_DIR"${DYLD_FRAMEWORK_PATH:+:$DYLD_FRAMEWORK_PATH}
DYLD_FRAMEWORK_PATH="$FRAMEWORKS_DIR":"$BUNDLE_DIR"Resources/Qt4/lib${DYLD_FRAMEWORK_PATH:+:$DYLD_FRAMEWORK_PATH}
export PYTHONEXECUTABLE
export PYTHONHOME
export DYLD_FRAMEWORK_PATH
export DYLD_LIBRARY_PATH="$BUNDLE_DIR"Resources/openbabel/lib/:"$BUNDLE_DIR"Resources/openbabel/lib/openbabel/2.2.3/:$ DYLD_LIBRARY_PATH
# LaunchServices passes the Carbon process identifier to the application with -psn paramter - we do not want it
if [[ "$1" == -psn* ]] ; then
shift
fi
echo "$0"
echo "$PYTHONEXECUTABLE"
echo "$@"
exec -a "$0" "$PYTHONEXECUTABLE" "$CANVAS_FILE" "$@"
答案 0 :(得分:2)
我想提出另一种解决方案。 它使用Cocoa设置应用程序名称, 无需摆弄符号链接。 在使用PyQt之前,请先在主模块中运行它。
if sys.platform.startswith('darwin'):
# Set app name, if PyObjC is installed
# Python 2 has PyObjC preinstalled
# Python 3: pip3 install pyobjc-framework-Cocoa
try:
from Foundation import NSBundle
bundle = NSBundle.mainBundle()
if bundle:
app_name = os.path.splitext(os.path.basename(sys.argv[0]))[0]
app_info = bundle.localizedInfoDictionary() or bundle.infoDictionary()
if app_info:
app_info['CFBundleName'] = app_name
except ImportError:
pass
答案 1 :(得分:1)
简短回答 - 这不是一件容易的事,因为你正在运行python解释器,它是自己的程序,有自己的默认标题。
答案很长 - 如果设置了环境变量PYTHONSTARTUP,Python解释器将在启动时执行shell脚本。您可以通过probalby设置窗口标题。
每次在启动之前必须做一些额外的环境变量时,要获得一个更改的标题。
答案 2 :(得分:1)
这是一个骗子,但我相信它有效,
创建一个指向python应用程序的sym链接。
ln -s / opt / local / bin / python / opt / local / bin / orange
然后调用orange而不是python来启动脚本
答案 3 :(得分:1)
事实证明,有一个更深层次的Info.plist捆绑在Contents/Frameworks/Python.framework/Versions/2.6/Resources/Python.app/Contents/Info.plist
,正在调用。