b50e5cf3a6
Before this you'd need to set the bundle name to launch the app via the Xcode debugger. With this it's set automatically. Change-Id: Ic84a6c8ba020580d5ff4afa9c104efbc2360b60e Reviewed-on: https://skia-review.googlesource.com/c/skia/+/334898 Commit-Queue: Jim Van Verth <jvanverth@google.com> Reviewed-by: John Stiles <johnstiles@google.com> Reviewed-by: Herb Derby <herb@google.com>
38 lines
1.3 KiB
Python
Executable File
38 lines
1.3 KiB
Python
Executable File
#!/usr/bin/env python2.7
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#
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# Copyright 2017 Google Inc.
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#
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# Use of this source code is governed by a BSD-style license that can be
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# found in the LICENSE file.
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import os
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import sys
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# Arguments to the script:
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# app path to binary to package, e.g. out/Debug/gen/dm
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# bundle_prefix the first part of the bundle ID, e.g. com.google (no trailing '.')
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# the app name will be appended to this to create the full bundle ID
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app,bundle_prefix = sys.argv[1:3]
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out, app = os.path.split(app)
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# Write a minimal Info.plist to name the package and point at the binary.
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with open(os.path.join(out, app + '_Info.plist'), 'w') as f:
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f.write('''
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<plist version="1.0">
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<dict>
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<key>CFBundleVersion</key> <string>0.1.0</string>
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<key>CFBundleShortVersionString</key> <string>0.1.0</string>
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<key>CFBundleExecutable</key> <string>{app}</string>
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<key>CFBundleIdentifier</key> <string>{bundle_prefix}.{app}</string>
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<key>CFBundlePackageType</key> <string>APPL</string>
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<key>LSRequiresIPhoneOS</key> <true/>
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<key>UIDeviceFamily</key> <array>
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<integer>1</integer>
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<integer>2</integer>
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</array>
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<key>UILaunchStoryboardName</key> <string>LaunchScreen</string>
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</dict>
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</plist>
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'''.format(app=app, bundle_prefix=bundle_prefix))
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