苹果硅Mac催化剂的xcodebuild构建

Question

我正在使用Xcode 12 beta 4构建一个支持苹果硅的框架，该项目在Xcode中构建得很好，目标是“任意Mac”。如果我使用命令

xcodebuild -project "project.xcodeproj" -configuration "Release" -scheme "MyScheme" -destination "generic/platform=macOS,variant=Mac Catalyst,name=Any Mac"

我知道错误了

xcodebuild: error: Unable to find a destination matching the provided destination specifier:
        { generic:1, platform:macOS, variant:Mac Catalyst, name:Any Mac }
    Unsupported device specifier option.
    The device “My Mac” does not support the following options: name
    Please supply only supported device specifier options.
...
Ineligible destinations for the "MyScheme" scheme:
        { platform:iOS, id:dvtdevice-DVTiPhonePlaceholder-iphoneos:placeholder, name:Any iOS Device }
        { platform:iOS Simulator, id:dvtdevice-DVTiOSDeviceSimulatorPlaceholder-iphonesimulator:placeholder, name:Any iOS Simulator Device }
        { platform:macOS, variant:Mac Catalyst, name:Any Mac }

我不知道为什么当目标在Xcode中正确构建时被列为不合格，或者为什么在启用泛型时，xcodebuild似乎使用“Mac”设备。我怎么能在苹果硅上用Apple构建Mac催化剂呢？

Answer 1

xcodebuild -project "project.xcodeproj“-configuration发布”-scheme "MyScheme“-destination”泛型/平台=macOS，variant=Mac催化剂，name=Any Mac“

这个命令现在可以在Xcode 12 beta 5中使用。