How to build a release version of an iOS framework in Xcode?

Question

Let\'s say I do the following:

1. Open Xcode 7
2. File | New | Project | Cocoa Touch Framework
3. Create \"TestFramework\" with the Swift language

Answer 1

An alternative to building a framework via the Xcode IDE is to build it from the command line.

You can produce a release build of your framework for iphoneos devices with the following command:

xcodebuild -workspace TestSDK.xcworkspace -scheme TestSDK -configuration Release -sdk iphoneos

You can change the value of the -configuration argument from Release to Debug in order to produce a debug build, or change the value of the -sdk argument from iphoneos to iphonesimulator in order to produce a build for Simulator devices.

Note that you may need to provide the -project argument instead of -workspace if your target is part of an Xcode project only and not part of an Xcode workspace. Run the xcodebuild -help command for the full list of xcodebuild options.

If you prefer to archive, you can do that from the command line also, as follows:

xcodebuild archive -workspace TestSDK.xcworkspace -scheme TestSDK -configuration Release -sdk iphoneos -archivePath "TestSDK_Release_iphoneos.xcarchive" SKIP_INSTALL=NO

Note that you can specify SKIP_INSTALL=NO as part of your project or target's Build Settings instead if you prefer.

Lastly, if you want to join up your iphoneos and iphonesimulator builds into a single binary, you can do that with the xcodebuild -create-xcframework command as follows:

xcodebuild -create-xcframework \
    -framework "TestSDK_Release_iphoneos.xcarchive/Products/Library/Frameworks/TestSDK.framework" \
    -framework "TestSDK_Release_iphonesimulator.xcarchive/Products/Library/Frameworks/TestSDK.framework" \
    -output "TestSDK.xcframework"

See here for the official guide to creating an XCFramework.