如何使Xcode Run Script x86_64兼容

Question

Hello hello社区！

所以这里是我的问题，welll不是一个真正的问题，但下面的运行脚本编译我的静态库，它的工程很棒！我唯一的问题是，它没有编译它的模拟器，我得到一个x86_64错误。我知道我可以编辑此代码以使其兼容可以有人告诉我我需要做什么？

# define output folder environment variable 
UNIVERSAL_OUTPUTFOLDER=${BUILD_DIR}/${CONFIGURATION}-universal 

# Step 1. Build Device and Simulator versions 
xcodebuild -target ${PROJECT_NAME} ONLY_ACTIVE_ARCH=NO -configuration ${CONFIGURATION} -sdk iphoneos BUILD_DIR="${BUILD_DIR}" 
BUILD_ROOT="${BUILD_ROOT}" xcodebuild -target ${PROJECT_NAME} -configuration  ${CONFIGURATION} -sdk iphonesimulator -arch i386 BUILD_DIR="${BUILD_DIR}"  BUILD_ROOT="${BUILD_ROOT}" 

# make sure the output directory exists 
mkdir -p "${UNIVERSAL_OUTPUTFOLDER}" 

# Step 2. Create universal binary file using lipo 
lipo -create -output "${UNIVERSAL_OUTPUTFOLDER}/lib${PROJECT_NAME}.a"  "${BUILD_DIR}/${CONFIGURATION}-iphoneos/lib${PROJECT_NAME}.a" "${BUILD_DIR}/${CONFIGURATION}-iphonesimulator/lib${PROJECT_NAME}.a" 

# Last touch. copy the header files. Just for convenience 
cp -R "${BUILD_DIR}/${CONFIGURATION}-iphoneos/include" "${UNIVERSAL_OUTPUTFOLDER}" 

Answer 1

在引用“构建框架”项目完成here。我能够发现，我可以通过添加额外的“-arch”参数来添加额外的构建。并只是使xcodebuild命令

xcodebuild -target ${PROJECT_NAME} -configuration  ${CONFIGURATION} -sdk iphonesimulator -arch i386 -arch x86_86 BUILD_DIR="${BUILD_DIR}"  BUILD_ROOT="${BUILD_ROOT}" 

它解决了我的问题！以防万一有人像我一样有一个缓慢的时刻。大声笑