iOS 框架缺少所需的架构 arm64

Question

我正在尝试在 iOS 中创建和使用框架。它以前可以工作，并且可以在模拟器上工作，但不能在较新的 iPhone 6 上工作。

我怀疑它与脚本有关，但据我所见（我大部分是为互联网复制脚本）它确实包含 arm64：

set -e
set +u
# Avoid recursively calling this script.
if [[ $SF_MASTER_SCRIPT_RUNNING ]]
then
exit 0
fi
set -u
export SF_MASTER_SCRIPT_RUNNING=1


# Constants
SF_TARGET_NAME=${PROJECT_NAME}
UNIVERSAL_OUTPUTFOLDER=${BUILD_DIR}/${CONFIGURATION}-universal

# Take build target
if [[ "$SDK_NAME" =~ ([A-Za-z]+) ]]
then
SF_SDK_PLATFORM=${BASH_REMATCH[1]}
else
echo "Could not find platform name from SDK_NAME: $SDK_NAME"
exit 1
fi

if [[ "$SF_SDK_PLATFORM" = "iphoneos" ]]
then
echo "Please choose iPhone simulator as the build target."
exit 1
fi

IPHONE_DEVICE_BUILD_DIR=${BUILD_DIR}/${CONFIGURATION}-iphoneos

# Build the other (non-simulator) platform
xcodebuild -project "${PROJECT_FILE_PATH}" -target "${TARGET_NAME}" -configuration "${CONFIGURATION}" -sdk iphoneos BUILD_DIR="${BUILD_DIR}" OBJROOT="${OBJROOT}" BUILD_ROOT="${BUILD_ROOT}" CONFIGURATION_BUILD_DIR="${IPHONE_DEVICE_BUILD_DIR}/arm64" SYMROOT="${SYMROOT}" ARCHS='arm64' VALID_ARCHS='arm64' $ACTION

xcodebuild -project "${PROJECT_FILE_PATH}" -target "${TARGET_NAME}" -configuration "${CONFIGURATION}" -sdk iphoneos BUILD_DIR="${BUILD_DIR}" OBJROOT="${OBJROOT}" BUILD_ROOT="${BUILD_ROOT}"  CONFIGURATION_BUILD_DIR="${IPHONE_DEVICE_BUILD_DIR}/armv7" SYMROOT="${SYMROOT}" ARCHS='armv7 armv7s' VALID_ARCHS='armv7 armv7s' $ACTION

# Copy the framework structure to the universal folder (clean it first)
rm -rf "${UNIVERSAL_OUTPUTFOLDER}"
mkdir -p "${UNIVERSAL_OUTPUTFOLDER}"
cp -R "${BUILD_DIR}/${CONFIGURATION}-iphonesimulator/${PROJECT_NAME}.framework" "${UNIVERSAL_OUTPUTFOLDER}/${PROJECT_NAME}.framework"

# Smash them together to combine all architectures
lipo -create  "${BUILD_DIR}/${CONFIGURATION}-iphonesimulator/${PROJECT_NAME}.framework/${PROJECT_NAME}" "${BUILD_DIR}/${CONFIGURATION}-iphoneos/arm64/${PROJECT_NAME}.framework/${PROJECT_NAME}" "${BUILD_DIR}/${CONFIGURATION}-iphoneos/armv7/${PROJECT_NAME}.framework/${PROJECT_NAME}" -output "${UNIVERSAL_OUTPUTFOLDER}/${PROJECT_NAME}.framework/${PROJECT_NAME}"

知道我需要做什么来编译 arm64 以便它可以在 iPhone 6+ 上运行吗？

我似乎收到了这个错误：

Please choose iPhone simulator as the build target. Command /bin/sh failed with exit code 1

这听起来与未运行的脚本有关。由于我的测试项目中没有任何脚本，所以我回到我的框架并尝试将其设置为 Run script only when installing 并重新构建我的框架，但这似乎仍然无法正常工作。

Answer 1

你不能，除非你有框架的源代码，并且可以重新编译它。 （无论如何，那是您最好的解决方案。）实际上您现在正在运行 64 位代码，并且您的框架仅针对 32 位代码进行编译。我确实编写了一个 hack，允许您使用以前版本的 arm 库，但我认为 hack 不会在跳转到更大的位大小后幸存下来。喜欢的可以试试：click for hack.

Answer 2

当您不为 iOS 模拟器构建时，您的脚本会生成该错误消息并退出（不生成任何库）。这是有问题的代码：

# Take build target
if [[ "$SDK_NAME" =~ ([A-Za-z]+) ]]
then
SF_SDK_PLATFORM=${BASH_REMATCH[1]}
else
echo "Could not find platform name from SDK_NAME: $SDK_NAME"
exit 1
fi

if [[ "$SF_SDK_PLATFORM" = "iphoneos" ]]
then
echo "Please choose iPhone simulator as the build target."
exit 1
fi

Answer 3

检查您的架构和有效架构，并将 $(inherited) 添加到其他链接器标志。