Question

我正在尝试在iOS中创建和使用框架。它以前工作，它可以在模拟器上运行，但不适用于较新的iPhone 6。

我怀疑它与脚本有关，但从我所看到的（我在大多数情况下复制了互联网的脚本）它确实有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}"

知道我需要做些什么来为arm64编译它才能在iPhone 6 +上运行？

我似乎得到了这个错误：

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，但这似乎仍然无效。

Answer 1

除非您拥有框架的源代码，否则不能重新编译它。（无论如何，这是你最好的解决方案。）你现在正在运行64位代码，而你的框架只针对32位代码进行编译。我确实编写了一个允许你使用以前版本的arm库的hack，但我不认为hack会在跳转到更大的位大小时存活下来。如果您愿意，可以尝试：click for hack.

Answer 2

检查您的架构和有效架构，并将 $（继承）添加到其他链接器标志。

Answer 3

您的脚本会生成该错误消息，并在您不为iOS模拟器构建时退出（不生成任何库）。这是令人反感的代码：

# 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

iOS框架缺少必需的架构arm64

3 个答案: