iOS framework missing required architecture arm64 - ios

Question

I'm trying to create and use a framework in iOS. It was working previously and it works on the simulator but not a newer iPhone 6.
I suspect it has to do with the script, but from what I can see (I copied the script for the internet for the most part) it does have arm64 in it:
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}"
Any idea what I need to do to get this to compile for arm64 so it'll run on an iPhone 6+?
I seem to get this error:
Please choose iPhone simulator as the build target.
Command /bin/sh failed with exit code 1
Which it sounds like has to do with a script not running. Since I don't have any scripts in my test project, I went back to my framework and tried setting it to Run script only when installing and re-built my framework, but that still does not appear to be working.

Answer 1

You can't, unless you have the source code to the framework, and can recompile it. (That is your best solution anyway.) It's really that you are now running 64 bit code, and your framework is compiled only for 32 bit code. I did write a hack that allowed you to use previous versions of arm libraries, but I don't think that hack will survive the jump to larger bit size. You can try it if you want: click for hack.

Answer 2

Your script generates that error message and quits (without generating any libraries) whenever you're not building for the iOS simulator. This is the offending code:
# 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

Check your architecture and valid architecture and add $(inherited) to Other Linker Flags.