下面是我专门用来从可执行文件中删除一个框架架构的脚本。

# Remove unused Framework architecture from "YourApp" framework.

FRAMEWORK_EXECUTABLE_PATH="${TARGET_BUILD_DIR}/${WRAPPER_NAME}/Frameworks/YourApp.framework/YourApp"

echo "$FRAMEWORK_EXECUTABLE_PATH"

cp "$FRAMEWORK_EXECUTABLE_PATH" "${FRAMEWORK_EXECUTABLE_PATH}_X86_64"

echo "Executing following command to remove x86_64 arch from YourApp framework executable"
echo "lipo -remove x86_64 \"$FRAMEWORK_EXECUTABLE_PATH\" -o \"${FRAMEWORK_EXECUTABLE_PATH}_X86_64\""

lipo -remove x86_64 "${FRAMEWORK_EXECUTABLE_PATH}_X86_64" -o "$FRAMEWORK_EXECUTABLE_PATH"

rm "${FRAMEWORK_EXECUTABLE_PATH}_X86_64"

将此脚本添加到项目目标的“构建阶段”中。请务必选中复选框:“仅在安装时运行脚本”

对我来说最简单的解决方法是

1-从嵌入式框架中移除框架。 2-添加框架作为链接框架

完成了!

您的框架包含ARM和x86代码，这允许您在设备或模拟器中使用它。如果你打算将你的应用程序提交到应用程序商店，运行以下脚本从二进制文件中剥离非活动代码。

1.在Project Navigator中选择您的目标，并单击项目编辑器顶部的Build Phases。

2.从“编辑器”菜单中选择“添加构建阶段”，然后选择“添加运行脚本构建阶段”(或单击“构建阶段”编辑器左上角的“+”按钮)。

3.展开刚刚添加的新Run Script构建阶段旁边的披露三角形。在脚本编辑器中，粘贴如下内容: bash

$ {BUILT_PRODUCTS_DIR} / $ {FRAMEWORKS_FOLDER_PATH} / / strip-frameworks.sh“YourframeworkName.framework”

问题是Buy框架包含模拟器(x86_64)和实际设备(ARM)的构建。

当然，你不允许向App Store提交不受支持的架构的二进制文件，所以解决方案是在提交之前“手动”从最终二进制文件中删除不需要的架构。

Daniel Kennett提出了一个很好的解决方案，并提供了这个脚本添加到构建阶段:

APP_PATH="${TARGET_BUILD_DIR}/${WRAPPER_NAME}"

# This script loops through the frameworks embedded in the application and
# removes unused architectures.
find "$APP_PATH" -name '*.framework' -type d | while read -r FRAMEWORK
do
    FRAMEWORK_EXECUTABLE_NAME=$(defaults read "$FRAMEWORK/Info.plist" CFBundleExecutable)
    FRAMEWORK_EXECUTABLE_PATH="$FRAMEWORK/$FRAMEWORK_EXECUTABLE_NAME"
    echo "Executable is $FRAMEWORK_EXECUTABLE_PATH"

    EXTRACTED_ARCHS=()

    for ARCH in $ARCHS
    do
        echo "Extracting $ARCH from $FRAMEWORK_EXECUTABLE_NAME"
        lipo -extract "$ARCH" "$FRAMEWORK_EXECUTABLE_PATH" -o "$FRAMEWORK_EXECUTABLE_PATH-$ARCH"
        EXTRACTED_ARCHS+=("$FRAMEWORK_EXECUTABLE_PATH-$ARCH")
    done

    echo "Merging extracted architectures: ${ARCHS}"
    lipo -o "$FRAMEWORK_EXECUTABLE_PATH-merged" -create "${EXTRACTED_ARCHS[@]}"
    rm "${EXTRACTED_ARCHS[@]}"

    echo "Replacing original executable with thinned version"
    rm "$FRAMEWORK_EXECUTABLE_PATH"
    mv "$FRAMEWORK_EXECUTABLE_PATH-merged" "$FRAMEWORK_EXECUTABLE_PATH"

done

我用过，效果很好。

编辑:请务必查看Varrry发布的修改后的脚本，因为这个脚本有一些小问题。

感谢以上所有的答案。下面是一个在swift 4.2和5中工作的脚本。将Your_Framework_Name字符串替换为框架的原始名称。

APP_PATH="${TARGET_BUILD_DIR}/${WRAPPER_NAME}"
FRAMEWORK_NAME="Your_Framework_Name.framework"
# Check if Framework is present.
FRAMEWORK_LOCATION=$(find "$APP_PATH" -name "$FRAMEWORK_NAME" -type d)
if [ -z $FRAMEWORK_LOCATION ]; then
echo "Couldn't find Your_Framework_Name.framework in $APP_PATH. Make sure 'Embed Frameworks' build phase is listed before the 'Strip Unused Architectures' build phase."
exit 1
fi
# This script strips unused architectures
find "$APP_PATH" -name "$FRAMEWORK_NAME" -type d | while read -r FRAMEWORK
do
FRAMEWORK_EXECUTABLE_NAME=$(defaults read "$FRAMEWORK/Info.plist" CFBundleExecutable)
FRAMEWORK_EXECUTABLE_PATH="$FRAMEWORK/$FRAMEWORK_EXECUTABLE_NAME"
echo "Executable is $FRAMEWORK_EXECUTABLE_PATH"
EXTRACTED_ARCHS=()
for ARCH in $ARCHS
do
echo "Extracting $ARCH from $FRAMEWORK_EXECUTABLE_NAME"
lipo -extract "$ARCH" "$FRAMEWORK_EXECUTABLE_PATH" -o "$FRAMEWORK_EXECUTABLE_PATH-$ARCH"
EXTRACTED_ARCHS+=("$FRAMEWORK_EXECUTABLE_PATH-$ARCH")
done
echo "Merging extracted architectures: ${ARCHS}"
lipo -o "$FRAMEWORK_EXECUTABLE_PATH-merged" -create "${EXTRACTED_ARCHS[@]}"
rm "${EXTRACTED_ARCHS[@]}"
echo "Replacing original executable with thinned version"
rm "$FRAMEWORK_EXECUTABLE_PATH"
mv "$FRAMEWORK_EXECUTABLE_PATH-merged" "$FRAMEWORK_EXECUTABLE_PATH"
done

我也有同样的问题。 即使在添加给定的运行脚本后，它也不能工作。 这是与Xcode相关的问题。 我使用的是Xcode 9.0版本，但最新版本是9.2。

所以我安装了最新的Xcode(9.2)，它工作。