如何为依赖于使用 cocoapods 添加的其他框架的所有体系结构构建 cocoa 触摸框架?
How to build cocoa touch framework for all architectures dependent on other frameworks added using cocoapods?
我想创建一个为模拟器和设备构建的输出框架的私有 cocoapod。
我创建了一个 Cocoa 触摸框架,我在其中使用 Cocoapods 集成了 CocoaLumberjack。
我想为所有可能的架构(模拟器和设备)构建框架。
默认构建设置,
'Build Active Architectures Only' is set to (Debug - Yes, Release - No).
一旦我将此调试设置设置为“否”,构建就会失败并出现以下链接器错误:
Undefined symbols for architecture i386:
"_OBJC_CLASS_$_DDASLLogger", referenced from:
objc-class-ref in MyManager.o
"_OBJC_CLASS_$_DDLog", referenced from:
objc-class-ref in MyManager.o
"_OBJC_CLASS_$_DDTTYLogger", referenced from:
objc-class-ref in MyManager.o
ld: symbol(s) not found for architecture i386
clang: error: linker command failed with exit code 1 (use -v to see invocation)
我知道 CocoaLumberjack 仅适用于调试版本中的活动架构。
所以我将“为调试构建活动架构”切换回“是”并成功构建框架。
为了构建所有架构的框架,我使用了在构建阶段添加的 运行 脚本,该脚本还声称将 ios-device 和 ios-simulator build 合并到一。这是脚本:
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}"
我选中 运行 脚本下方的 'Run script only when installing' 复选框。现在我为通用 iOS 设备构建框架。
现在我单击项目层次结构中的 Products 组和 select MyFramework.framework 文件,右键单击并 select 显示在 finder 中。它在 finder 中打开 ~/Library/Developer/Xcode/DerivedData/MyFramework-dlpsipmxkqmemwgqrfeovlzgyhca/Build/Products/Debug-iphoneos/MyFramework.framework。
Now, I create a new tag 'MyFramework-v0.0.1' containing the commit where I added MyFramework.framework file.
我转到 ~/.cocoapods/repos/MyRepoName/ 并创建一个 podspec 文件 MyFramework.podspec,如下所示:
Pod::Spec.new do |s|
s.name = "MyFramework"
s.version = "0.0.1"
s.summary = "A pod for MyFramework"
s.description = "A pod designed for MyFramework"
s.homepage = "My_Private_Repo_Path"
s.license = { :type => "MIT", :file => "FILE_LICENSE" }
s.authors = { "My_Username" => "my_email_address"
}
s.platform = :ios, "8.0"
s.ios.deployment_target = "8.0"
s.source = { :git => "My_Private_Repo_Path", :tag => 'MyFramework-v'+String(s.version) }
s.requires_arc = true
s.vendored_frameworks = "MyFramework.framework"
s.dependency "CocoaLumberjack"
end
现在当我 运行 在终端中输入以下命令时:
pod repo push MyRepoName MyFramework.podspec
我收到以下错误:
Validating spec
-> MyFramework (0.0.1)
- ERROR | [iOS] xcodebuild: Returned an unsuccessful exit code. You can use ` --verbose` for more information.
- NOTE | [iOS] xcodebuild: ld: warning: ignoring file MyFramework/MyFramework.framework/MyFramework, missing required architecture i386 in file MyFramework/MyFramework.framework/MyFramework (2 slices)
- NOTE | [iOS] xcodebuild: ld: warning: ignoring file MyFramework/MyFramework.framework/MyFramework, missing required architecture x86_64 in file MyFramework/MyFramework.framework/MyFramework (2 slices)
- NOTE | [iOS] xcodebuild: fatal error: lipo: -remove's specified would result in an empty fat file
[!] The `MyFramework.podspec` specification does not validate.
如何为所有设备和模拟器构建 cocoa 触摸框架,它依赖于使用 cocoapods 添加的另一个框架 (CocoaLumberjack)?我需要创建一个输出框架的私有 pod。
基本上我发现我只需要遵循这些简单的步骤。
1. Create a cocoa touch framework.
2. Set bitcode enabled to No.
3. Select your target and choose edit schemes. Select Run and choose Release from Info tab.
4. No other setting required.
5. Now build the framework for any simulator as simulator runs on x86 architecture.
6. Click on Products group in Project Navigator and find the .framework file.
7. Right click on it and click on Show in finder. Copy and paste it in any folder, I personally prefer the name 'simulator'.
8. Now build the framework for Generic iOS Device and follow the steps 6 through
9. Just rename the folder to 'device' instead of 'simulator'.
10. Copy the device .framework file and paste in any other directory. I prefer the immediate super directory of both.
所以目录结构现在变成:
- Desktop
- device
- MyFramework.framework
- simulator
- MyFramework.framework
- MyFramework.framework
现在打开终端并 cd 到桌面。现在开始输入以下命令:
lipo -create 'device/MyFramework.framework/MyFramework' 'simulator/MyFramework.framework/MyFramework' -output 'MyFramework.framework/MyFramework'
就是这样。在这里,我们合并 MyFramework.framework 中存在的 MyFramework 二进制文件的模拟器和设备版本。我们得到了一个通用框架,可以构建所有架构,包括模拟器和设备。
现在,为此框架创建 pod 没有任何区别。它就像一个魅力。另请注意,也有 运行 脚本可用于实现相同的功能,但我花了很多时间来寻找正确的脚本。所以我建议你使用这个方法。
XCode 11
首先请注意,您不能使用具有模拟器支持的胖框架(x84_64 arch)发布到 AppStore,因此您需要制作两个胖框架:一个用于使用 ARM arch 发布(仅限设备)和一个用于调试 - ARM 和 x86_64 架构。
您可以将下一个脚本放到项目的文件夹中,以从命令行制作胖框架:
- 生成文件
BUILD_DIR = build
BUILD = @sh build.sh ${BUILD_DIR}
default:
@echo "Build framework makefile"
@echo "usage: make (release | debug | all | rebuild | clean)"
release:
${BUILD} Release YourTargetScheme # <- your target scheme
debug:
${BUILD} Debug YourTargetScheme
clean:
rm -r ${BUILD_DIR}
all: release debug
rebuild: clean all
- build.sh
# Debug
set -x
# Params
BUILD_DIR=
CONFIGURATION=
SCHEME=
WORKSPACE=YourWorkspace.xcworkspace # <- your workspace file
DERIVED_DATA_PATH=$BUILD_DIR/DerivedData
# Destinations
IPNONEOS="generic/platform=iOS"
IPNONESIMULATOR="platform=iOS Simulator,name=iPhone 8"
# Build
if [ $CONFIGURATION = "Release" ]; then
xcodebuild \
-quiet \
-showBuildTimingSummary \
-workspace $WORKSPACE \
-configuration $CONFIGURATION \
-scheme $SCHEME \
-derivedDataPath $DERIVED_DATA_PATH \
-destination "$IPNONEOS"
else
xcodebuild \
-quiet \
-showBuildTimingSummary \
-workspace $WORKSPACE \
-configuration $CONFIGURATION \
-scheme $SCHEME \
-derivedDataPath $DERIVED_DATA_PATH \
-destination "$IPNONEOS" \
-destination "$IPNONESIMULATOR"
fi
# Move
FRAMEWORK=$SCHEME.framework
FRAMEWORK_PATH=$BUILD_DIR/$CONFIGURATION/$FRAMEWORK
mkdir $BUILD_DIR/$CONFIGURATION
rm -r $FRAMEWORK_PATH
if [ $CONFIGURATION = "Release" ]; then
mv $DERIVED_DATA_PATH/Build/Products/Release-iphoneos/$FRAMEWORK $FRAMEWORK_PATH
else
mv $DERIVED_DATA_PATH/Build/Products/Debug-iphoneos/$FRAMEWORK $FRAMEWORK_PATH
BINARY_FILE=$FRAMEWORK_PATH/$SCHEME
ARMV7=$FRAMEWORK_PATH/armv7
ARM64=$FRAMEWORK_PATH/arm64
x86_64=$FRAMEWORK_PATH/x86_64
lipo $BINARY_FILE -extract armv7 -o $ARMV7
lipo $BINARY_FILE -extract arm64 -o $ARM64
cp $DERIVED_DATA_PATH/Build/Products/Debug-iphonesimulator/$FRAMEWORK/$SCHEME $x86_64
lipo -create $ARMV7 $ARM64 $x86_64 -o $BINARY_FILE
# Clean
rm -rf $ARMV7 $ARM64 $x86_64
fi
运行 在您的项目文件夹中使用以下命令之一来构建所需的框架:
make all # Debug and Release frameworks
make release # Release only for devices and AppStore (armv7 and arm64 archs)
make debug # Debug with simulator support (armv7, arm64 and x86_64 archs)
然后您可以在项目文件夹内的 build
目录中找到您的胖框架。
我想创建一个为模拟器和设备构建的输出框架的私有 cocoapod。
我创建了一个 Cocoa 触摸框架,我在其中使用 Cocoapods 集成了 CocoaLumberjack。 我想为所有可能的架构(模拟器和设备)构建框架。
默认构建设置,
'Build Active Architectures Only' is set to (Debug - Yes, Release - No).
一旦我将此调试设置设置为“否”,构建就会失败并出现以下链接器错误:
Undefined symbols for architecture i386:
"_OBJC_CLASS_$_DDASLLogger", referenced from:
objc-class-ref in MyManager.o
"_OBJC_CLASS_$_DDLog", referenced from:
objc-class-ref in MyManager.o
"_OBJC_CLASS_$_DDTTYLogger", referenced from:
objc-class-ref in MyManager.o
ld: symbol(s) not found for architecture i386
clang: error: linker command failed with exit code 1 (use -v to see invocation)
我知道 CocoaLumberjack 仅适用于调试版本中的活动架构。
所以我将“为调试构建活动架构”切换回“是”并成功构建框架。
为了构建所有架构的框架,我使用了在构建阶段添加的 运行 脚本,该脚本还声称将 ios-device 和 ios-simulator build 合并到一。这是脚本:
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}"
我选中 运行 脚本下方的 'Run script only when installing' 复选框。现在我为通用 iOS 设备构建框架。 现在我单击项目层次结构中的 Products 组和 select MyFramework.framework 文件,右键单击并 select 显示在 finder 中。它在 finder 中打开 ~/Library/Developer/Xcode/DerivedData/MyFramework-dlpsipmxkqmemwgqrfeovlzgyhca/Build/Products/Debug-iphoneos/MyFramework.framework。
Now, I create a new tag 'MyFramework-v0.0.1' containing the commit where I added MyFramework.framework file.
我转到 ~/.cocoapods/repos/MyRepoName/ 并创建一个 podspec 文件 MyFramework.podspec,如下所示:
Pod::Spec.new do |s|
s.name = "MyFramework"
s.version = "0.0.1"
s.summary = "A pod for MyFramework"
s.description = "A pod designed for MyFramework"
s.homepage = "My_Private_Repo_Path"
s.license = { :type => "MIT", :file => "FILE_LICENSE" }
s.authors = { "My_Username" => "my_email_address"
}
s.platform = :ios, "8.0"
s.ios.deployment_target = "8.0"
s.source = { :git => "My_Private_Repo_Path", :tag => 'MyFramework-v'+String(s.version) }
s.requires_arc = true
s.vendored_frameworks = "MyFramework.framework"
s.dependency "CocoaLumberjack"
end
现在当我 运行 在终端中输入以下命令时:
pod repo push MyRepoName MyFramework.podspec
我收到以下错误:
Validating spec
-> MyFramework (0.0.1)
- ERROR | [iOS] xcodebuild: Returned an unsuccessful exit code. You can use ` --verbose` for more information.
- NOTE | [iOS] xcodebuild: ld: warning: ignoring file MyFramework/MyFramework.framework/MyFramework, missing required architecture i386 in file MyFramework/MyFramework.framework/MyFramework (2 slices)
- NOTE | [iOS] xcodebuild: ld: warning: ignoring file MyFramework/MyFramework.framework/MyFramework, missing required architecture x86_64 in file MyFramework/MyFramework.framework/MyFramework (2 slices)
- NOTE | [iOS] xcodebuild: fatal error: lipo: -remove's specified would result in an empty fat file
[!] The `MyFramework.podspec` specification does not validate.
如何为所有设备和模拟器构建 cocoa 触摸框架,它依赖于使用 cocoapods 添加的另一个框架 (CocoaLumberjack)?我需要创建一个输出框架的私有 pod。
基本上我发现我只需要遵循这些简单的步骤。
1. Create a cocoa touch framework.
2. Set bitcode enabled to No.
3. Select your target and choose edit schemes. Select Run and choose Release from Info tab.
4. No other setting required.
5. Now build the framework for any simulator as simulator runs on x86 architecture.
6. Click on Products group in Project Navigator and find the .framework file.
7. Right click on it and click on Show in finder. Copy and paste it in any folder, I personally prefer the name 'simulator'.
8. Now build the framework for Generic iOS Device and follow the steps 6 through
9. Just rename the folder to 'device' instead of 'simulator'.
10. Copy the device .framework file and paste in any other directory. I prefer the immediate super directory of both.
所以目录结构现在变成:
- Desktop
- device
- MyFramework.framework
- simulator
- MyFramework.framework
- MyFramework.framework
现在打开终端并 cd 到桌面。现在开始输入以下命令:
lipo -create 'device/MyFramework.framework/MyFramework' 'simulator/MyFramework.framework/MyFramework' -output 'MyFramework.framework/MyFramework'
就是这样。在这里,我们合并 MyFramework.framework 中存在的 MyFramework 二进制文件的模拟器和设备版本。我们得到了一个通用框架,可以构建所有架构,包括模拟器和设备。
现在,为此框架创建 pod 没有任何区别。它就像一个魅力。另请注意,也有 运行 脚本可用于实现相同的功能,但我花了很多时间来寻找正确的脚本。所以我建议你使用这个方法。
XCode 11
首先请注意,您不能使用具有模拟器支持的胖框架(x84_64 arch)发布到 AppStore,因此您需要制作两个胖框架:一个用于使用 ARM arch 发布(仅限设备)和一个用于调试 - ARM 和 x86_64 架构。
您可以将下一个脚本放到项目的文件夹中,以从命令行制作胖框架:
- 生成文件
BUILD_DIR = build
BUILD = @sh build.sh ${BUILD_DIR}
default:
@echo "Build framework makefile"
@echo "usage: make (release | debug | all | rebuild | clean)"
release:
${BUILD} Release YourTargetScheme # <- your target scheme
debug:
${BUILD} Debug YourTargetScheme
clean:
rm -r ${BUILD_DIR}
all: release debug
rebuild: clean all
- build.sh
# Debug
set -x
# Params
BUILD_DIR=
CONFIGURATION=
SCHEME=
WORKSPACE=YourWorkspace.xcworkspace # <- your workspace file
DERIVED_DATA_PATH=$BUILD_DIR/DerivedData
# Destinations
IPNONEOS="generic/platform=iOS"
IPNONESIMULATOR="platform=iOS Simulator,name=iPhone 8"
# Build
if [ $CONFIGURATION = "Release" ]; then
xcodebuild \
-quiet \
-showBuildTimingSummary \
-workspace $WORKSPACE \
-configuration $CONFIGURATION \
-scheme $SCHEME \
-derivedDataPath $DERIVED_DATA_PATH \
-destination "$IPNONEOS"
else
xcodebuild \
-quiet \
-showBuildTimingSummary \
-workspace $WORKSPACE \
-configuration $CONFIGURATION \
-scheme $SCHEME \
-derivedDataPath $DERIVED_DATA_PATH \
-destination "$IPNONEOS" \
-destination "$IPNONESIMULATOR"
fi
# Move
FRAMEWORK=$SCHEME.framework
FRAMEWORK_PATH=$BUILD_DIR/$CONFIGURATION/$FRAMEWORK
mkdir $BUILD_DIR/$CONFIGURATION
rm -r $FRAMEWORK_PATH
if [ $CONFIGURATION = "Release" ]; then
mv $DERIVED_DATA_PATH/Build/Products/Release-iphoneos/$FRAMEWORK $FRAMEWORK_PATH
else
mv $DERIVED_DATA_PATH/Build/Products/Debug-iphoneos/$FRAMEWORK $FRAMEWORK_PATH
BINARY_FILE=$FRAMEWORK_PATH/$SCHEME
ARMV7=$FRAMEWORK_PATH/armv7
ARM64=$FRAMEWORK_PATH/arm64
x86_64=$FRAMEWORK_PATH/x86_64
lipo $BINARY_FILE -extract armv7 -o $ARMV7
lipo $BINARY_FILE -extract arm64 -o $ARM64
cp $DERIVED_DATA_PATH/Build/Products/Debug-iphonesimulator/$FRAMEWORK/$SCHEME $x86_64
lipo -create $ARMV7 $ARM64 $x86_64 -o $BINARY_FILE
# Clean
rm -rf $ARMV7 $ARM64 $x86_64
fi
运行 在您的项目文件夹中使用以下命令之一来构建所需的框架:
make all # Debug and Release frameworks
make release # Release only for devices and AppStore (armv7 and arm64 archs)
make debug # Debug with simulator support (armv7, arm64 and x86_64 archs)
然后您可以在项目文件夹内的 build
目录中找到您的胖框架。