根据构建配置嵌入不同的框架 [英] Embed different framework based on build configuration

问题描述

我有4种构建配置:

• 调试
• 发布
• WhiteLabelDebug
• WhiteLabelRelease

Debug和Release应该链接&嵌入我的自定义LightTheme.framework而不是WhiteLabelTheme.framework.

Debug and Release should link & embed my custom LightTheme.framework and not the WhiteLabelTheme.framework.

WhiteLabelDebug和WhiteLabelRelease应该链接并嵌入我的自定义WhiteLabelTheme.framework而不是LightTheme.framework.

WhiteLabelDebug and WhiteLabelRelease should link and embed my custom WhiteLabelTheme.framework and not the LightTheme.framework.

这两个框架都有一些自定义代码，本地化的字符串，颜色，字体和图像.我们使用不同的方案来构建每个配置，并将其作为完全不同的应用程序提交到App Store.但是，目前我们的品牌应用程序和带有白色标签的应用程序都有图像，字体，颜色，字符串等两组，这确实会损害文件的大小.

Both of these frameworks have some custom code, localized strings, colors, fonts, and images. We use a different scheme to build each config and submit it to the App Store as a totally different app. However currently both our branded app, and the white label one have both sets of images, fonts, colors, strings, etc. Which really hurts file size.

我可以轻松地将两个框架都弱链接到应用程序目标，但是我不知道如何仅基于当前配置嵌入框架.

I can easily weak link both frameworks into the app target, but I can't figure out how to only embed a framework based on the current configuration.

推荐答案

转到目标构建阶段，并在最后一步中添加New Run Script Phase.

Go to your target Build Phases and add New Run Script Phase as the last step.

脚本如下所示:

if [[ ${CONFIGURATION} == "WhiteLabelRelease" ]]; then
    rm -R "${TARGET_BUILD_DIR}/${FRAMEWORKS_FOLDER_PATH}/LightTheme.framework"
elif [[ ${CONFIGURATION} == "Release" ]]; then
    rm -R "${TARGET_BUILD_DIR}/${FRAMEWORKS_FOLDER_PATH}/WhiteLabelTheme.framework"
fi

在构建任何* Release版本(以及归档以供提交)时，脚本将删除相应的框架.调试时会将它们保留在适当的位置.

When building any of your *Release builds (and Archiving for submission), the script will remove a corresponding framework. It will leave them in place when debugging.

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