xcode xcodebuild：找到输出应用程序文件的位置
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xcodebuild: find where the output app file is
提问by hzxu
Is there anyway to specify where to put the output file (ie .app) for xcodebuild? By default it is in:
无论如何要指定 xcodebuild 的输出文件（即 .app）放在哪里？默认情况下，它位于：
/Users/myusername/Library/Developer/Xcode/DerivedData/Project-crkyjjbuqqnqqvfxehsjsarzlbbr/Build/Products/Release-iphoneos/Project.app
/Users/myusername/Library/Developer/Xcode/DerivedData/Project-crkyjjbuqqnqqvfxehsjsarzlbbr/Build/Products/Release-iphoneos/Project.app
As I need to find the output Project.app file, if there is no way to specify the path, can I use a script to look for that "/Users/myusername/Library/Developer/Xcode/DerivedData/" folder and inside it find the one with latest modified date and prefix "Project-"? What does the script look like?
由于我需要找到输出 Project.app 文件，如果无法指定路径，我可以使用脚本来查找“/Users/myusername/Library/Developer/Xcode/DerivedData/”文件夹及其内部找到具有最新修改日期和前缀“Project-”的那个？剧本是什么样的？
Thanks
谢谢
回答by Nathaniel Irons
Override the CONFIGURATION_BUILD_DIR environment variable in an xcodebuild argument. For instance, this command would put its build products under "/tmp/sportsball/":
覆盖 xcodebuild 参数中的 CONFIGURATION_BUILD_DIR 环境变量。例如，此命令会将其构建产品放在“/tmp/sportsball/”下：
xcodebuild -workspace Sportsball.xcworkspace -scheme "Sportsball Debug" -configuration Debug clean build CONFIGURATION_BUILD_DIR=/tmp/sportsball/
xcodebuild -workspace Sportsball.xcworkspace -scheme "Sportsball Debug" -configuration Debug clean build CONFIGURATION_BUILD_DIR=/tmp/sportsball/
