从命令行使用xcodebuild构建可可应用后，该如何运行?
问题描述：
我正在使用这样的Xcode项目从命令行构建Mac OS X Cocoa应用程序:
I'm building a Mac OS X Cocoa application from the command line using an Xcode project like this:
xcodebuild -scheme MyApp -configuration Debug
构建完成后如何运行?
答
我写了一个脚本来做到这一点:
I wrote a script to do this:
#!/bin/bash
x=$( xcodebuild -showBuildSettings -project MyApp.xcodeproj | grep ' BUILD_DIR =' | sed -e 's/.*= *//' )
DYLD_FRAMEWORK_PATH=$x/Debug DYLD_LIBRARY_PATH=$x/Debug $x/Debug/MyApp.app/Contents/MacOS/MyApp
(我是通过从Xcode运行应用程序，然后在进程上运行ps -wwE -p来查看其环境变量来解决这个问题的.)
(I figured this out by running the application from Xcode and then ps -wwE -p on the process to see its environment variables.)