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[英]How to get the app path name from an Xcode project on the command line?
[英]How to run an Xcode built app from the command line (bash)
继我的问题如何从bash运行xcode项目?
我在构建目录中找到了构建的* .app,但我正在试图弄清楚如何让xcode打开它,因为它不能仅仅作为Mac OS X程序运行。
有任何想法吗?
感谢ch在这篇文章: iPhone的XCode测试自动化
我遇到了同样的问题,试图从构建我的xcode项目启动我的调试输出。 我只是尝试以下
open ./prog.app&
似乎可以做到这一点。
我使用bash文件解决了我的问题,该文件使用rsync来同步工作目录,然后使用AppleScript来构建和运行项目。
sync.sh:
#!/bin/bash
# script for moving development files into xcode for building
developmentDirectory="/"
xcodeDirectory="/"
echo 'Synchronising...'
rsync -r $developmentDirectory $xcodeDirectory \
--exclude='.DS_Store' --exclude='.*' --exclude='utils/' --exclude='photos'
echo 'Synchronising Done at:'
echo $(date)
buildandrun:
set projectName to "projectName"
# AppleScript uses : for / in directory paths
set projectDir to "Users:username:Documents:" & projectName & ":" & projectName & ".xcodeproj"
tell application "Xcode"
open projectDir
tell project projectName
clean
build
(* for some reasons, debug will hang even the debug process has completed.
The try block is created to suppress the AppleEvent timeout error
*)
try
debug
end try
end tell
quit
end tell
然后,最后,我在.bash_profile中使用名为run.sh的shell脚本:
run.sh:
#!/bin/bash
bash utils/sync.sh
osascript utils/buildandrun
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