我想在Xcode中建立一个python 3项目,但是在要求我为该方案选择可执行文件的步骤中,我失败了。请参阅Xcode 4+中的Python答案中的步骤16 ?。
我使用“哪个python3”来找到可执行文件:
$ which python3
/usr/local/bin/python3
但是Xcode不允许我通过单击它从“ / usr / local / bin”中选择python3:
If I try following the suggestion of trojanfoe below to uncover the symbolic link through: ls -l /usr/local/bin/python3 I get to "../Cellar/python3/3.5.1/bin/python3" which in turn is a symbolic link pointing to "../Frameworks/Python.framework/Versions/3.5/bin/python3" which finally points at "Python3.5" in that directory. However, my Xcode also is not willing to let me select this file:
I'm running Xcode 7.3.1 in OS X 10.11.4. I've installed python 3.5.1 with home-brew.
In the file selection window right-click on python3.5 and select quick look. Close the quick look window and python3.5 will become selectable. (images attached show workflow referencing the newer python3.6 - it is the same for all python releases).
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