通过符号链接启动正确的python程序
我最近安装了 ActivePython 2.7。所以我在 /usr/local/bin/python 中有一个指向 2.7 目录的符号链接。问题是，当我在命令行中输入“python”时，它会在 /usr/bin/python 中启动 MacPython 2.6。我不明白为什么符号链接不优先。 /usr/local/bin 在我的路径上。任何帮助将不胜感激。
~ $ which python
/usr/bin/python
~ $ find /usr -iname python
/usr/bin/python
/usr/local/bin/python
路径看起来像这样：
/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/local/git/bin:/opt/PalmPDK/bin:/opt/PalmSDK/Current/bin:/usr/X11/bin
I recently installed ActivePython 2.7. So I've got a symbolic link in /usr/local/bin/python which points to the 2.7 directory. Trouble is, when I type "python" at the command line, it starts up MacPython 2.6 in /usr/bin/python. I don't understand why the symbolic link is not taking precedence. /usr/local/bin is on my PATH. Any help would be appreciated.
~ $ which python
/usr/bin/python
~ $ find /usr -iname python
/usr/bin/python
/usr/local/bin/python
Path looks like this:
/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/usr/local/git/bin:/opt/PalmPDK/bin:/opt/PalmSDK/Current/bin:/usr/X11/bin
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当您在命令行中键入“python”时，shell 从左到右检查 $PATH 中的值，并调用它首先找到的程序。像这样检查当前的 $PATH 设置：
如果它同时包含 /usr/bin 和 /usr/local/bin ，则以先到者为准。
When you type “python” on the command line, the shell checks the values in your $PATH from left to right, and invokes the program it finds first. Check your current $PATH setting like this:
If it contains both /usr/bin and /usr/local/bin then whichever comes first will take precedence.
在您的 PATH 变量中， /usr/bin 可能位于 /usr/local/bin/ 目录之前。在这种情况下，您应该使用完整路径才能工作。
in your PATH variable, /usr/bin probabbly is before the /usr/local/bin/ directory. you should use the full path in order to work in this case.