bash - bash:从Unix路径中获取字段5和7？

给定如下路径：

/data/mirrors/third-party/centos/5/projectA/x86_64
/data/mirrors/third-party/centos/5/projectA/i386
/data/mirrors/third-party/centos/5/projectA/noarch
/data/mirrors/third-party/centos/4/projectB/x86_64
/data/mirrors/third-party/centos/4/projectB/i386
/data/mirrors/third-party/centos/4/projectB/noarch
/data/mirrors/third-party/centos/4/projectC/x86_64
/data/mirrors/third-party/centos/4/projectC/i386
/data/mirrors/third-party/centos/4/projectC/noarch

如何使用Bash shell命令从字段5和7（“5”和“x86_64”）获取值？
到目前为止，我有类似的东西，但我正在寻找更优雅的东西，而且不需要捕捉“垃圾”：

cd /data/mirrors/third-party/centos/5/project/x86_64
echo `pwd` | tr '/' ' ' | while read junk1 junk2 junk3 junk4 version junk5 arch; do
    echo version=$version arch=$arch
done
version=5 arch=x86_64

最佳答案

您可以使用IFS和数组将目录拆分为其组件：

#!/bin/bash
saveIFS=$IFS
IFS='/'
dirs=($(pwd))
IFS=$saveIFS
version=${dirs[5]}
arch=${dirs[7]}