Question

鉴于这样的道路:

/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

我如何利用Bash的炮兵指挥从第5和第7场(5和x86_64) values取价值?

我有这样的一些东西,但我只想找一些更令人费解的事,而不需要抓起 ju子。

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

Answer 1

您可使用IFS,也可使用一个阵列将名录分为其组成部分:

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

Answer 2

对我来说,这项工作是:

pwd | awk -F /   {print "version=" $6 " arch=" $8}

Answer 3

> p=$(pwd)
> echo $p
/data/mirrors/third-party/centos/5/projectA/x86_64

> basename ${p}
x86_64

> basename ${p%/*/*}
5

你们也可以使用这样的东西:

echo `expr match "$p"  <regular-expression> `

......也许有人会帮助我经常表达这种说法。

Answer 4

审判

echo `pwd` | cut -d /  -f6,8 | tr  /

显示外地

或显示环形和arch

echo `pwd` | cut -d /  -f6,8 | sed -e  s/(.*)/(.*)/version=1 arch=2/

友情链接