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获得当前目录名称的 4 种方法

galeki posted @ 2007年11月08日 22:52 in Bash 和它的朋友们 with tags shell 目录 , 3347 阅读

翻译自: http://www.linux.com/feature/120291

在写 shell 脚本的时候,有时我们需要得到当前目录的名称,pwd 命令可以返回当前目录的完整路径,而不是目录名,如果要得到目录名,有下面 4 种方法:

basename

使用 basename 是最简单的方法:

basename /usr/local/bin
bin

链接 basename 和pwd,就可获得当前目录:

cd /usr/local/bin
basename `pwd`
bin

使用 ## 拆分变量

${var##pattern} 可以从 $var 变量中,去掉可以匹配 pattern 最大长度的字符串,所以,要获得目录名,可以:

cd /var/log/squid
echo ${PWD##*/}
squid

$PWD 是保存目录完整路径的环境变量,pattern 就是 */,也就是去掉了前面所有的上级目录。

使用 awk 和 rev

cd /usr/share/cups/data
pwd | rev | awk –F \/ '{print $1}' | rev
data

把结果分步显示,你就知道原理了:

pwd
/usr/share/cups/data
pwd | rev
atad/supc/erahs/rsu/
pwd | rev | awk –F \/ '{print $1}'
atad
pwd | rev | awk –F \/ '{print $1}' | rev
data

-F 选项是指定分隔符,目录之间分隔符为 /。

使用 sed

cd /home/smith/music
pwd | sed 's,^\(.*/\)\?\([^/]*\),\2,'
music

第一个 pattern,^\(.*/\)?,匹配 / 符号前所有的东西,第二个 [^/]* ,匹配剩余部分不包含 / 的部分。

 

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