Question

我在文件夹中列出了一系列的mp3

/mnt/media/Music1/DJ_Mixes_01-71/DJ aaa/01.mp3
/mnt/media/Music1/DJ_Mixes_01-71/DJ 02/01.mp3
/mnt/media/Music1/DJ_Mixes_01-71/DJ Chemist/01.mp3

有没有办法可以使用sed来获取文件夹名称

DJ aaa
DJ 02
DJ Chemist

Answer 1

您还可以将参数展开与子字符串提取一起用作“dirname / basename”的替代方法;以下是从数据文件中读取所有目录名称的快速示例：

#!/bin/bash

printf "\n The following directories were isolated:\n\n"

while read -r line || test -n "$line" ; do

    pname="${line%/*}"      # remove filename from line
    lastd="${pname##*/}"    # remove up to last '/'

    printf " %-12s  from  %s\n" "$lastd" "$line"

done <"$1"

printf "\n"

exit 0

<强>输入

$ cat dat/mp3dirs.txt
/mnt/media/Music1/DJ_Mixes_01-71/DJ aaa/01.mp3
/mnt/media/Music1/DJ_Mixes_01-71/DJ 02/01.mp3
/mnt/media/Music1/DJ_Mixes_01-71/DJ Chemist/01.mp3

<强>输出：

$ ./lastdir.sh dat/mp3dirs.txt

 The following directories were isolated:

 DJ aaa        from  /mnt/media/Music1/DJ_Mixes_01-71/DJ aaa/01.mp3
 DJ 02         from  /mnt/media/Music1/DJ_Mixes_01-71/DJ 02/01.mp3
 DJ Chemist    from  /mnt/media/Music1/DJ_Mixes_01-71/DJ Chemist/01.mp3

Answer 2

将basename与dirname结合使用：

 f="/mnt/media/Music1/DJ_Mixes_01-71/DJ aaa/01.mp3"
 d=$(dirname "$f")
 echo "$(basename "$d")"

为您提供DJ aaa

Answer 3

通过sed：

sed -r 's/.*\/(.*)\/.*$/\1/' file

通过awk：

awk -F/ '{print $(NF-1)}' file

通过grep：

grep -Po '.*\/\K.*(?=\/.*$)' file

通过剪切：

cut -d'/' -f6 file            # for fixed number of fields

或

rev file |cut -d'/' -f2 |rev  # for any length of fields

通过cuts：

cuts -2 file

Answer 4

您可以尝试如下。 \(和\)将标记从DJ到/

之前的最后一个字符的模式

sed 's#.*\(DJ [[:alnum:]]\{1,\}\)/.*#\1#g'

结果

DJ aaa
DJ 02
DJ Chemist

使用bash提取文件夹名称

4 个答案: