Shell编程:从命令输出中选择随机行 [英] Shell programming: Select random line from command output

问题描述

我正在尝试创建一个简单的Shell脚本，该脚本涉及从当前工作目录中选择一个随机目录，然后导航到该目录.

I'm trying to create a simple Shell script which involves selecting a random directory from the current working directory, and navigating to it.

谁能说明如何列出所有目录，然后从该列表中随机选择一个?

我试图避免将所有目录都列出到一个文本文件中，而只是从该文件中选择一条随机行(这很简单).

I'm trying to avoid listing all directories to a text file and simply selecting a random line from that file (which is simple enough).

我的最初尝试包括使用 ls -d */命令仅列出目录.此命令在输入到终端时有效，但是返回错误:

My initial attempts included using the ls -d */ command to list only directories. This command worked when it was entered into the terminal, however returned the error:

ls: */: No such file or directory

当我尝试将其实现到此脚本中时:

when I tried to implement it into this script:

DIR_LIST=` ls -d */`

echo "$DIR_LIST"

推荐答案

查找.-maxdepth 1型d！-小路 .|shuf -n 1

没有 shuf 版本:

# To exclude hidden directories, use -name like so:
# find ... -name ".*"

# A version of find using only POSIX options:
# This might be slower for large directories (untested). Will need to be modified to list anything other than current directory.
# dirs="$(find . -type d ! -path "./*/*" ! -name ".")"

# Includes hidden directories.
dirs="$(find . -maxdepth 1 -type d ! -path .)"

numlines="$(printf "%s\n" "${dirs}" | wc -l)"
lineno="$((${RANDOM} % ${numlines} + 1))"
random_line="$(printf "%s\n" "${dirs}" | sed -n "${lineno}{p;q}")"

echo "Your selected directory: ${random_line}"

编辑:基于注释的改进代码.

Improved code based on comments.

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