如何改组包含空格的文件名数组？ [英] How can I shuffle an array of filenames that have spaces in them?

问题描述

我有一个文件名数组，其中可能包含空格。我正在使用 shuf 命令，但它使用文件名中的空格作为分隔符，并在改组时拆分文件名。有没有办法解决这个问题，还是我必须放弃 shuf 命令？有什么建议吗？

I have an array of filenames that may contain spaces in them. I am using the shuf command but it uses the spaces in the filenames as a delimiter and breaks up the filenames when it shuffles. Is there a way around this or do I have to abandon the shuf command? Any suggestions?

#!/bin/bash

vids=()

vids+=("file with spaces.txt")

for arr in "${vids[@]}"; do
    echo -e "$arr\n"
done

vids=( $(shuf -e "${vids[@]}") )    #shuffle contents of array

for arr in "${vids[@]}"; do
    echo -e "$arr\n"
done

exit 0

输出：

file with spaces.txt

file

with

spaces.txt

推荐答案

您的方法不起作用的原因是，shell在 $（...），并且无法将换行符视为分隔符。您可以使用 mapfile 将行读入数组（在Bash 4+中）：

The reason your approach doesn't work is that the shell applies word splitting to the output of a command inside $(...), and there's no way to treat newlines as a separator. You could either use mapfile to read lines into an array (in Bash 4+):

mapfile -t vids < <(shuf -e "${vids[@]}")

Bash您可以在循环的同时使用老式的：

Or in older versions of Bash you could use good old-fashioned while loop:

vids2=()
while read -r item; do
    vids2+=("$item")
done < <(shuf -e "${vids[@]}")

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