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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