Question

我正在尝试将result=$(ls -Q)返回的所有项目放到数组中。

如果我回复result，它看起来像这样："folder2" "folder space" "test 123"。请注意，某些项目包含空格。

如何将每个项目放在数组中的引号（引号必须保留在那里）。我想要的输出是：

echo ${ARRAY[0]}  --> "folder2"
echo ${ARRAY[1]}  --> "folder space"
echo ${ARRAY[2]}  --> "test 123"

我怎样才能做到这一点？

Answer 1

像这样使用filename expansion：

ARRAY=(*)

示例：

mkdir test
cd test
# Create file with space in name
touch 'foo bar'
# Create file with newline in name
touch 'hello
world'

files=(*)

for file in "${files[@]}" ; do
    echo "name: ${file}"
done

输出：

name: foo bar                                                                                                                         
name: hello                                                                                                                           
world

Answer 2

不要解析ls的输出。即使使用-Q，它也已经坏了。在我的有毒文件名目录中，我得到了这个：

res=(*)
for f in "${res[@]}"; do echo "file: $f"; done 

file: 1 file.tar.gz
file: 2 file.tar.gz
file: 3 file.tar.gz
file: 4 file.tar.gz
file: 5 file.tar.gz
file: abc3<foo.mp3
file: bb bb
file: dd dd
file: ff ff
file: file-mit"Quote"
file: m d.sh
file: regex=*a*
file: zeilen
umbruch.zip

我的第一个回答和评论是错误的：

ls -Q
"1 file.tar.gz"  "3 file.tar.gz"  "5 file.tar.gz"  "bb bb"  "ff ff"              "m d.sh"     "zeilen\numbruch.zip"
"2 file.tar.gz"  "4 file.tar.gz"  "abc3<foo.mp3"   "dd dd"  "file-mit\"Quote\""  "regex=*a*"

result=($(ls -Q))
for f in ${result[@]}; do echo "file $f"; done 

file "1
file file.tar.gz"
file "2
file file.tar.gz"
file "3
file file.tar.gz"
file "4
file file.tar.gz"
file "5
file file.tar.gz"
file "abc3<foo.mp3"
file "bb
file bb"
file "dd
file dd"
file "ff
file ff"
file "file-mit\"Quote\""
file "m
file d.sh"
file "regex=*a*"
file "zeilen\numbruch.zip"

每个字符串都带有引号，作为数组中的自有元素

2 个答案: