string 在 Bash 中将分隔的字符串读入数组
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Reading a delimited string into an array in Bash
提问by Nikola Novak
I have a variable which contains a space-delimited string:
我有一个包含空格分隔字符串的变量：
line="1 1.50 string"
I want to split that string with space as a delimiter and store the result in an array, so that the following:
我想用空格作为分隔符分割该字符串并将结果存储在数组中，以便以下内容：
echo ${arr[0]}
echo ${arr[1]}
echo ${arr[2]}
outputs
产出
1
1.50
string
Somewhere I found a solution which doesn't work:
我在某处找到了一个不起作用的解决方案：
arr=$(echo ${line})
If I run the echo statements above after this, I get:
如果我在此之后运行上面的 echo 语句，我会得到：
1 1.50 string
[empty line]
[empty line]
I also tried
我也试过
IFS=" "
arr=$(echo ${line})
with the same result. Can someone help, please?
结果相同。有人可以帮忙吗？
回答by kev
In order to convert a string into an array, please use
要将字符串转换为数组，请使用
arr=($line)
or
或者
read -a arr <<< $line
It is crucial not to use quotes since this does the trick.
重要的是不要使用引号，因为这可以解决问题。
回答by Randy
Try this:
尝试这个：
arr=(`echo ${line}`);
回答by Isaac
In: arr=( $line ). The "split" comes associated with "glob".
Wildcards (*,?and []) will be expanded to matching filenames.
在：arr=( $line )。“分裂”与“glob”相关联。
通配符（*,?和[]）将扩展为匹配的文件名。
The correct solution is only slightly more complex:
正确的解决方案只是稍微复杂一点：
IFS=' ' read -a arr <<< "$line"
No globbing problem; the split character is set in $IFS, variables quoted.
没有通配问题；拆分字符设置在$IFS，引用变量。
回答by David Anderson
If you need parameter expansion, then try:
如果您需要参数扩展，请尝试：
eval "arr=($line)"
For example, take the following code.
例如，采用以下代码。
line='a b "c d" "*" *'
eval "arr=($line)"
for s in "${arr[@]}"; do
echo "$s"
done
If the current directory contained the files a.txt, b.txtand c.txt, then executing the code would produce the following output.
如果当前目录包含文件a.txt,b.txt和c.txt，则执行代码将产生以下输出。
a
b
c d
*
a.txt
b.txt
c.txt
回答by Hitesh
line="1 1.50 string"
arr=$( $line | tr " " "\n")
for x in $arr
do
echo "> [$x]"
done
