I have a variable which contains a space-delimited string:
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:
echo ${arr[0]}
echo ${arr[1]}
echo ${arr[2]}
outputs
1
1.50
string
Somewhere I found a solution which doesn't work:
arr=$(echo ${line})
If I run the echo statements above after this, I get:
1 1.50 string
[empty line]
[empty line]
I also tried
IFS=" "
arr=$(echo ${line})
with the same result. Can someone help, please?
set -f; arr=($string); set +f seems to be faster than read -r -a <<< $string.
shellcheck SC2206: Quote to prevent word splitting/globbing, or split robustly with mapfile or read -a.
In order to convert a string into an array, create an array from the string, letting the string get split naturally according to the IFS (Internal Field Separator) variable, which is the space char by default:
arr=($line)
or pass the string to the stdin of the read command using the herestring (<<<) operator:
read -a arr <<< "$line"
For the first example, it is crucial not to use quotes around $line since that is what allows the string to get split into multiple elements.
See also: https://github.com/koalaman/shellcheck/wiki/SC2206
In: arr=( $line ). The "split" comes associated with "glob".
Wildcards (*,? and []) will be expanded to matching filenames.
The correct solution is only slightly more complex:
IFS=' ' read -a arr <<< "$line"
No globbing problem; the split character is set in $IFS, variables quoted.
arr=($line) in the accepted answer suffers from globbing issues. For example, try: line="twinkling *"; arr=($line); declare -p arr.
<<< but it may be a good idea to still use double quotes for consistency and readability.
Try this:
arr=(`echo ${line}`);
line='*'
GNU Make 4.2.1 solution, but it doesn't did the job
sh shell. That shell has no arrays. The question is about arrays. Can't give you an answer compatible with both requirements. Unless you claim that the positional arguments are the only array in sh and, then, this: set -- $line; printf '%s\n' "$@" would work. Note that glob characters are still a problem in this case.
If you need parameter expansion, then try:
eval "arr=($line)"
For example, take the following code.
line='a b "c d" "*" *'
eval "arr=($line)"
for s in "${arr[@]}"; do
echo "$s"
done
If the current directory contained the files a.txt, b.txt and c.txt, then executing the code would produce the following output.
a
b
c d
*
a.txt
b.txt
c.txt
GNU Make 4.2.1 solution, but is not that
line="1 1.50 string"
arr=$( $line | tr " " "\n")
for x in $arr
do
echo "> [$x]"
done
tr is superfluous but it should loop over "${arr[@]}" instead, not $arr
Follow WeChat
Success story sharing
Want to stay one step ahead of the latest teleworks?
Subscribe Now相似问题
- How to check if a string contains a substring in Bash
- How do I read / convert an InputStream into a String in Java?
- How to insert an item into an array at a specific index (JavaScript)
- How do I split a string on a delimiter in Bash?
- Sort array of objects by string property value
- How to convert a string to lower case in Bash
- How to concatenate string variables in Bash
- Loop through an array of strings in Bash?
- How to split a string into an array in Bash?
for i in ${arr[@]}; do echo $i; doneecho ${arr[@]}$linehas globbing characters in it.mkdir x && cd x && touch A B C && line="*" arr=($line); echo ${#arr[@]}gives 3declare -a "arr=($line)"will ignoreIFSdelimiters inside quoted stringsline='*',read -a arr <<<$linealways work, but onlyarr=($line)fails.