How does splitting string to array by 'read' with IFS word separator in bash generated extra space element?

问题

With only one character for IFS, it works fine:

shell@kernel: ~> l="2.4.3"; IFS="." read -a la <<< "$l"; for ((i = 0; i < ${#la[@]}; ++i)) do echo ${la[$i]}; done;
2
4
3

While there are two characters for IFS, extra space element generated

shell@kernel: ~> l="2->4->3"; IFS="->" read -a la <<< "$l"; for ((i = 0; i < ${#la[@]}; ++i)) do echo ${la[$i]}; done;
2

4

3
shell@kernel: ~> l="2..4..3"; IFS=".." read -a la <<< "$l"; for ((i = 0; i < ${#la[@]}; ++i)) do echo ${la[$i]}; done;
2

4

3

How can I get rid of the extra space element in the array?

回答1:

Continuing from the comment, you can either test for an empty element before storing the value in the array, or you can deal with the empty value when you echo it. Frankly, its simpler to do the latter, e.g.

l="2->4->3"; IFS="->" read -a la <<< "$l"; \
for ((i = 0; i < ${#la[@]}; ++i)) do \
[ -n "${la[i]}" ] && echo ${la[$i]}; done

Output

2
4
3

回答2:

i would just use sed to replace the separator

see, reusing your script + sed

bash$ l="2->4->3"
bash$ read -a la <<< "$(echo $l | sed 's/->/ /g')"
bash$ for ((i = 0; i < ${#la[@]}; ++i)) do echo ${la[$i]}; done
2
4
3

but I think I would do it entirely differently

bash$ l="2->4->3"
bash$ for x in `echo $l | sed 's/->/ /g'`; do echo $x; done
2
4
3

hope this helps

回答3:

You could transform the separator -> into a single character with:

l="2->4->3"

IFS="-" read -a la <<< "${l//->/-}"

printf '%s' "${la[@]}"

If there is the risk that the string may contain additional - then use a character that is improbable that will be in the string:

IFS="┵" read -a la <<< "${l//->/┵}"

来源：https://stackoverflow.com/questions/40011867/how-does-splitting-string-to-array-by-read-with-ifs-word-separator-in-bash-gen