问如何在bash shell中将一个字符串拆分成至少由一个空格分隔的多个字符串？

EN

sentence="This is   a sentence."
for word in $sentence
do
    echo $word
done

This
is
a
sentence.

sentence="this is a story"
stringarray=($sentence)

echo ${stringarray[0]}

for i in "${stringarray[@]}"
do
  :
  # do whatever on $i
done

for i in $sentence
do
  :
  # do whatever on $i
done

var="string    to  split"
read -ra arr <<<"$var"

var="string    to  split"
read -ra arr -d '' <<<"$var"

touch NOPE
var="* a  *"
read -ra arr <<<"$var"
for a in "${arr[@]}"; do echo "[$a]"; done

[*]
[a]
[*]

IFS=: read -ra arr < <(grep "^$USER:" /etc/passwd)
for a in "${arr[@]}"; do echo "[$a]"; done

[tino]
[x]
[1000]
[1000]
[Valentin Hilbig]
[/home/tino]
[/bin/bash]

IFS=: read -ra arr <<<' split  :   this    '
for a in "${arr[@]}"; do echo "[$a]"; done

[ split  ]
[   this    ]

var=$'\n\nthis is\n\n\na test\n\n'
IFS=$'\n' read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[@]}"; do let i++; echo "$i [$a]"; done

1 [this is]
2 [a test]

unset IFS
var=$'\n\nthis is\n\n\na test\n\n'
read -ra arr -d '' <<<"$var"
i=0; for a in "${arr[@]}"; do let i++; echo "$i [$a]"; done

1 [this]
2 [is]
3 [a]
4 [test]

case "$var" in
'')                empty_var;;                # variable is empty
*' '*)             have_space "$var";;        # have SPC
*[[:space:]]*)     have_whitespace "$var";;   # have whitespaces like TAB
*[^-+.,A-Za-z0-9]*) have_nonalnum "$var";;    # non-alphanum-chars found
*[-+.,]*)          have_punctuation "$var";;  # some punctuation chars found
*)                 default_case "$var";;      # if all above does not match
esac

case "$var" in (*' '*) true;; (*) false;; esac