如何迭代包含bash空格的列表

Question

你能告诉我如何迭代列表中的项目可以包含空格吗？

x=("some word", "other word", "third word")
for word in $x ; do
    echo -e "$word\n"
done

如何强制输出:

some word
other word
third word

代替:

some
word
(...)
third
word

Answer 1

要正确循环项目,您需要使用${var[@]}.你需要引用它来确保带有空格的项目不被拆分:"${var[@]}".

全部一起:

x=("some word" "other word" "third word")
for word in "${x[@]}" ; do
  echo -e "$word\n"
done

或者,萨纳(感谢Charles Duffy)printf:

x=("some word" "other word" "third word")
for word in "${x[@]}" ; do
  printf '%s\n\n' "$word"
done