我如何在没有换行符的情况下“回显”内容？

Question

I have the following code:

for x in "${array[@]}"
do
  echo "$x"
done

The results are something like this (I sort these later in some cases):

1
2
3
4
5

Is there a way to print it as 1 2 3 4 5 instead? Without adding a newline every time?

Answer 1

Yes. Use the -n option:

echo -n "$x"

From help echo :

-n do not append a newline

This would strips off the last newline too, so if you want you can add a final newline after the loop:

for ...; do ...; done; echo

---

Note:

This is not portable among various implementations of echo builtin/external executable. The portable way would be to use printf instead:

printf '%s' "$x"

Answer 2

printf '%s\n' "${array[@]}" | sort | tr '\n' ' '

printf '%s\n' -- 比echo更健壮，并且为了sort的缘故，您希望在此处使用换行符"${array[@]}" -- 对您的特定数组而言，引号是不必要的，但这是一种很好的做法，因为您不需要通常需要在那里进行分词和全局扩展

Answer 3

You don't need a for loop to sort numbers from an array.

Use process substitution like this:

sort <(printf "%s\n" "${array[@]}")

To remove new lines, use:

sort <(printf "%s\n" "${array[@]}") | tr '\n' ' '

Answer 4

You can also do it this way:

array=(1 2 3 4 5)
echo "${array[@]}"

Answer 5

如果出于某种原因， -n不能为您解决此问题，您还可以将\c添加到要回显的内容的末尾：

echo "$x\c"