在 Bash 中的单独行上打印数组元素？
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原文地址: http://stackoverflow.com/questions/15691942/
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Print array elements on separate lines in Bash?
提问by Axel Bregnsbo
How do I print the array element of a Bash array on separate lines? This one works, but surely there is a better way:
如何在单独的行上打印 Bash 数组的数组元素？这个有效，但肯定有更好的方法：
$ my_array=(one two three)
$ for i in ${my_array[@]}; do echo $i; done
one
two
three
Tried this one but it did not work:
试过这个，但没有用：
$ IFS=$'\n' echo ${my_array[*]}
one two three
回答by Gilles Quenot
Try doing this :
尝试这样做：
$ printf '%s\n' "${my_array[@]}"
The difference between $@and $*:
之间的差异$@和$*：
Unquoted, the results are unspecified. In Bash, both expand to separate args and then wordsplit and globbed.
Quoted,
"$@"expands each element as a separate argument, while"$*"expands to the args merged into one argument:"$1c$2c..."(wherecis the first char ofIFS).
未引用，结果未指定。在 Bash 中，两者都扩展为单独的 args，然后进行分词和全局化。
引用，
"$@"将每个元素扩展为单独的参数，而"$*"扩展为合并为一个参数的 args:（"$1c$2c..."其中c是 的第一个字符IFS）。
You almost always want "$@". Same goes for "${arr[@]}".
你几乎总是想要"$@"。也一样"${arr[@]}"。
Always quote them!
总是引用它们！
回答by perreal
Just quote the argument to echo:
只需引用 echo 的参数：
( IFS=$'\n'; echo "${my_array[*]}" )
the sub shell helps restoring the IFS after use
子外壳有助于在使用后恢复 IFS
回答by Steven Penny
Using for:
使用为：
for each in "${alpha[@]}"
do
echo "$each"
done
Using history; note this will fail if your values contain !:
使用历史；请注意，如果您的值包含!以下内容，这将失败：
history -p "${alpha[@]}"
Using basename; note this will fail if your values contain /:
使用basename; 请注意，如果您的值包含/以下内容，这将失败：
basename -a "${alpha[@]}"
Using shuf; note that results might not come out in order:
使用shuf；请注意，结果可能不会按顺序出现：
shuf -e "${alpha[@]}"
回答by 0x00
Another useful variant is pipe to tr:
另一个有用的变体是 pipe to tr：
echo "${my_array[@]}" | tr ' ' '\n'
echo "${my_array[@]}" | tr ' ' '\n'
This looks simple and compact
这看起来简单而紧凑
回答by wuxmedia
I tried the answers here in a giant for...if loop, but didn't get any joy - so I did it like this, maybe messy but did the job:
我在一个巨大的 for...if 循环中尝试了这里的答案，但没有得到任何乐趣 - 所以我这样做了，也许很乱，但完成了工作：
# EXP_LIST2 is iterated
# imagine a for loop
EXP_LIST="List item"
EXP_LIST2="$EXP_LIST2 \n $EXP_LIST"
done
echo -e $EXP_LIST2
although that added a space to the list, which is fine - I wanted it indented a bit. Also presume the "\n" could be printed in the original $EP_LIST.
尽管这在列表中添加了一个空格，这很好 - 我希望它缩进一点。还假设可以在原始 $EP_LIST 中打印“\n”。
