替换bash数组中的空元素

本文介绍了替换bash数组中的空元素的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

想象一下我创建了一个像这样的数组:

Imagine I created an array like this:

IFS="|" read -ra ARR <<< "zero|one|||four"

现在

echo ${#ARR[@]} > 5 echo "${ARR[@]}" > zero one four echo "${ARR[0]}" > zero echo "${ARR[2]}" > # Nothing, because it is empty

问题是如何将空元素替换为另一个字符串?

The question is how can I replace the empty elements with another string?

我尝试过

${ARR[@]///other} ${ARR[@]//""/other}

他们都没有工作.

我希望将此作为输出

zero one other other four

推荐答案

要使shell扩展起作用，您需要遍历其元素并对每个元素进行替换:

To have the shell expansion behave, you need to loop through its elements and perform the replacement on each one of them:

$ IFS="|" read -ra ARR <<< "zero|one|||four" $ for i in "${ARR[@]}"; do echo "${i:-other}"; done zero one other other four

位置:

$ {parameter:-word}

如果参数未设置或为null，则替换单词的扩展名.否则，将替换参数的值.

If parameter is unset or null, the expansion of word is substituted. Otherwise, the value of parameter is substituted.

要将它们存储在新数组中，只需添加+=( element ):

To store them in a new array, just do so by appending with +=( element ):

$ new=() $ for i in "${ARR[@]}"; do new+=("${i:-other}"); done $ printf "%s\n" "${new[@]}" zero one other other four