在Bash中，如何将数字列表转换为数字范围? [英] In Bash, how to convert number list into ranges of numbers?

问题描述

目前，我从命令中获得了数字的排序输出:

Currently I have a sorted output of numbers from a command:

18,19,62,161,162,163,165

18,19,62,161,162,163,165

我想将这些数字列表压缩为单个数字或数字范围的列表

I would like to condense these number lists into a list of single numbers or ranges of numbers

18-19,62,161-163,165

18-19,62,161-163,165

我考虑过尝试对bash中的数组进行排序并读取下一个数字以查看其是否为+1 ...重击:

I thought about trying to sort through the array in bash and read the next number to see if it is +1... I have a PHP function that does essentially the same thing, but I'm having trouble transposing it to Bash:

foreach ($missing as $key => $tag) {
    $next = $missing[$key+1];
    if (!isset($first)) {
        $first = $tag;
    }
    if($next != $tag + 1) {
        if($first == $tag) {
            echo '<tr><td>'.$tag.'</td></tr>';
        } else {
            echo '<tr><td>'.$first.'-'.$tag.'</td></tr>';
        }
        unset($first);
    }
}

我在想bash中可能有一个衬里可以做到这一点，但我的Google搜寻工作即将结束....

I'm thinking there's probably a one-liner in bash that could do this but my Googling is coming up short....

更新: 谢谢@Karoly Horvath提供的快速答案，我曾经用它来完成我的项目.我肯定会对任何更简单的解决方案感兴趣.

UPDATE: Thank you @Karoly Horvath for a quick answer which I used to finish my project. I'd sure be interested in any simpler solutions out there.

推荐答案

是的，shell进行变量替换，如果未设置prev，则该行变为:

Yes, shell does variable substitution, if prev is not set, that line becomes:

if [ -ne $n+1] 

这是一个有效的版本:

numbers="18,19,62,161,162,163,165"

echo $numbers, | sed "s/,/\n/g" | while read num; do
    if [[ -z $first ]]; then
        first=$num; last=$num; continue;
    fi
    if [[ num -ne $((last + 1)) ]]; then
        if [[ first -eq last ]]; then echo $first; else echo $first-$last; fi
        first=$num; last=$num
    else
        : $((last++))
    fi
done | paste -sd ","

18-19,62,161-163,165

这篇关于在Bash中，如何将数字列表转换为数字范围?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！