如果你知道字段的数量，你可以用-k,n,n，得到一个超级简单的解

echo '2.4.5
2.8
2.4.5.1
2.10.2' | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -k 4,4 -g

2.4.5
2.4.5.1
2.8
2.10.2

下面是另一个纯bash版本，比公认的答案要小得多。它只检查版本是否小于或等于“最小版本”，并且它将按字典顺序检查字母数字序列，这通常会给出错误的结果(举个常见的例子，“snapshot”不晚于“release”)。它将工作的主要/次要。

is_number() {
    case "$BASH_VERSION" in
        3.1.*)
            PATTERN='\^\[0-9\]+\$'
            ;;
        *)
            PATTERN='^[0-9]+$'
            ;;
    esac

    [[ "$1" =~ $PATTERN ]]
}

min_version() {
    if [[ $# != 2 ]]
    then
        echo "Usage: min_version current minimum"
        return
    fi

    A="${1%%.*}"
    B="${2%%.*}"

    if [[ "$A" != "$1" && "$B" != "$2" && "$A" == "$B" ]]
    then
        min_version "${1#*.}" "${2#*.}"
    else
        if is_number "$A" && is_number "$B"
        then
            [[ "$A" -ge "$B" ]]
        else
            [[ ! "$A" < "$B" ]]
        fi
    fi
}

我遇到并解决了这个问题，添加了一个额外的(更短更简单的)答案…

首先注意，扩展shell比较失败了，你可能已经知道了…

    if [[ 1.2.0 < 1.12.12 ]]; then echo true; else echo false; fi
    false

使用sort -t'。'-g(或者kanaka提到的sort -V)来排序版本和简单的bash字符串比较，我找到了一个解决方案。输入文件包含列3和列4中的版本，我想对它们进行比较。这将遍历列表，确定匹配项或其中一个大于另一个。希望这仍然可以帮助那些希望使用bash尽可能简单地做到这一点的人。

while read l
do
    #Field 3 contains version on left to compare (change -f3 to required column).
    kf=$(echo $l | cut -d ' ' -f3)
    #Field 4 contains version on right to compare (change -f4 to required column).
    mp=$(echo $l | cut -d ' ' -f4)

    echo 'kf = '$kf
    echo 'mp = '$mp

    #To compare versions m.m.m the two can be listed and sorted with a . separator and the greater version found.
    gv=$(echo -e $kf'\n'$mp | sort -t'.' -g | tail -n 1)

    if [ $kf = $mp ]; then 
        echo 'Match Found: '$l
    elif [ $kf = $gv ]; then
        echo 'Karaf feature file version is greater '$l
    elif [ $mp = $gv ]; then
        echo 'Maven pom file version is greater '$l
   else
       echo 'Comparison error '$l
   fi
done < features_and_pom_versions.tmp.txt

感谢Barry的博客给出了排序的想法…… 裁判:http://bkhome.org/blog/?viewDetailed=02199

$ for OVFTOOL_VERSION in "4.2.0" "4.2.1" "5.2.0" "3.2.0" "4.1.9" "4.0.1" "4.3.0" "4.5.0" "4.2.1" "30.1.0" "4" "5" "4.1" "4.3"
> do
>   if [ $(echo "$OVFTOOL_VERSION 4.2.0" | tr " " "\n" | sort --version-sort | head -n 1) = 4.2.0 ]; then 
>     echo "$OVFTOOL_VERSION is >= 4.2.0"; 
>   else 
>     echo "$OVFTOOL_VERSION is < 4.2.0"; 
>   fi
> done
4.2.0 is >= 4.2.0
4.2.1 is >= 4.2.0
5.2.0 is >= 4.2.0
3.2.0 is < 4.2.0
4.1.9 is < 4.2.0
4.0.1 is < 4.2.0
4.3.0 is >= 4.2.0
4.5.0 is >= 4.2.0
4.2.1 is >= 4.2.0
30.1.0 is >= 4.2.0
4 is < 4.2.0
5 is >= 4.2.0
4.1 is < 4.2.0
4.3 is >= 4.2.0

我希望这对某些人有用(使用awk)

  #!/bin/bash

  available_version=1.2.3 # or online version
  this_version=1.2.1

  if [[ "ok" == "$(echo | awk "(${available_version} > ${this_version}) { print \"ok\"; }")" ]]; then
    echo "Notice, new version is available"
  elif [[ "ok" == "$(echo | awk "(${available_version} == ${this_version}) { print \"ok\"; }")" ]]; then
    echo "OK, version is up to date"
  else
    echo "Warning, the current version is ahead of the online version!"
  fi

我实现了一个函数，返回与Dennis Williamson相同的结果，但使用更少的行数。它最初执行一个健全性检查，导致1..0从他的测试中失败(我认为应该是这样)，但他所有的其他测试都通过了这段代码:

#!/bin/bash
version_compare() {
    if [[ $1 =~ ^([0-9]+\.?)+$ && $2 =~ ^([0-9]+\.?)+$ ]]; then
        local l=(${1//./ }) r=(${2//./ }) s=${#l[@]}; [[ ${#r[@]} -gt ${#l[@]} ]] && s=${#r[@]}

        for i in $(seq 0 $((s - 1))); do
            [[ ${l[$i]} -gt ${r[$i]} ]] && return 1
            [[ ${l[$i]} -lt ${r[$i]} ]] && return 2
        done

        return 0
    else
        echo "Invalid version number given"
        exit 1
    fi
}