当Bash变得太复杂时，就把它输送到python中!

vercomp(){ echo "$1" "$2" | python3 -c "import re, sys; arr = lambda x: list(map(int, re.split('[^0-9]+', x))); x, y = map(arr, sys.stdin.read().split()); exit(not x >= y)"; }

比较两个版本号的例子:

vercomp 2.8 2.4.5 && echo ">=" || echo "<"

这个python一行代码比较左边版本号和右边版本号，如果左边版本号等于或更高，则退出0。它还处理2.4.5rc3这样的版本

分解后，这是可读的代码:

import re, sys

# Convert a version string into a list "2.4.5" -> [2, 4, 5]
arr = lambda x: list(map(int, re.split('[^0-9]+', x)))

# Read the version numbers from stdin and apply the above function to them
x, y = map(arr, sys.stdin.read().split())

# Exit 0 if the left number is greater than the right
exit(not x >= y)

下面是对顶部答案(Dennis的)的改进，它更简洁，并使用了不同的返回值方案，以便通过单个比较轻松实现<=和>=。它还比较不是[0-9]的第一个字符之后的所有内容。]因此1.0rc1 < 1.0rc2。

# Compares two tuple-based, dot-delimited version numbers a and b (possibly
# with arbitrary string suffixes). Returns:
# 1 if a<b
# 2 if equal
# 3 if a>b
# Everything after the first character not in [0-9.] is compared
# lexicographically using ASCII ordering if the tuple-based versions are equal.
compare_versions() {
    if [[ $1 == "$2" ]]; then
        return 2
    fi
    local IFS=.
    local i a=(${1%%[^0-9.]*}) b=(${2%%[^0-9.]*})
    local arem=${1#${1%%[^0-9.]*}} brem=${2#${2%%[^0-9.]*}}
    for ((i=0; i<${#a[@]} || i<${#b[@]}; i++)); do
        if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
            return 1
        elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
            return 3
        fi
    done
    if [ "$arem" '<' "$brem" ]; then
        return 1
    elif [ "$arem" '>' "$brem" ]; then
        return 3
    fi
    return 2
}

为了解决@gammazero的评论，一个(我认为)与语义版本兼容的更长的版本是:

# Compares two dot-delimited decimal-element version numbers a and b that may
# also have arbitrary string suffixes. Compatible with semantic versioning, but
# not as strict: comparisons of non-semver strings may have unexpected
# behavior.
#
# Returns:
# 1 if a<b
# 2 if equal
# 3 if a>b
compare_versions() {
    local LC_ALL=C

    # Optimization
    if [[ $1 == "$2" ]]; then
        return 2
    fi

    # Compare numeric release versions. Supports an arbitrary number of numeric
    # elements (i.e., not just X.Y.Z) in which unspecified indices are regarded
    # as 0.
    local aver=${1%%[^0-9.]*} bver=${2%%[^0-9.]*}
    local arem=${1#$aver} brem=${2#$bver}
    local IFS=.
    local i a=($aver) b=($bver)
    for ((i=0; i<${#a[@]} || i<${#b[@]}; i++)); do
        if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
            return 1
        elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
            return 3
        fi
    done

    # Remove build metadata before remaining comparison
    arem=${arem%%+*}
    brem=${brem%%+*}

    # Prelease (w/remainder) always older than release (no remainder)
    if [ -n "$arem" -a -z "$brem" ]; then
        return 1
    elif [ -z "$arem" -a -n "$brem" ]; then
        return 3
    fi

    # Otherwise, split by periods and compare individual elements either
    # numerically or lexicographically
    local a=(${arem#-}) b=(${brem#-})
    for ((i=0; i<${#a[@]} && i<${#b[@]}; i++)); do
        local anns=${a[i]#${a[i]%%[^0-9]*}} bnns=${b[i]#${b[i]%%[^0-9]*}}
        if [ -z "$anns$bnns" ]; then
            # Both numeric
            if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
                return 1
            elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
                return 3
            fi
        elif [ -z "$anns" ]; then
            # Numeric comes before non-numeric
            return 1
        elif [ -z "$bnns" ]; then
            # Numeric comes before non-numeric
            return 3
        else
            # Compare lexicographically
            if [[ ${a[i]} < ${b[i]} ]]; then
                return 1
            elif [[ ${a[i]} > ${b[i]} ]]; then
                return 3
            fi
        fi
    done

    # Fewer elements is earlier
    if (( ${#a[@]} < ${#b[@]} )); then
        return 1
    elif (( ${#a[@]} > ${#b[@]} )); then
        return 3
    fi

    # Must be equal!
    return 2
}

可能没有普遍正确的方法来实现这一点。如果您正在尝试比较Debian包系统中的版本，请尝试dpkg——compare-versions <first> <relation> <second>。

如果它只是想知道一个版本是否比另一个版本低，我会检查sort——version-sort是否会改变我的版本字符串的顺序:

    string="$1
$2"
    [ "$string" == "$(sort --version-sort <<< "$string")" ]

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

function version { echo "$@" | awk -F. '{ printf("%d%03d%03d%03d\n", $1,$2,$3,$4); }'; }

这样用:

if [ $(version $VAR) -ge $(version "6.2.0") ]; then
    echo "Version is up to date"
fi

(来自https://apple.stackexchange.com/a/123408/11374)