另一种方法(@joynes的修改版本)比较问题中问到的虚线版本 (即“1.2”、“2.3.4”、“1.0”、“1.10.1”等)。 最大数量的位置必须事先知道。该方法期望最多3个版本位置。

expr $(printf "$1\n$2" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != $2

使用示例:

expr $(printf "1.10.1\n1.7" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != "1.7"

返回:1，因为1.10.1大于1.7

expr $(printf "1.10.1\n1.11" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != "1.11"

返回:0，因为1.10.1比1.11低

你可以递归地拆分。和下面的算法进行比较，从这里开始。如果版本相同则返回10，如果版本1大于版本2则返回11，否则返回9。

#!/bin/bash
do_version_check() {

   [ "$1" == "$2" ] && return 10

   ver1front=`echo $1 | cut -d "." -f -1`
   ver1back=`echo $1 | cut -d "." -f 2-`

   ver2front=`echo $2 | cut -d "." -f -1`
   ver2back=`echo $2 | cut -d "." -f 2-`

   if [ "$ver1front" != "$1" ] || [ "$ver2front" != "$2" ]; then
       [ "$ver1front" -gt "$ver2front" ] && return 11
       [ "$ver1front" -lt "$ver2front" ] && return 9

       [ "$ver1front" == "$1" ] || [ -z "$ver1back" ] && ver1back=0
       [ "$ver2front" == "$2" ] || [ -z "$ver2back" ] && ver2back=0
       do_version_check "$ver1back" "$ver2back"
       return $?
   else
           [ "$1" -gt "$2" ] && return 11 || return 9
   fi
}    

do_version_check "$1" "$2"

源

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
}

GNU排序有一个选项:

printf '2.4.5\n2.8\n2.4.5.1\n' | sort -V

给:

2.4.5
2.4.5.1
2.8

我实现了另一个比较器函数。这一个有两个特定的要求:(i)我不希望函数失败使用返回1，但echo代替;(ii)当我们从git存储库中检索版本时，版本“1.0”应该大于“1.0.2”，这意味着“1.0”来自trunk。

function version_compare {
  IFS="." read -a v_a <<< "$1"
  IFS="." read -a v_b <<< "$2"

  while [[ -n "$v_a" || -n "$v_b" ]]; do
    [[ -z "$v_a" || "$v_a" -gt "$v_b" ]] && echo 1 && return
    [[ -z "$v_b" || "$v_b" -gt "$v_a" ]] && echo -1 && return

    v_a=("${v_a[@]:1}")
    v_b=("${v_b[@]:1}")
  done

  echo 0
}

请随意评论并提出改进建议。

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
}