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
}

下面是一个不需要任何外部工具的纯Bash版本:

#!/bin/bash
vercomp () {
    if [[ $1 == $2 ]]
    then
        return 0
    fi
    local IFS=.
    local i ver1=($1) ver2=($2)
    # fill empty fields in ver1 with zeros
    for ((i=${#ver1[@]}; i<${#ver2[@]}; i++))
    do
        ver1[i]=0
    done
    for ((i=0; i<${#ver1[@]}; i++))
    do
        if [[ -z ${ver2[i]} ]]
        then
            # fill empty fields in ver2 with zeros
            ver2[i]=0
        fi
        if ((10#${ver1[i]} > 10#${ver2[i]}))
        then
            return 1
        fi
        if ((10#${ver1[i]} < 10#${ver2[i]}))
        then
            return 2
        fi
    done
    return 0
}

testvercomp () {
    vercomp $1 $2
    case $? in
        0) op='=';;
        1) op='>';;
        2) op='<';;
    esac
    if [[ $op != $3 ]]
    then
        echo "FAIL: Expected '$3', Actual '$op', Arg1 '$1', Arg2 '$2'"
    else
        echo "Pass: '$1 $op $2'"
    fi
}

# Run tests
# argument table format:
# testarg1   testarg2     expected_relationship
echo "The following tests should pass"
while read -r test
do
    testvercomp $test
done << EOF
1            1            =
2.1          2.2          <
3.0.4.10     3.0.4.2      >
4.08         4.08.01      <
3.2.1.9.8144 3.2          >
3.2          3.2.1.9.8144 <
1.2          2.1          <
2.1          1.2          >
5.6.7        5.6.7        =
1.01.1       1.1.1        =
1.1.1        1.01.1       =
1            1.0          =
1.0          1            =
1.0.2.0      1.0.2        =
1..0         1.0          =
1.0          1..0         =
EOF

echo "The following test should fail (test the tester)"
testvercomp 1 1 '>'

运行测试:

$ . ./vercomp
The following tests should pass
Pass: '1 = 1'
Pass: '2.1 < 2.2'
Pass: '3.0.4.10 > 3.0.4.2'
Pass: '4.08 < 4.08.01'
Pass: '3.2.1.9.8144 > 3.2'
Pass: '3.2 < 3.2.1.9.8144'
Pass: '1.2 < 2.1'
Pass: '2.1 > 1.2'
Pass: '5.6.7 = 5.6.7'
Pass: '1.01.1 = 1.1.1'
Pass: '1.1.1 = 1.01.1'
Pass: '1 = 1.0'
Pass: '1.0 = 1'
Pass: '1.0.2.0 = 1.0.2'
Pass: '1..0 = 1.0'
Pass: '1.0 = 1..0'
The following test should fail (test the tester)
FAIL: Expected '>', Actual '=', Arg1 '1', Arg2 '1'

如果你有coreutils-7 (Ubuntu Karmic，而不是Jaunty)，那么你的排序命令应该有一个-V选项(版本排序)，你可以使用它来进行比较:

verlte() {
    [  "$1" = "`echo -e "$1\n$2" | sort -V | head -n1`" ]
}

verlt() {
    [ "$1" = "$2" ] && return 1 || verlte $1 $2
}

verlte 2.5.7 2.5.6 && echo "yes" || echo "no" # no
verlt 2.4.10 2.4.9 && echo "yes" || echo "no" # no
verlt 2.4.8 2.4.10 && echo "yes" || echo "no" # yes
verlte 2.5.6 2.5.6 && echo "yes" || echo "no" # yes
verlt 2.5.6 2.5.6 && echo "yes" || echo "no" # no

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>。

### the answer is does we second argument is higher
function _ver_higher {
        ver=`echo -ne "$1\n$2" |sort -Vr |head -n1`
        if [ "$2" == "$1" ]; then
                return 1
        elif [ "$2" == "$ver" ]; then
                return 0
        else
                return 1
        fi
}

if _ver_higher $1 $2; then
        echo higher
else
        echo same or less
fi

它非常简单和小。

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。