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

$ 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

您可以通过版本命令行查看版本约束

$ version ">=1.0, <2.0" "1.7"
$ go version | version ">=1.9"

Bash脚本示例:

#!/bin/bash

if `version -b ">=9.0.0" "$(gcc --version)"`; then
  echo "gcc version satisfies constraints >=9.0.0"
else
  echo "gcc version doesn't satisfies constraints >=9.0.0"
fi

我使用嵌入式Linux (Yocto)与BusyBox。BusyBox排序没有-V选项(但BusyBox expr匹配可以做正则表达式)。所以我需要一个Bash版本的比较，它适用于这个约束。

我做了以下(类似于Dennis Williamson的回答)来比较使用“自然排序”类型的算法。它将字符串分成数字部分和非数字部分;它以数字方式比较数字部分(因此10大于9)，并以纯ASCII方式比较非数字部分。

ascii_frag() {
    expr match "$1" "\([^[:digit:]]*\)"
}

ascii_remainder() {
    expr match "$1" "[^[:digit:]]*\(.*\)"
}

numeric_frag() {
    expr match "$1" "\([[:digit:]]*\)"
}

numeric_remainder() {
    expr match "$1" "[[:digit:]]*\(.*\)"
}

vercomp_debug() {
    OUT="$1"
    #echo "${OUT}"
}

# return 1 for $1 > $2
# return 2 for $1 < $2
# return 0 for equal
vercomp() {
    local WORK1="$1"
    local WORK2="$2"
    local NUM1="", NUM2="", ASCII1="", ASCII2=""
    while true; do
        vercomp_debug "ASCII compare"
        ASCII1=`ascii_frag "${WORK1}"`
        ASCII2=`ascii_frag "${WORK2}"`
        WORK1=`ascii_remainder "${WORK1}"`
        WORK2=`ascii_remainder "${WORK2}"`
        vercomp_debug "\"${ASCII1}\" remainder \"${WORK1}\""
        vercomp_debug "\"${ASCII2}\" remainder \"${WORK2}\""

        if [ "${ASCII1}" \> "${ASCII2}" ]; then
            vercomp_debug "ascii ${ASCII1} > ${ASCII2}"
            return 1
        elif [ "${ASCII1}" \< "${ASCII2}" ]; then
            vercomp_debug "ascii ${ASCII1} < ${ASCII2}"
            return 2
        fi
        vercomp_debug "--------"

        vercomp_debug "Numeric compare"
        NUM1=`numeric_frag "${WORK1}"`
        NUM2=`numeric_frag "${WORK2}"`
        WORK1=`numeric_remainder "${WORK1}"`
        WORK2=`numeric_remainder "${WORK2}"`
        vercomp_debug "\"${NUM1}\" remainder \"${WORK1}\""
        vercomp_debug "\"${NUM2}\" remainder \"${WORK2}\""

        if [ -z "${NUM1}" -a -z "${NUM2}" ]; then
            vercomp_debug "blank 1 and blank 2 equal"
            return 0
        elif [ -z "${NUM1}" -a -n "${NUM2}" ]; then
            vercomp_debug "blank 1 less than non-blank 2"
            return 2
        elif [ -n "${NUM1}" -a -z "${NUM2}" ]; then
            vercomp_debug "non-blank 1 greater than blank 2"
            return 1
        fi

        if [ "${NUM1}" -gt "${NUM2}" ]; then
            vercomp_debug "num ${NUM1} > ${NUM2}"
            return 1
        elif [ "${NUM1}" -lt "${NUM2}" ]; then
            vercomp_debug "num ${NUM1} < ${NUM2}"
            return 2
        fi
        vercomp_debug "--------"
    done
}

它可以比较更复杂的版本号，例如

1.2-r3和1.2-r4 1.2 r3 vs 1.2r4

请注意，对于Dennis Williamson的回答中的一些极端情况，它不会返回相同的结果。特别是:

1            1.0          <
1.0          1            >
1.0.2.0      1.0.2        >
1..0         1.0          >
1.0          1..0         <

但这些都是极端情况，我认为结果仍然是合理的。

这里是另一个没有任何外部调用的纯bash解决方案:

#!/bin/bash

function version_compare {

IFS='.' read -ra ver1 <<< "$1"
IFS='.' read -ra ver2 <<< "$2"

[[ ${#ver1[@]} -gt ${#ver2[@]} ]] && till=${#ver1[@]} || till=${#ver2[@]}

for ((i=0; i<${till}; i++)); do

    local num1; local num2;

    [[ -z ${ver1[i]} ]] && num1=0 || num1=${ver1[i]}
    [[ -z ${ver2[i]} ]] && num2=0 || num2=${ver2[i]}

    if [[ $num1 -gt $num2 ]]; then
        echo ">"; return 0
    elif
       [[ $num1 -lt $num2 ]]; then
        echo "<"; return 0
    fi
done

echo "="; return 0
}

echo "${1} $(version_compare "${1}" "${2}") ${2}"

还有更简单的解决方案，如果你确定所讨论的版本在第一个点后不包含前导零:

#!/bin/bash

function version_compare {

local ver1=${1//.}
local ver2=${2//.}


    if [[ $ver1 -gt $ver2 ]]; then
        echo ">"; return 0
    elif    
       [[ $ver1 -lt $ver2 ]]; then
        echo "<"; return 0
    fi 

echo "="; return 0
}

echo "${1} $(version_compare "${1}" "${2}") ${2}"

这适用于像1.2.3 vs 1.3.1 vs 0.9.7这样的版本，但不适用于其他版本 1.2.3 vs 1.2.3.0或1.01.1 vs 1.1.1

哇……这是一个老问题，但我认为这是一个相当优雅的答案。首先，使用shell参数展开(参见shell参数展开)将每个点分隔的版本转换为自己的数组。

v1="05.2.3"     # some evil examples that work here
v2="7.001.0.0"

declare -a v1_array=(${v1//./ })
declare -a v2_array=(${v2//./ })

现在，这两个数组将版本号按优先级顺序作为数值字符串。上面的许多解决方案都是从这里开始的，但它们都源于这样的观察:版本字符串只是一个具有任意基数的整数。我们可以测试找到第一个不相等的数字(就像strcmp对字符串中的字符所做的那样)。

compare_version() {
  declare -a v1_array=(${1//./ })
  declare -a v2_array=(${2//./ })

  while [[ -nz $v1_array ]] || [[ -nz $v2_array ]]; do
    let v1_val=${v1_array:-0}  # this will remove any leading zeros
    let v2_val=${v2_array:-0}
    let result=$((v1_val-v2_val))

    if (( result != 0 )); then
      echo $result
      return
    fi

    v1_array=("${v1_array[@]:1}") # trim off the first "digit". it doesn't help
    v2_array=("${v2_array[@]:1}")
  done

  # if we get here, both the arrays are empty and neither has been numerically
  # different, which is equivalent to the two versions being equal

  echo 0
  return
}

如果第一个版本小于第二个版本，则返回负数，如果相等则返回零，如果第一个版本大于第二个版本则返回正数。一些输出:

$ compare_version 1 1.2
-2
$ compare_version "05.1.3" "5.001.03.0.0.0.1"
-1
$ compare_version "05.1.3" "5.001.03.0.0.0"
0
$ compare_version "05.1.3" "5.001.03.0"
0
$ compare_version "05.1.3" "5.001.30.0"
-27
$ compare_version "05.2.3" "7.001.0.0"
-2
$ compare_version "05.1.3" "5.001.30.0"
-27
$ compare_version "7.001.0.0" "05.1.3"
2

堕落的例子，”。2”或“3.0.”不起作用(未定义的结果)，并且如果'. 2 '旁边出现非数字字符。它可能会失败(还没有测试)，但肯定是没有定义的。因此，这应该与一个消毒函数或适当的检查有效的格式相匹配。此外，我相信通过一些调整，可以在没有太多额外负担的情况下使其更加健壮。