我的观点是:

vercomp () {
    if [[ "${1}" == "${2}" ]]; then
        echo '0'
        return
    fi
    echo "${1}" | sed 's/\([0-9]\+\)\./\1\n/g' | {
        _RES_=-1
        for _VB_ in $(echo "${2}" | sed 's/\([0-9]\+\)\./\1\n/g'); do
            if ! read -r _VA_ || [[ "${_VB_}" -gt "${_VA_}" ]]; then
                _RES_=1
                break
            fi
        done
        read -r _VA_ && echo '-1' || echo "${_RES_}"
    }
}

语法:

vercomp VERSION_A VERSION_B

打印:

-1如果VERSION_A是最近的版本 如果两个版本相等，则为0 如果VERSION_B是最近的版本，则为1

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

你们都给出了复杂的解决方案。这里有一个更简单的例子。

function compare_versions {
    local a=${1%%.*} b=${2%%.*}
    [[ "10#${a:-0}" -gt "10#${b:-0}" ]] && return 1
    [[ "10#${a:-0}" -lt "10#${b:-0}" ]] && return 2
    a=${1:${#a} + 1} b=${2:${#b} + 1}
    [[ -z $a && -z $b ]] || compare_versions "$a" "$b"
}

用法:compare_versions <ver_a> <ver_b>

返回代码1表示第一个版本大于第二个版本，2表示小于第二个版本，0表示两者相等。

也是一个非递归的版本:

function compare_versions {
    local a=$1 b=$2 x y

    while [[ $a || $b ]]; do
        x=${a%%.*} y=${b%%.*}
        [[ "10#${x:-0}" -gt "10#${y:-0}" ]] && return 1
        [[ "10#${x:-0}" -lt "10#${y:-0}" ]] && return 2
        a=${a:${#x} + 1} b=${b:${#y} + 1}
    done

    return 0
}

与其编写冗长的代码使您的生活过于复杂，不如使用一些已经存在的东西。很多时候，当bash不够用时，python可以提供帮助。你仍然可以很容易地从bash脚本调用它(额外的好处:从bash到python的变量替换):

VERSION1=1.2.3
VERSION2=1.2.4

cat <<EOF | python3 | grep -q True
from packaging import version
print(version.parse("$VERSION1") > version.parse("$VERSION2"))
EOF

if [ "$?" == 0 ];  then
   echo "$VERSION1 is greater than $VERSION2"
else
   echo "$VERSION2 is greater or equal than $VERSION1"
fi

这里有更多信息:如何比较Python中的版本号?

这个怎么样?似乎有用?

checkVersion() {
subVer1=$1
subVer2=$2

[ "$subVer1" == "$subVer2" ] && echo "Version is same"
echo "Version 1 is $subVer1"
testVer1=$subVer1
echo "Test version 1 is $testVer1"
x=0
while [[ $testVer1 != "" ]]
do
  ((x++))
  testVer1=`echo $subVer1|cut -d "." -f $x`
  echo "testVer1 now is $testVer1"
  testVer2=`echo $subVer2|cut -d "." -f $x`
  echo "testVer2 now is $testVer2"
  if [[ $testVer1 -gt $testVer2 ]]
  then
    echo "$ver1 is greater than $ver2"
    break
  elif [[ "$testVer2" -gt "$testVer1" ]]
  then
    echo "$ver2 is greater than $ver1"
    break
  fi
  echo "This is the sub verion for first value $testVer1"
  echo "This is the sub verion for second value $testVer2"
done
}

ver1=$1
ver2=$2
checkVersion "$ver1" "$ver2"

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
}