是否有办法在bash上比较这些字符串,例如:2.4.5和2.8和2.4.5.1?


当前回答

我希望这对某些人有用(使用awk)

  #!/bin/bash

  available_version=1.2.3 # or online version
  this_version=1.2.1

  if [[ "ok" == "$(echo | awk "(${available_version} > ${this_version}) { print \"ok\"; }")" ]]; then
    echo "Notice, new version is available"
  elif [[ "ok" == "$(echo | awk "(${available_version} == ${this_version}) { print \"ok\"; }")" ]]; then
    echo "OK, version is up to date"
  else
    echo "Warning, the current version is ahead of the online version!"
  fi

其他回答

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

$ 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
function version_compare () {
  function sub_ver () {
    local len=${#1}
    temp=${1%%"."*} && indexOf=`echo ${1%%"."*} | echo ${#temp}`
    echo -e "${1:0:indexOf}"
  }
  function cut_dot () {
    local offset=${#1}
    local length=${#2}
    echo -e "${2:((++offset)):length}"
  }
  if [ -z "$1" ] || [ -z "$2" ]; then
    echo "=" && exit 0
  fi
  local v1=`echo -e "${1}" | tr -d '[[:space:]]'`
  local v2=`echo -e "${2}" | tr -d '[[:space:]]'`
  local v1_sub=`sub_ver $v1`
  local v2_sub=`sub_ver $v2`
  if (( v1_sub > v2_sub )); then
    echo ">"
  elif (( v1_sub < v2_sub )); then
    echo "<"
  else
    version_compare `cut_dot $v1_sub $v1` `cut_dot $v2_sub $v2`
  fi
}

### Usage:

version_compare "1.2.3" "1.2.4"
# Output: <

功劳归于@Shellman

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。

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