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
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" "$2"
源
其他回答
你们都给出了复杂的解决方案。这里有一个更简单的例子。
function compare_versions {
local a=${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:${#y} + 1}
done
return 0
}
function version { echo "$@" | awk -F. '{ printf("%d%03d%03d%03d\n", $1,$2,$3,$4); }'; }
这样用:
if [ $(version $VAR) -ge $(version "6.2.0") ]; then
echo "Version is up to date"
fi
(来自https://apple.stackexchange.com/a/123408/11374)
这也是一个纯bash解决方案,因为printf是bash内置的。
function ver()
# Description: use for comparisons of version strings.
# $1 : a version string of form 1.2.3.4
# use: (( $(ver 1.2.3.4) >= $(ver 1.2.3.3) )) && echo "yes" || echo "no"
{
printf "%02d%02d%02d%02d" ${1//./ }
}
### 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。
我使用一个函数来规范化这些数字,然后比较它们。
for循环用于将版本字符串中的八进制数转换为十进制数,例如:1.08→1 8,1.0030→1 30,2021-02-03→2021 2 3…
(用bash 5.0.17测试
#!/usr/bin/env bash
v() {
printf "%04d%04d%04d%04d%04d" $(for i in ${1//[^0-9]/ }; do printf "%d " $((10#$i)); done)
}
while read -r test; do
set -- $test
printf "$test "
eval "if [[ $(v $1) $3 $(v $2) ]] ; then echo true; else echo false; fi"
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 ==
1 1 >
1.2.3~rc2 1.2.3~rc4 >
1.2.3~rc2 1.2.3~rc4 ==
1.2.3~rc2 1.2.3~rc4 <
1.2.3~rc2 1.2.3~rc4 !=
1.2.3~rc2 1.2.3+rc4 <
2021-11-23-rc1 2021-11-23-rc1.1 <
2021-11-23-rc1 2021-11-23-rc1-rf1 <
2021-01-03-rc1 2021-01-04 <
5.0.17(1)-release 5.0.17(2)-release <
EOF
结果:
1 1 == true
2.1 2.2 < true
3.0.4.10 3.0.4.2 > true
4.08 4.08.01 < true
3.2.1.9.8144 3.2 > true
3.2 3.2.1.9.8144 < true
1.2 2.1 < true
2.1 1.2 > true
5.6.7 5.6.7 == true
1.01.1 1.1.1 == true
1.1.1 1.01.1 == true
1 1.0 == true
1.0 1 == true
1.0.2.0 1.0.2 == true
1..0 1.0 == true
1.0 1..0 == true
1 1 > false
1.2.3~rc2 1.2.3~rc4 > false
1.2.3~rc2 1.2.3~rc4 == false
1.2.3~rc2 1.2.3~rc4 < true
1.2.3~rc2 1.2.3~rc4 != true
1.2.3~rc2 1.2.3+rc4 < true
2021-11-23-rc1 2021-11-23-rc1.1 < true
2021-11-23-rc1 2021-11-23-rc1-rf1 < true
2021-01-03-rc1 2021-01-04 < true
5.0.17(1)-release 5.0.17(2)-release < true