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

### 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。

其他回答

$ 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

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//./ }
}

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

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'

当Bash变得太复杂时,就把它输送到python中!

vercomp(){ echo "$1" "$2" | python3 -c "import re, sys; arr = lambda x: list(map(int, re.split('[^0-9]+', x))); x, y = map(arr, sys.stdin.read().split()); exit(not x >= y)"; }

比较两个版本号的例子:

vercomp 2.8 2.4.5 && echo ">=" || echo "<"

这个python一行代码比较左边版本号和右边版本号,如果左边版本号等于或更高,则退出0。它还处理2.4.5rc3这样的版本

分解后,这是可读的代码:

import re, sys

# Convert a version string into a list "2.4.5" -> [2, 4, 5]
arr = lambda x: list(map(int, re.split('[^0-9]+', x)))

# Read the version numbers from stdin and apply the above function to them
x, y = map(arr, sys.stdin.read().split())

# Exit 0 if the left number is greater than the right
exit(not x >= y)