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
$ 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
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
}
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" "$ver2"
如果你知道字段的数量,你可以用-k,n,n,得到一个超级简单的解
echo '2.4.5
2.8
2.4.5.1
2.10.2' | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -k 4,4 -g
2.4.5
2.4.5.1
2.8
2.10.2
function version { echo "$@" | 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)
我实现了一个函数,返回与Dennis Williamson相同的结果,但使用更少的行数。它最初执行一个健全性检查,导致1..0从他的测试中失败(我认为应该是这样),但他所有的其他测试都通过了这段代码:
#!/bin/bash
version_compare() {
if [[ $1 =~ ^([0-9]+\.?)+$ && $2 =~ ^([0-9]+\.?)+$ ]]; then
local l=(${1//./ }) r=(${2//./ }) s=${#l[@]}; [[ ${#r[@]} -gt ${#l[@]} ]] && s=${#r[@]}
for i in $(seq 0 $((s - 1))); do
[[ ${l[$i]} -gt ${r[$i]} ]] && return 1
[[ ${l[$i]} -lt ${r[$i]} ]] && return 2
done
return 0
else
echo "Invalid version number given"
exit 1
fi
}