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
其他回答
当Bash变得太复杂时,就把它输送到python中!
vercomp(){ echo "$1" "$2" | python3 -c "import re, sys; arr = lambda x: list(map(int, re.split('[^0-9]+', x))); x, y = map(arr, sys.stdin.read().split()); exit(not x >= y)"; }
比较两个版本号的例子:
vercomp 2.8 2.4.5 && echo ">=" || echo "<"
这个python一行代码比较左边版本号和右边版本号,如果左边版本号等于或更高,则退出0。它还处理2.4.5rc3这样的版本
分解后,这是可读的代码:
import re, sys
# Convert a version string into a list "2.4.5" -> [2, 4, 5]
arr = lambda x: list(map(int, re.split('[^0-9]+', x)))
# Read the version numbers from stdin and apply the above function to them
x, y = map(arr, sys.stdin.read().split())
# Exit 0 if the left number is greater than the right
exit(not x >= y)
function version_compare () {
function sub_ver () {
local len=${#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 $v2`
fi
}
### Usage:
version_compare "1.2.3" "1.2.4"
# Output: <
功劳归于@Shellman
下面是另一个纯bash版本,比公认的答案要小得多。它只检查版本是否小于或等于“最小版本”,并且它将按字典顺序检查字母数字序列,这通常会给出错误的结果(举个常见的例子,“snapshot”不晚于“release”)。它将工作的主要/次要。
is_number() {
case "$BASH_VERSION" in
3.1.*)
PATTERN='\^\[0-9\]+\$'
;;
*)
PATTERN='^[0-9]+$'
;;
esac
[[ "$1" =~ $PATTERN ]]
}
min_version() {
if [[ $# != 2 ]]
then
echo "Usage: min_version current minimum"
return
fi
A="${1%%.*}"
B="${2%%.*}"
if [[ "$A" != "$1" && "$B" != "$2" && "$A" == "$B" ]]
then
min_version "${1#*.}" "${2#*.}"
else
if is_number "$A" && is_number "$B"
then
[[ "$A" -ge "$B" ]]
else
[[ ! "$A" < "$B" ]]
fi
fi
}
### 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 $2; then
echo higher
else
echo same or less
fi
它非常简单和小。
与其编写冗长的代码使您的生活过于复杂,不如使用一些已经存在的东西。很多时候,当bash不够用时,python可以提供帮助。你仍然可以很容易地从bash脚本调用它(额外的好处:从bash到python的变量替换):
VERSION1=1.2.3
VERSION2=1.2.4
cat <<EOF | python3 | grep -q True
from packaging import version
print(version.parse("$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?