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
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
其他回答
感谢Dennis的解决方案，我们可以扩展它以允许比较运算符'>'，'<'，'='，'=='，'<='和'>='。
# compver ver1 '=|==|>|<|>=|<=' ver2
compver() {
local op
vercomp $1 $3
case $? in
0) op='=';;
1) op='>';;
2) op='<';;
esac
[[ $2 == *$op* ]] && return 0 || return 1
}
然后我们可以在表达式中使用比较运算符，比如:
compver 1.7 '<=' 1.8
compver 1.7 '==' 1.7
compver 1.7 '=' 1.7
并且只测试结果的真/假，比如:
if compver $ver1 '>' $ver2; then
echo "Newer"
fi
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
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。
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
如果它只是想知道一个版本是否比另一个版本低，我会检查sort——version-sort是否会改变我的版本字符串的顺序:
string="$1
$2"
[ "$string" == "$(sort --version-sort <<< "$string")" ]
