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
下面是对顶部答案(Dennis的)的改进，它更简洁，并使用了不同的返回值方案，以便通过单个比较轻松实现<=和>=。它还比较不是[0-9]的第一个字符之后的所有内容。]因此1.0rc1 < 1.0rc2。
# Compares two tuple-based, dot-delimited version numbers a and b (possibly
# with arbitrary string suffixes). Returns:
# 1 if a<b
# 2 if equal
# 3 if a>b
# Everything after the first character not in [0-9.] is compared
# lexicographically using ASCII ordering if the tuple-based versions are equal.
compare_versions() {
if [[ $1 == "$2" ]]; then
return 2
fi
local IFS=.
local i a=(${1%%[^0-9.]*}) b=(${2%%[^0-9.]*})
local arem=${1#${1%%[^0-9.]*}} brem=${2#${2%%[^0-9.]*}}
for ((i=0; i<${#a[@]} || i<${#b[@]}; i++)); do
if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
return 1
elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
return 3
fi
done
if [ "$arem" '<' "$brem" ]; then
return 1
elif [ "$arem" '>' "$brem" ]; then
return 3
fi
return 2
}
为了解决@gammazero的评论，一个(我认为)与语义版本兼容的更长的版本是:
# Compares two dot-delimited decimal-element version numbers a and b that may
# also have arbitrary string suffixes. Compatible with semantic versioning, but
# not as strict: comparisons of non-semver strings may have unexpected
# behavior.
#
# Returns:
# 1 if a<b
# 2 if equal
# 3 if a>b
compare_versions() {
local LC_ALL=C
# Optimization
if [[ $1 == "$2" ]]; then
return 2
fi
# Compare numeric release versions. Supports an arbitrary number of numeric
# elements (i.e., not just X.Y.Z) in which unspecified indices are regarded
# as 0.
local aver=${1%%[^0-9.]*} bver=${2%%[^0-9.]*}
local arem=${1#$aver} brem=${2#$bver}
local IFS=.
local i a=($aver) b=($bver)
for ((i=0; i<${#a[@]} || i<${#b[@]}; i++)); do
if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
return 1
elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
return 3
fi
done
# Remove build metadata before remaining comparison
arem=${arem%%+*}
brem=${brem%%+*}
# Prelease (w/remainder) always older than release (no remainder)
if [ -n "$arem" -a -z "$brem" ]; then
return 1
elif [ -z "$arem" -a -n "$brem" ]; then
return 3
fi
# Otherwise, split by periods and compare individual elements either
# numerically or lexicographically
local a=(${arem#-}) b=(${brem#-})
for ((i=0; i<${#a[@]} && i<${#b[@]}; i++)); do
local anns=${a[i]#${a[i]%%[^0-9]*}} bnns=${b[i]#${b[i]%%[^0-9]*}}
if [ -z "$anns$bnns" ]; then
# Both numeric
if ((10#${a[i]:-0} < 10#${b[i]:-0})); then
return 1
elif ((10#${a[i]:-0} > 10#${b[i]:-0})); then
return 3
fi
elif [ -z "$anns" ]; then
# Numeric comes before non-numeric
return 1
elif [ -z "$bnns" ]; then
# Numeric comes before non-numeric
return 3
else
# Compare lexicographically
if [[ ${a[i]} < ${b[i]} ]]; then
return 1
elif [[ ${a[i]} > ${b[i]} ]]; then
return 3
fi
fi
done
# Fewer elements is earlier
if (( ${#a[@]} < ${#b[@]} )); then
return 1
elif (( ${#a[@]} > ${#b[@]} )); then
return 3
fi
# Must be equal!
return 2
}
其他回答
function version { echo "$@" | 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
这里有一个支持修订的纯Bash解决方案(例如。“1.0-r1”)，答案是基于丹尼斯·威廉姆森(Dennis Williamson)发布的答案。可以很容易地修改它以支持'-RC1'之类的东西，或者通过更改正则表达式从更复杂的字符串中提取版本。
有关实现的详细信息，请参考代码内注释和/或启用包含的调试代码:
#!/bin/bash
# Compare two version strings [$1: version string 1 (v1), $2: version string 2 (v2)]
# Return values:
# 0: v1 == v2
# 1: v1 > v2
# 2: v1 < v2
# Based on: https://stackoverflow.com/a/4025065 by Dennis Williamson
function compare_versions() {
# Trivial v1 == v2 test based on string comparison
[[ "$1" == "$2" ]] && return 0
# Local variables
local regex="^(.*)-r([0-9]*)$" va1=() vr1=0 va2=() vr2=0 len i IFS="."
# Split version strings into arrays, extract trailing revisions
if [[ "$1" =~ ${regex} ]]; then
va1=(${BASH_REMATCH[1]})
[[ -n "${BASH_REMATCH[2]}" ]] && vr1=${BASH_REMATCH[2]}
else
va1=($1)
fi
if [[ "$2" =~ ${regex} ]]; then
va2=(${BASH_REMATCH[1]})
[[ -n "${BASH_REMATCH[2]}" ]] && vr2=${BASH_REMATCH[2]}
else
va2=($2)
fi
# Bring va1 and va2 to same length by filling empty fields with zeros
(( ${#va1[@]} > ${#va2[@]} )) && len=${#va1[@]} || len=${#va2[@]}
for ((i=0; i < len; ++i)); do
[[ -z "${va1[i]}" ]] && va1[i]="0"
[[ -z "${va2[i]}" ]] && va2[i]="0"
done
# Append revisions, increment length
va1+=($vr1)
va2+=($vr2)
len=$((len+1))
# *** DEBUG ***
#echo "TEST: '${va1[@]} (?) ${va2[@]}'"
# Compare version elements, check if v1 > v2 or v1 < v2
for ((i=0; i < len; ++i)); do
if (( 10#${va1[i]} > 10#${va2[i]} )); then
return 1
elif (( 10#${va1[i]} < 10#${va2[i]} )); then
return 2
fi
done
# All elements are equal, thus v1 == v2
return 0
}
# ---------- everything below this line is just for testing ----------
# Test compare_versions [$1: version string 1, $2: version string 2, $3: expected result]
function test_compare_versions() {
local op
compare_versions "$1" "$2"
case $? in
0) op="==" ;;
1) op=">" ;;
2) op="<" ;;
esac
if [[ "$op" == "$3" ]]; then
echo -e "\e[1;32mPASS: '$1 $op $2'\e[0m"
else
echo -e "\e[1;31mFAIL: '$1 $3 $2' (result: '$1 $op $2')\e[0m"
fi
}
echo -e "\nThe following tests should pass:"
while read -r test; do
test_compare_versions $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 ==
1.0-r1 1.0-r3 <
1.0-r9 2.0 <
3.0-r15 3.0-r9 >
...-r1 ...-r2 <
2.0-r1 1.9.8.21-r2 >
1.0 3.8.9.32-r <
-r -r3 <
-r3 -r >
-r3 -r3 ==
-r -r ==
0.0-r2 0.0.0.0-r2 ==
1.0.0.0-r2 1.0-r2 ==
0.0.0.1-r7 -r9 >
0.0-r0 0 ==
1.002.0-r6 1.2.0-r7 <
001.001-r2 1.1-r2 ==
5.6.1-r0 5.6.1 ==
EOF
echo -e "\nThe following tests should fail:"
while read -r test; do
test_compare_versions $test
done << EOF
1 1 >
3.0.5-r5 3..5-r5 >
4.9.21-r3 4.8.22-r9 <
1.0-r 1.0-r1 ==
-r 1.0-r >
-r1 0.0-r1 <
-r2 0-r2 <
EOF
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//./ }
}
