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
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
其他回答
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
}
对于旧版本/busybox排序。简单的形式提供了粗略的结果,往往奏效。
sort -n
这是特别有用的版本,其中包含alpha符号,如
10.c.3
10.a.4
2.b.5
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
另一种方法(@joynes的修改版本)比较问题中问到的虚线版本 (即“1.2”、“2.3.4”、“1.0”、“1.10.1”等)。 最大数量的位置必须事先知道。该方法期望最多3个版本位置。
expr $(printf "$1\n$2" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != $2
使用示例:
expr $(printf "1.10.1\n1.7" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != "1.7"
返回:1,因为1.10.1大于1.7
expr $(printf "1.10.1\n1.11" | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -g | sed -n 2p) != "1.11"
返回:0,因为1.10.1比1.11低
您可以通过版本命令行查看版本约束
$ version ">=1.0, <2.0" "1.7"
$ go version | version ">=1.9"
Bash脚本示例:
#!/bin/bash
if `version -b ">=9.0.0" "$(gcc --version)"`; then
echo "gcc version satisfies constraints >=9.0.0"
else
echo "gcc version doesn't satisfies constraints >=9.0.0"
fi