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
这在版本中最多为4个字段。
$ function ver { printf "%03d%03d%03d%03d" $(echo "$1" | tr '.' ' '); }
$ [ $(ver 10.9) -lt $(ver 10.10) ] && echo hello
hello
其他回答
### 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 $2; then
echo higher
else
echo same or less
fi
它非常简单和小。
哇……这是一个老问题,但我认为这是一个相当优雅的答案。首先,使用shell参数展开(参见shell参数展开)将每个点分隔的版本转换为自己的数组。
v1="05.2.3" # some evil examples that work here
v2="7.001.0.0"
declare -a v1_array=(${v1//./ })
declare -a v2_array=(${v2//./ })
现在,这两个数组将版本号按优先级顺序作为数值字符串。上面的许多解决方案都是从这里开始的,但它们都源于这样的观察:版本字符串只是一个具有任意基数的整数。我们可以测试找到第一个不相等的数字(就像strcmp对字符串中的字符所做的那样)。
compare_version() {
declare -a v1_array=(${1//./ })
declare -a v2_array=(${2//./ })
while [[ -nz $v1_array ]] || [[ -nz $v2_array ]]; do
let v1_val=${v1_array:-0} # this will remove any leading zeros
let v2_val=${v2_array:-0}
let result=$((v1_val-v2_val))
if (( result != 0 )); then
echo $result
return
fi
v1_array=("${v1_array[@]:1}") # trim off the first "digit". it doesn't help
v2_array=("${v2_array[@]:1}")
done
# if we get here, both the arrays are empty and neither has been numerically
# different, which is equivalent to the two versions being equal
echo 0
return
}
如果第一个版本小于第二个版本,则返回负数,如果相等则返回零,如果第一个版本大于第二个版本则返回正数。一些输出:
$ compare_version 1 1.2
-2
$ compare_version "05.1.3" "5.001.03.0.0.0.1"
-1
$ compare_version "05.1.3" "5.001.03.0.0.0"
0
$ compare_version "05.1.3" "5.001.03.0"
0
$ compare_version "05.1.3" "5.001.30.0"
-27
$ compare_version "05.2.3" "7.001.0.0"
-2
$ compare_version "05.1.3" "5.001.30.0"
-27
$ compare_version "7.001.0.0" "05.1.3"
2
堕落的例子,”。2”或“3.0.”不起作用(未定义的结果),并且如果'. 2 '旁边出现非数字字符。它可能会失败(还没有测试),但肯定是没有定义的。因此,这应该与一个消毒函数或适当的检查有效的格式相匹配。此外,我相信通过一些调整,可以在没有太多额外负担的情况下使其更加健壮。
我遇到并解决了这个问题,添加了一个额外的(更短更简单的)答案…
首先注意,扩展shell比较失败了,你可能已经知道了…
if [[ 1.2.0 < 1.12.12 ]]; then echo true; else echo false; fi
false
使用sort -t'。'-g(或者kanaka提到的sort -V)来排序版本和简单的bash字符串比较,我找到了一个解决方案。输入文件包含列3和列4中的版本,我想对它们进行比较。这将遍历列表,确定匹配项或其中一个大于另一个。希望这仍然可以帮助那些希望使用bash尽可能简单地做到这一点的人。
while read l
do
#Field 3 contains version on left to compare (change -f3 to required column).
kf=$(echo $l | cut -d ' ' -f3)
#Field 4 contains version on right to compare (change -f4 to required column).
mp=$(echo $l | cut -d ' ' -f4)
echo 'kf = '$kf
echo 'mp = '$mp
#To compare versions m.m.m the two can be listed and sorted with a . separator and the greater version found.
gv=$(echo -e $kf'\n'$mp | sort -t'.' -g | tail -n 1)
if [ $kf = $mp ]; then
echo 'Match Found: '$l
elif [ $kf = $gv ]; then
echo 'Karaf feature file version is greater '$l
elif [ $mp = $gv ]; then
echo 'Maven pom file version is greater '$l
else
echo 'Comparison error '$l
fi
done < features_and_pom_versions.tmp.txt
感谢Barry的博客给出了排序的想法…… 裁判:http://bkhome.org/blog/?viewDetailed=02199
我希望这对某些人有用(使用awk)
#!/bin/bash
available_version=1.2.3 # or online version
this_version=1.2.1
if [[ "ok" == "$(echo | awk "(${available_version} > ${this_version}) { print \"ok\"; }")" ]]; then
echo "Notice, new version is available"
elif [[ "ok" == "$(echo | awk "(${available_version} == ${this_version}) { print \"ok\"; }")" ]]; then
echo "OK, version is up to date"
else
echo "Warning, the current version is ahead of the online version!"
fi
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。