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
如果你知道字段的数量,你可以用-k,n,n,得到一个超级简单的解
echo '2.4.5
2.8
2.4.5.1
2.10.2' | sort -t '.' -k 1,1 -k 2,2 -k 3,3 -k 4,4 -g
2.4.5
2.4.5.1
2.8
2.10.2
其他回答
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
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。
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
可能没有普遍正确的方法来实现这一点。如果您正在尝试比较Debian包系统中的版本,请尝试dpkg——compare-versions <first> <relation> <second>。
下面是一个不需要任何外部工具的纯Bash版本:
#!/bin/bash
vercomp () {
if [[ $1 == $2 ]]
then
return 0
fi
local IFS=.
local i ver1=($1) ver2=($2)
# fill empty fields in ver1 with zeros
for ((i=${#ver1[@]}; i<${#ver2[@]}; i++))
do
ver1[i]=0
done
for ((i=0; i<${#ver1[@]}; i++))
do
if [[ -z ${ver2[i]} ]]
then
# fill empty fields in ver2 with zeros
ver2[i]=0
fi
if ((10#${ver1[i]} > 10#${ver2[i]}))
then
return 1
fi
if ((10#${ver1[i]} < 10#${ver2[i]}))
then
return 2
fi
done
return 0
}
testvercomp () {
vercomp $1 $2
case $? in
0) op='=';;
1) op='>';;
2) op='<';;
esac
if [[ $op != $3 ]]
then
echo "FAIL: Expected '$3', Actual '$op', Arg1 '$1', Arg2 '$2'"
else
echo "Pass: '$1 $op $2'"
fi
}
# Run tests
# argument table format:
# testarg1 testarg2 expected_relationship
echo "The following tests should pass"
while read -r test
do
testvercomp $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 =
EOF
echo "The following test should fail (test the tester)"
testvercomp 1 1 '>'
运行测试:
$ . ./vercomp
The following tests should pass
Pass: '1 = 1'
Pass: '2.1 < 2.2'
Pass: '3.0.4.10 > 3.0.4.2'
Pass: '4.08 < 4.08.01'
Pass: '3.2.1.9.8144 > 3.2'
Pass: '3.2 < 3.2.1.9.8144'
Pass: '1.2 < 2.1'
Pass: '2.1 > 1.2'
Pass: '5.6.7 = 5.6.7'
Pass: '1.01.1 = 1.1.1'
Pass: '1.1.1 = 1.01.1'
Pass: '1 = 1.0'
Pass: '1.0 = 1'
Pass: '1.0.2.0 = 1.0.2'
Pass: '1..0 = 1.0'
Pass: '1.0 = 1..0'
The following test should fail (test the tester)
FAIL: Expected '>', Actual '=', Arg1 '1', Arg2 '1'