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
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
其他回答
我实现了另一个比较器函数。这一个有两个特定的要求:(i)我不希望函数失败使用返回1,但echo代替;(ii)当我们从git存储库中检索版本时,版本“1.0”应该大于“1.0.2”,这意味着“1.0”来自trunk。
function version_compare {
IFS="." read -a v_a <<< "$1"
IFS="." read -a v_b <<< "$2"
while [[ -n "$v_a" || -n "$v_b" ]]; do
[[ -z "$v_a" || "$v_a" -gt "$v_b" ]] && echo 1 && return
[[ -z "$v_b" || "$v_b" -gt "$v_a" ]] && echo -1 && return
v_a=("${v_a[@]:1}")
v_b=("${v_b[@]:1}")
done
echo 0
}
请随意评论并提出改进建议。
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//./ }
}
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
$ for OVFTOOL_VERSION in "4.2.0" "4.2.1" "5.2.0" "3.2.0" "4.1.9" "4.0.1" "4.3.0" "4.5.0" "4.2.1" "30.1.0" "4" "5" "4.1" "4.3"
> do
> if [ $(echo "$OVFTOOL_VERSION 4.2.0" | tr " " "\n" | sort --version-sort | head -n 1) = 4.2.0 ]; then
> echo "$OVFTOOL_VERSION is >= 4.2.0";
> else
> echo "$OVFTOOL_VERSION is < 4.2.0";
> fi
> done
4.2.0 is >= 4.2.0
4.2.1 is >= 4.2.0
5.2.0 is >= 4.2.0
3.2.0 is < 4.2.0
4.1.9 is < 4.2.0
4.0.1 is < 4.2.0
4.3.0 is >= 4.2.0
4.5.0 is >= 4.2.0
4.2.1 is >= 4.2.0
30.1.0 is >= 4.2.0
4 is < 4.2.0
5 is >= 4.2.0
4.1 is < 4.2.0
4.3 is >= 4.2.0
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