bash 判断 Ubuntu 版本是否大于 18.04

[sudoy]

就是想通过 .sh 文件判断 Ubuntu 版本是否高于 18.04 。搜了半天没找到合适的，下面这个还是错的。哪位大佬帮忙改下

VER=$(lsb_release --release | cut -f2)

if [ "$VER > 18.04" | bc ]; then
        echo "Ubuntu version is greater than 18"
else
        echo "Ubuntu version is less than 18"
exit 0
fi

[love]

python -c "print($VER > 18.04)" | grep True && {
echo OK
}

[Xusually]

#cat ver.sh

function version { echo "$@" | awk -F. '{ printf("%d%03d%03d%03d\n", $1,$2,$3,$4); }'; }

VER=$(lsb_release --release | cut -f2)

if [ $(version $VER) -ge $(version "18.04") ]; then
echo "Ubuntu version is greater than 18"
else
echo "Ubuntu version is less than 18"
fi

# chmod +x ver.sh
# ./ver.sh
Ubuntu version is greater than 18

[miscnote]

just my way:
perl -e 'open $fd,"/etc/issue";$str=<$fd>;($ver)=$str=~/(\d+\.\d+\.\d+)/;print "true" if $ver>18.04'

[Xusually]

# lsb_release --release
Release: 20.04

[sudoy]

@Xusually 感谢！我测试了这个方案可行的！

[sudoy]

@love @miscnote 两位这个方法好像是要依赖于 python 和 perl，我想要纯 bash 脚本的

[taolu]

test $(echo "$VER > 18.04" | bc) -eq 1 && echo "Ubuntu version is greater than 18" || echo "Ubuntu version is less than 18"