bash 判断 Ubuntu 版本是否大于 18.04

資深大佬 : sudoy 5

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

大佬有話說 (7)

資深大佬 : love

python -c “print($VER > 18.04)” | grep True && {
echo OK
}
資深大佬 : Xusually

#cat ver.sh

function version { echo “[email protected]” | awk -F. ‘{ printf(“%d%03d%03d%03dn”, $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
資深大佬 : miscnote

just my way:
perl -e ‘open $fd,”/etc/issue”;$str=<$fd>;($ver)=$str=~/(d+.d+.d+)/;print “true” if $ver>18.04’
資深大佬 : Xusually

# lsb_release –release
Release: 20.04
主資深大佬 : sudoy

@Xusually 感谢！我测试了这个方案可行的！
主資深大佬 : 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
資深大佬 : taolu

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