bash 判断 Ubuntu 版本是否大于 18.04

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这是一个创建于 1362 天前的主题，其中的信息可能已经有所发展或是发生改变。

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

第 1 条附言 · 2021-03-02 13:52:55 +08:00

感谢 @@Xusually 的解决方案，亲测可行，代码如下：

# Check if Ubuntu version is greater than 18.04
echo "Checking Ubuntu version..."
VER=$(lsb_release --release | cut -f2)
function version { echo "$@" | awk -F. '{ printf("%d%03d%03d%03d\n", $1,$2,$3,$4); }'; }

if [ $(version $VER) -ge $(version "18.04") ]; then
        echo "Ubuntu version is greater than 18.04"
else
        echo "Ubuntu version is less than 18.04"
exit 0
fi

7 条回复 • 2021-03-02 13:52:33 +08:00

love

2021-03-02 13:30:00 +08:00

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

Xusually

2021-03-02 13:34:34 +08:00

#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