如何创建检查网站可用性的python函数？

我目前编写的代码如下：

def Ws_availbality(url):
    check = print(urllib.request.urlopen("https://www.youtube.com/").getcode())
    if check == 200:
        print ("Website is running")
    else:
        print("The website is currently down")

Ws_availbality('https://www.youtube.com/')

---

问题是，尽管它返回200，但我无法执行if语句。它只执行else语句。我也对不同的功能方法持开放态度

您必须将打印内容移到其他位置：

def Ws_availbality(url):
    check = urllib.request.urlopen("https://www.youtube.com/").getcode()
    print(check)
    if check == 200:
        print ("Website is running")
    else:
        print("The website is currently down")

Ws_availbality('https://www.youtube.com/')

print返回None，所以check是None。print做什么？