在python中组合或缩短if语句

Question

我定义了以下功能以检查新书的数量。

def getNewSellerNumber(isbn):
    res=requests.get('http://www.amazon.com/dp/'+isbn)
    soup = bs4.BeautifulSoup(res.text,'html.parser')
    elements=soup.select('#mediaOlp > div > div > div > div.a-fixed-right-grid-col.accordion-row-left-content.a-col-left > div:nth-of-type(2) > div > span:nth-of-type(1) > a')

    if elements == []:
        elements2=soup.select('#mediaOlp > div > div > div.a-fixed-right-grid-col.accordion-row-left-content.a-col-left > div:nth-of-type(2) > div > span:nth-of-type(1) > a')
        if elements2 == []:
            elements3=soup.select('span > span:nth-of-type(3) > span.olp-new.olp-link > a')
            if elements3 ==[]:
                return None
            else:
                return elements3[0].text
        else:
            return elements2[0].text
    else:
        return elements[0].text

它工作正常但我将来可能会添加另一个CSS选择器。这意味着我的If statemenst将会成长。为了存储CSS选择器，我创建了一个列表如下，但这次，它在每个条件中都返回None，因为其中一个选择器总是错误的。

def getSellers(isbn):
    res = requests.get('http://www.amazon.com/dp/'+isbn)
    soup = bs4.BeautifulSoup(res.text,'html.parser')
    selectors = ['#mediaOlp > div > div > div > div.a-fixed-right-grid-col.accordion-row-left-content.a-col-left > div:nth-of-type(2) > div > span:nth-of-type(1) > a','#mediaOlp > div > div > div.a-fixed-right-grid-col.accordion-row-left-content.a-col-left > div:nth-of-type(2) > div > span:nth-of-type(1) > a','span > span:nth-of-type(3) > span.olp-new.olp-link > a' ]
    for i in selectors:
        elements = soup.select(i)
        if elements ==[]:
            return None
        else:
            print(elements[0].text)

如何组合或缩短这些if语句？

Answer 1

循环后移动返回无

for i in selectors:
    elements = soup.select(i)
    if elements:
       return elements[0].text
return None