如何获取div tag中的所有li标记

Question

我正在抓取一个网站来获取公司和产品的详细信息。 它有一个div标记，其中有li标记，我想得到div标记中的所有li标记。 我使用的是python3.5.1和beauthoulsoup

我的代码：

from bs4 import BeautifulSoup
import urllib.request
import re
r = urllib.request.urlopen('http://i.cantonfair.org.cn/en/ExpExhibitorList.aspx?k=glassware')
soup = BeautifulSoup(r, "html.parser")

links = soup.find_all("a", href=re.compile(r"expexhibitorlist\.aspx\?categoryno=[0-9]+"))
linksfromcategories = ([link["href"] for link in links])

string = "http://i.cantonfair.org.cn/en/"
linksfromcategories = [string + x for x in linksfromcategories]

for link in linksfromcategories:
    response = urllib.request.urlopen(link)
    soup2 = BeautifulSoup(response, "html.parser")
    links2 = soup2.find_all("a", href=re.compile(r"\ExpExhibitorList\.aspx\?categoryno=[0-9]+"))
    linksfromsubcategories = ([link["href"] for link in links2])
    linksfromsubcategories = [string + x for x in linksfromsubcategories]
    for link in linksfromsubcategories:
        response = urllib.request.urlopen(link)
        soup3 = BeautifulSoup(response, "html.parser")
        links3 = soup3.find_all("a", href=re.compile(r"\ExpExhibitorList\.aspx\?categoryno=[0-9]+"))
        linksfromsubcategories2 = ([link["href"] for link in links3])
        linksfromsubcategories2 = [string + x for x in linksfromsubcategories2]
        for link in linksfromsubcategories2:
            response2 = urllib.request.urlopen(link)
            soup4 = BeautifulSoup(response2, "html.parser")
            companylink = soup4.find_all("a", href=re.compile(r"\expCompany\.aspx\?corpid=[0-9]+"))
            companylink = ([link["href"] for link in companylink])
            companylink = [string + x for x in companylink]
            for link in companylink:
                response3 = urllib.request.urlopen(link)
                soup5 = BeautifulSoup(response3, "html.parser")
                companydetail = soup5.find_all("div", id="contact")
                for element in companydetail:
                    companyname = element.a[0].get_text()
                    print (companyname)
                    companyaddress = element.a[1].get_text()
                    print (companyaddress)And I am getting error

我犯了错误

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Answer 1

companydetail是ResultSet。也就是说，它是一个包含许多元素的iterable对象（比如list或set）。发生此错误是因为您试图对此ResultSet对象调用.find_all()。您应该像这样遍历这个对象，并对ResultSet中的元素调用find_all()：

for d in companydetail:
    lis = d.find_all('li')

或者使用列表理解获取companydetail中所有li的列表：

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