如何获取div标签中的所有li标签

我在网站上获取公司和产品详细信息。它有div标签，里面有li标签，我想要获得div标签中的所有li标签。我使用python 3.5.1和BeautifulSoup如何获取div标签中的所有li标签

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

而且我得到错误

Traceback (most recent call last): 
    File "D:\python\phase3.py", line 54, in <module> 
    lis = companydetail.find_all('li') 
AttributeError: 'ResultSet' object has no attribute 'find_all'

来源

2016-02-26 Aman Kumar

它说的：

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

或者使用列表中理解的companydetail得到所有li s的列表第54行有错误，但是您只包含了37行，其中没有一行包含抛出错误的代码。 – wpercy

companydetail是ResultSet。也就是说，它是一个包含许多元素的可迭代对象（如list或set）。由于您尝试在此ResultSet对象上尝试调用.find_all()，因此发生此错误。你应该通过这个对象像下面这样迭代，并呼吁元素find_all()在ResultSet：

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

来源

2016-02-26 19:51:34 wpercy

好的这个作品但我得到两次列表。为什么？ –

你是什么意思“获得两次列表”？ – wpercy

像从李atg我得到公司的详细信息，如姓名和电子邮件ID，但该名称和电子邮件ID是两次。可能是我刮了两次URL？ –

如何获取div标签中的所有li标签

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