Question

我正在抓取一个网站来获取公司和产品详情。它有div标签，其中有li标签，我想获得div标签内的所有li标签。我使用的是python 3.5.1和BeautifulSoup

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

我收到错误

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'

Answer 1

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.find_all()对象上调用ResultSet，因此发生错误。您应该像这样迭代这个对象，并在find_all()中的元素上调用ResultSet：

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

或者使用列表理解来获取li中所有companydetail的列表：

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

如何在div标签中获取所有li标签

1 个答案: