CODEWITHSUNDEEP
pythonweb-scrapingbeautifulsoup
Aman Kumar
Aman Kumar

Reputation: 1592

How to scrape url from list using python

I want to scrape the url present in the list. Basically I am scraping a website in I am scraping a link from that I am finding particular link an scraping those links and I search for another particular links a scrape it. My code:

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]
subcatlinks = list()
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])
  subcatlinks.append(linksfromsubcategories)
responses = urllib.request.urlopen(subcatlinks)
soup3 = BeautifulSoup(responses, "html.parser")
print (soup3)

And I am getting the error 

Traceback (most recent call last):
  File "D:\python\phase2.py", line 46, in <module>
    responses = urllib.request.urlopen(subcatlinks)
  File "C:\Users\amanp\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 162, in urlopen
    return opener.open(url, data, timeout)
  File "C:\Users\amanp\AppData\Local\Programs\Python\Python35-32\lib\urllib\request.py", line 456, in open
    req.timeout = timeout
AttributeError: 'list' object has no attribute 'timeout'

Upvotes: 0

Views: 958

Answers (1)

Josh Rumbut
Josh Rumbut

Reputation: 2710

You can only pass in one link at a time to urllib.request.urlopen as opposed to a whole list of them.

So you'll need another loop like this:

for link in subcatlinks:
    response = urllib.request.urlopen(link)
    soup3 = BeautifulSoup(response, "html.parser")
    print(soup3)

Upvotes: 1

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