Question

我有一个简单的BeautifulSoup抓取工具，它根据添加的函数数量返回深度为2或更多的服务器链接：

import requests
from bs4 import BeautifulSoup

def spider():
    address = "http://dog.carnivore.mammal.xyz"
    pageFull = requests.get(address)
    pageText = pageFull.text
    soup = BeautifulSoup(pageText, "html.parser")
    for link in soup.findAll("a"):
        href = link.get("href")
        print(href)
        depth2(href)

def depth2(address):
    pageFull = requests.get(address)
    pageText = pageFull.text
    soup = BeautifulSoup(pageText, "html.parser")
    for name in soup.findAll("a"):
        href = name.get("href")
        print(href)
        depth3(href)

def depth3(address):
    etc...

spider()

这当然可以进一步扩展，但我正在寻找的是扩展蜘蛛函数的方法，使其能够递归地继续在新的链接上继续，没有新的深度方法 - 但只有新的深度仍然在服务器上，在这种情况下 mammal.xyz。如何完成这样的递归循环？

Answer 1

from bs4 import BeautifulSoup
import requests

from urllib.parse import urlparse, urljoin

MAX_DEPTH = 4

already_visited = set()


def _get_uri(base_url, href):
    if href.startswith('#'):
        print('discarding anchor {}'.format(href))
        return None

    # We use urljoin to deal with relative hrefs (e.g. <a href="../about.html">)
    # urljoin resolves them using the base_url we provide
    absolute_url = urljoin(base_url, href)

    # Here we take the hostname and we strip www because we want to consider
    # www.test.com and test.com as the same location
    base_url_host = urlparse(base_url).hostname.replace('www.', '')
    absolute_url_host = urlparse(absolute_url).hostname.replace('www.', '')

    # Skip if the href is external (outside of the starting domain)
    # Note that we just stripped www so that www.foo.com and foo.com are considered ==
    if absolute_url_host != base_url_host:
        print('Skipping {}'.format(href))
        return None

    return absolute_url


def spider(start_address):
    _crawl_page(start_address, '/', [])


def _crawl_page(start_address, uri, followed_path):
    path = followed_path[:]
    if len(path) > MAX_DEPTH:
        return

    # For better debugging, we keep track of the current recursion path by creating a list
    # of the hrefs visited by this recursion branch. Instead of using curr_deep, we can just check
    # the length of the visited list, as we append items as we visit them.
    path.append(uri)

    print('Visiting: {}, path: {}'.format(uri, path))

    # Update a set with the current href, so that we don't visit twice the same uri.
    # In practice hrefs should be normalised here to be safe, it's not trivial to deal with
    # all the possible edge cases consistently.
    already_visited.add(uri)

    pageFull = requests.get(start_address)
    pageText = pageFull.text
    soup = BeautifulSoup(pageText, "html.parser")

    for link in soup.findAll("a"):
        uri = _get_uri(start_address, link.get("href"))

        if uri and uri not in already_visited:
            # Recursively visit the links. 
            # Note this is a depth-first search (DFS) visit of a tree. 
            _crawl_page(start_address, uri, path)


spider("http://www.sheldonbrown.com/index.html")

Python：蜘蛛递归循环

1 个答案: