获取 Urls 列表，然后在 Python 3.5.1 中从所有 Urls 中查找特定文本

Question

所以我有这段代码，它将以列表格式 url 提供我需要的

import requests
from bs4 import BeautifulSoup

offset = 0
links = []
with requests.Session() as session:
while True:
    r = session.get("http://rayleighev.deviantart.com/gallery/44021661/Reddit?offset=%d" % offset)
    soup = BeautifulSoup(r.content, "html.parser")
    new_links = soup.find_all("a", {'class' : "thumb"})

    # no more links - break the loop
    if not new_links:
        break

    # denotes the number of gallery pages gone through at one time (# of pages times 24 equals the number below)
    links.extend(new_links)
    print(len(links))
    offset += 24

    #denotes the number of gallery pages(# of pages times 24 equals the number below)
    if offset == 48:
        break

for link in links:
    print(link.get("href"))

之后，我尝试从所有 url 中获取不同的文本，并且所有这些文本在每个文本上都位于相对相同的位置。但是，每当我运行下面的下半部分时，我都会收到一大块 html 文本和一些错误，我不确定如何修复它或者是否还有其他错误，并且最好更简单，从每个 url.

获取文本的方法

import urllib.request
import re

for link in links:
    url = print("%s" % link) 

headers = {}
headers['User-Agent'] = 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_1) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/39.0.2171.95 Safari/537.36'
req = urllib.request.Request(url, headers = headers)
resp = urllib.request.urlopen(req)
respData = resp.read()

paragraphs = re.findall(r'</a><br /><br />(.*?)</div>', str(respData))

if paragraphs != None:
    paragraphs = re.findall(r'<br /><br />(.*?)</span>', str(respData))

if paragraphs != None:
    paragraphs = re.findall(r'<br /><br />(.*?)</span></div>', str(respData))

for eachP in paragraphs:
    print(eachP)

title = re.findall(r'<title>(.*?)</title>', str(respData))
for eachT in title:
    print(eachT)

Answer 1

您的代码：

for link in links:
    url = print("%s" % link)

将 None 分配给 url。也许你的意思是：

for link in links:
    url = "%s" % link.get("href")

也没有理由使用 urllib 来获取网站内容，您可以像以前一样使用请求，方法是更改：

req = urllib.request.Request(url, headers = headers)
resp = urllib.request.urlopen(req)
respData = resp.read()

至

req = requests.get(url, headers=headers)
soup = BeautifulSoup(req.content, "html.parser")

现在您可以通过以下方式获取标题和段落：

title = soup.find('div', {'class': 'dev-title-container'}).h1.text
paragraph = soup.find('div', {'class': 'text block'}).text

获取 Urls 列表，然后在 Python 3.5.1 中从所有 Urls 中查找特定文本

Getting a list of Urls and then finding specific text from all of them in Python 3.5.1

python

python-3.x

beautifulsoup

python-requests

urllib