如何识别并关注 link，然后使用 BeautifulSoup 从新网页打印数据

Question

我正在尝试 (1) 从网页中获取标题，(2) 打印标题，(3) 按照 link 进入下一页，(4) 从下一页中获取标题页，以及 (5) 从下一页开始打印标题。

步骤(1)和(4)是相同的功能，步骤(2)和(5)是相同的功能。唯一的区别是功能 (4) 和 (5) 在下一页执行。

#Imports
from urllib.request import urlopen
from bs4 import BeautifulSoup
import re


##Internet
#Link to webpage 
web_page = urlopen("http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=31&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/%22deep+learning%22")
#Soup object
soup = BeautifulSoup(web_page, 'html.parser')

我对第 1 步和第 2 步没有任何问题。我的代码能够获取标题并有效地打印它。第 1 步和第 2 步：

##Get Data
def get_title():
    #Patent Number
    Patent_Number = soup.title.text
    print(Patent_Number)

get_title()

我得到的输出正是我想要的：

#Print Out
United States Patent: 10530579

我在执行第 3 步时遇到问题。对于第 (3) 步，我已经能够识别正确的 link，但无法按照它进入下一页。我正在识别我想要的 link，图像标签上方的 'href'。

Picture of link to follow.

以下代码是我对步骤 3、4 和 5 的工作草稿：

#Get
def get_link():
    ##Internet
    #Link to webpage 
    html = urlopen("http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=31&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/%22deep+learning%22")
    #Soup object
    soup = BeautifulSoup(html, 'html.parser')
    #Find image
    ##image = <img valign="MIDDLE" src="/netaicon/PTO/nextdoc.gif" border="0" alt="[NEXT_DOC]">
    #image = soup.find("img", valign="MIDDLE")
    image = soup.find("img", valign="MIDDLE", alt="[NEXT_DOC]")
    #Get new link
    new_link = link.attrs['href']
    print(new_link)

get_link()

我得到的输出：

#Print Out
##/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=32&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/"deep+learning"

输出正是我想要遵循的 link。简而言之，我尝试编写的函数会将 new_link 变量作为新网页打开，并在新网页上执行（1）和（2）中执行的相同功能。生成的输出将是两个标题而不是一个（一个用于网页，一个用于新网页）。

本质上，我需要写一个：

urlopen(new_link)

函数，而不是：

print(new_link)

函数。然后，在新网页上执行步骤 4 和 5。但是，我无法弄清楚如何打开新页面并获取标题。一个问题是 new_link 不是 url，而是我要单击的 link。

Answer 1

此函数打印下一页的标题，而不是 print(new_link)。

def get_link():
    ##Internet
    #Link to webpage 
    html = urlopen("http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=31&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/%22deep+learning%22")
    #Soup object
    soup = BeautifulSoup(html, 'html.parser')
    #Find image
    image = soup.find("img", valign="MIDDLE", alt="[NEXT_DOC]")
    #Follow link
    link = image.parent
    new_link = link.attrs['href']
    new_page = urlopen('http://patft.uspto.gov/'+new_link)
    soup = BeautifulSoup(new_page, 'html.parser')
    #Patent Number
    Patent_Number = soup.title.text
    print(Patent_Number)

get_link()

添加“http://patft.uspto.gov/”加上 new_link - 将 link 变为有效的 url。然后，我可以打开 url，导航到页面并检索标题。

Answer 2

您可以使用一些正则表达式来提取和格式化 link（以防它发生变化），整个示例代码如下：

# The first link
url = "http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=31&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/%22deep+learning%22"

# Test loop (to grab 5 records)
for _ in range(5):
    web_page = urlopen(url)
    soup = BeautifulSoup(web_page, 'html.parser')

    # step 1 & 2 - grabbing and printing title from a webpage
    print(soup.title.text) 

    # step 4 - getting the link from the page
    next_page_link = soup.find('img', {'alt':'[NEXT_DOC]'}).find_parent('a').get('href')

    # extracting the link (determining the prefix (http or https) and getting the site data (everything until the first /))
    match = re.compile("(?P<prefix>http(s)?://)(?P<site>[^/]+)(?:.+)").search(url)
    if match:
        prefix = match.group('prefix')
        site = match.group('site')

    # formatting the link to the next page
    url = '%s%s%s' % (prefix, site, next_page_link)

    # printing the link just for debug purpose
    print(url)

    # continuing with the loop

Answer 3

尽管您找到了解决方案以防万一有人尝试类似的方法。不建议在所有情况下使用我的以下解决方案。在这种情况下，因为所有页面的 url 仅页码不同。我们可以动态生成这些然后批量请求如下。您可以更改 r 的上限，直到页面存在为止。

from urllib.request import urlopen
from bs4 import BeautifulSoup
import pandas as pd

head = "http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r="  # no trailing /
trail = """&f=G&l=50&co1=AND&d=PTXT&s1=("deep+learning".CLTX.+or+"deep+learning".DCTX.)&OS=ACLM/"deep+learning"""

final_url = []
news_data = []
for r in range(32,38): #change the upper range as per requirement
    final_url.append(head + str(r) + trail)
for url in final_url:
    try:
        page = urlopen(url)
        soup = BeautifulSoup(page, 'html.parser')   
        patentNumber = soup.title.text
        news_articles = [{'page_url':  url,
                     'patentNumber':  patentNumber}
                    ]
        news_data.extend(news_articles)     
    except Exception as e:
        print(e)
        print("continuing....")
        continue
df =  pd.DataFrame(news_data)

Answer 4

借此机会清理您的代码。我删除了不必要的 re 导入并简化了您的功能：

from urllib.request import urlopen
from bs4 import BeautifulSoup


def get_soup(web_page):
    web_page = urlopen(web_page)
    return BeautifulSoup(web_page, 'html.parser')

def get_title(soup):
    return soup.title.text  # Patent Number

def get_next_link(soup):
    return soup.find("img", valign="MIDDLE", alt="[NEXT_DOC]").parent['href']

base_url = 'http://patft.uspto.gov'
web_page = base_url + '/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=31&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/%22deep+learning%22'

soup = get_soup(web_page)

get_title(soup)
> 'United States Patent: 10530579'

get_next_link(soup)
> '/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=32&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/"deep+learning"'

soup = get_soup(base_url + get_next_link(soup))
get_title(soup)
> 'United States Patent: 10529534'

get_next_link(soup)
> '/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=1&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=33&f=G&l=50&co1=AND&d=PTXT&s1=(%22deep+learning%22.CLTX.+or+%22deep+learning%22.DCTX.)&OS=ACLM/"deep+learning"'

如何识别并关注 link，然后使用 BeautifulSoup 从新网页打印数据

How to identify and follow a link, then print data from a new webpage with BeautifulSoup

html

python

beautifulsoup

web-scraping

urlopen