1

我正在尝试从 Google Scholar 的搜索结果中抓取 PDF 链接。我试图根据 URL 的变化设置页面计数器，但是在前八个输出链接之后，我得到了重复的链接作为输出。

#!/usr/bin/env python
from mechanize import Browser
from BeautifulSoup import BeautifulSoup
from bs4 import BeautifulSoup
import urllib2
import requests


#modifying the url as per page
urlCounter = 0
while urlCounter <=30:
    urlPart1 = "http://scholar.google.com/scholar?start="
    urlPart2 = "&q=%22entity+resolution%22&hl=en&as_sdt=0,4"
    url = urlPart1 + str(urlCounter) + urlPart2
    page = urllib2.Request(url,None,{"User-Agent":"Mozilla/5.0 (X11; U; Linux i686) Gecko/20071127 Firefox/2.0.0.11"})
    resp = urllib2.urlopen(page)
    html = resp.read()
    soup = BeautifulSoup(html)
    urlCounter = urlCounter + 10

    recordCount = 0
    while recordCount <=9:
        recordPart1 = "gs_ggsW"
        finRecord = recordPart1 + str(recordCount)
        recordCount = recordCount+1

    #printing the links
        for link in soup.find_all('div', id = finRecord):
            linkstring = str(link)
            soup1 = BeautifulSoup(linkstring)
        for link in soup1.find_all('a'):
            print(link.get('href'))
4

2 回答 2

1

更改代码中的以下行：

finRecord = recordPart1 + str(recordCount)

到

finRecord = recordPart1 + str(recordCount+urlCounter-10)

真正的问题：第一页的 div id 是 gs_ggsW[0-9]，但第二页的 id 是 gs_ggsW[10-19]。如此美丽的汤在第二页上找不到链接。

Python 的变量作用域可能会使来自其他语言（如 Java）的人感到困惑。执行下面的for循环后，变量link仍然存在。因此，该链接引用到第一页上的最后一个链接。

for link in soup1.find_all('a'):
    print(link.get('href'))

更新：

谷歌可能不提供部分论文的pdf下载链接，所以不能用id来匹配每篇论文的链接。您可以使用 css 选择器将所有链接匹配在一起。

soup = BeautifulSoup(html)
urlCounter = urlCounter + 10
for link in soup.select('div.gs_ttss a'):
    print(link.get('href'))
于 2013-11-01T07:35:24.923 回答
0

查看SelectorGadget Chrome 扩展程序，CSS通过单击浏览器中的所需元素来获取选择器。

在线 IDE中用于提取 PDF 的代码和示例：

from bs4 import BeautifulSoup
import requests, lxml

params = {
    "q": "entity resolution", # search query
    "hl": "en"                # language
}

# https://requests.readthedocs.io/en/master/user/quickstart/#custom-headers
headers = {
    "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3538.102 Safari/537.36 Edge/18.19582",
}

html = requests.get("https://scholar.google.com/scholar", params=params, headers=headers, timeout=30)
soup = BeautifulSoup(html.text, "lxml")

for pdf_link in soup.select(".gs_or_ggsm a"):
  pdf_file_link = pdf_link["href"]
  print(pdf_file_link)


# output from the first page:
'''
https://linqs.github.io/linqs-website/assets/resources/getoor-vldb12-slides.pdf
http://ilpubs.stanford.edu:8090/859/1/2008-7.pdf
https://drum.lib.umd.edu/bitstream/handle/1903/4241/umi-umd-4070.pdf;sequence=1
https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.169.9535&rep=rep1&type=pdf
https://arxiv.org/pdf/1208.1927
https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.77.6875&rep=rep1&type=pdf
http://da.qcri.org/ntang/pubs/vldb18-deeper.pdf
'''

或者，您可以使用来自 SerpApi的Google Scholar Organic Results API来实现相同的目的。这是一个带有免费计划的付费 API。

主要区别在于您只需要从结构化 JSON 中获取数据，而不是弄清楚如何从 HTML 中提取数据，如何绕过搜索引擎的块。

要集成的代码：

from serpapi import GoogleSearch

params = {
    "api_key": "YOUR_API_KEY",   # SerpApi API key
    "engine": "google_scholar",  # Google Scholar organic reuslts
    "q": "entity resolution",    # search query
    "hl": "en"                   # language
}

search = GoogleSearch(params)
results = search.get_dict()

for pdfs in results["organic_results"]:
    for link in pdfs.get("resources", []):
        pdf_link = link["link"]
        print(pdf_link)


# output:
'''
https://linqs.github.io/linqs-website/assets/resources/getoor-vldb12-slides.pdf
http://ilpubs.stanford.edu:8090/859/1/2008-7.pdf
https://drum.lib.umd.edu/bitstream/handle/1903/4241/umi-umd-4070.pdf;sequence=1
https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.169.9535&rep=rep1&type=pdf
https://arxiv.org/pdf/1208.1927
https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.77.6875&rep=rep1&type=pdf
http://da.qcri.org/ntang/pubs/vldb18-deeper.pdf
'''

如果你想从自然搜索结果中获取更多数据，我有一篇专门的Scrape Google Scholar with Python博客文章。

免责声明，我为 SerpApi 工作。

于 2022-02-09T14:23:16.730 回答