网页抓取 Google Domains

Question

我试图从前 100 个结果中获取域列表：例如：abc.com/xxxx/dddd 域应该是：abc.com我正在使用以下代码：import timefrom bs4 import BeautifulSoupimport requestssearch=input("What do you want to ask: ")search=search.replace(" ","+")link="https://www.google.com/search?q="+searchprint(link)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'}source=requests.get(link, headers=headers).textsoup=BeautifulSoup(source,"html.parser")soup=BeautifulSoup(source,"html.parser")但是，我不知道如何仅选择域，也不知道如何指定 100 个结果。当我写soup.text我只得到：'te - Pesquisa Google(function(){window.google={kEI:\'jsCaXM3AHM6g5OUP4eyT2A0\',kEXPI:\'31\',authuser:0,kscs:\'c9c918f0_jsCaXM3AHM6g5OUP4eyT2A0\',kGL:\'BR\'};google.sn=\'web\';google.kHL=\'pt-BR\';})();(function(){google.lc=[];google.li=0;google.getEI=function(a){for(var b;a&&(!a.getAttribute||!(b=a.getAttribute("eid")));)a=a.parentNode;return b||google.kEI};google.getLEI=function(a){for(var b=null;a&&(!a.getAttribute||!(b=a.getAttribute("leid")));)a=a.parentNode;return b};google.https=function(){return"https:"==window.location.protocol};google.ml=function(){return null};google.time=function()

Answer 1

获得 100 个结果

您必须逐页抓取，直到它有 100 个结果。假设 要废弃的关键字beautiful+girls URL 适用于像这样的第 2 页https://www.google.com/search?q=beautiful+girls&start=10

仅获取域

首先，您必须使用“srg”类获取所有 div（查看源代码后，我看到所有链接都在此）

srg_divs = soup.findAll("div", {"class": "srg"})

然后你会发现所有的标签

out = ''

for div in srg_divs:

    links = div.find_all('a', href=True)

    for a in links:

        # url to domain

        parsed_uri = urlparse(a['href'])

        domain = '{uri.netloc}'.format(uri=parsed_uri)

        # exclude googleusercontent.com

        if 'googleusercontent' in domain or domain == '':

            continue

        out += domain + '\n'