从网站抓取电子邮件

Question

我尝试了其他帖子的多次迭代，但似乎没有任何帮助或满足我的需求。

我有一个要遍历的URL列表，并提取包含电子邮件地址的所有相关URL。然后，我想将URL和电子邮件地址存储到csv文件中。

例如，如果我去了10torr.com，该程序应在主URL中找到每个站点（即：10torr.com/about）并提取所有电子邮件。

下面是5个示例网站的列表，这些网站在通过我的代码运行时当前处于数据框格式。它们保存在变量small_site下。

一个有用的答案将包括使用下面列出的名为get_info()的用户定义函数。将网站硬编码到Spider本身是不可行的选择，因为许多其他具有不同网站列表长度的人都会使用它。

    Website
    http://10torr.com/
    https://www.10000drops.com/
    https://www.11wells.com/
    https://117westspirits.com/
    https://www.onpointdistillery.com/

下面是我正在运行的代码。蜘蛛程序似乎正在运行，但是我的csv文件中没有输出。

import os
import pandas as pd
import re
import scrapy
from scrapy.crawler import CrawlerProcess
from scrapy.linkextractors.lxmlhtml import LxmlLinkExtractor

small_site = site.head()


#%% Start Spider
class MailSpider(scrapy.Spider):

    name = 'email'

    def parse(self, response):

        links = LxmlLinkExtractor(allow=()).extract_links(response)
        links = [str(link.url) for link in links]
        links.append(str(response.url))

        for link in links:
            yield scrapy.Request(url=link, callback=self.parse_link) 

    def parse_link(self, response):

        for word in self.reject:
            if word in str(response.url):
                return

        html_text = str(response.text)
        mail_list = re.findall('\w+@\w+\.{1}\w+', html_text)

        dic = {'email': mail_list, 'link': str(response.url)}
        df = pd.DataFrame(dic)

        df.to_csv(self.path, mode='a', header=False)
        df.to_csv(self.path, mode='a', header=False)


#%% Preps a CSV File
def ask_user(question):
    response = input(question + ' y/n' + '\n')
    if response == 'y':
        return True
    else:
        return False
def create_file(path):
    response = False
    if os.path.exists(path):
        response = ask_user('File already exists, replace?')
        if response == False: return 

    with open(path, 'wb') as file: 
        file.close()


#%% Defines function that will extract emails and enter it into CSV
def get_info(url_list, path, reject=[]):

    create_file(path)
    df = pd.DataFrame(columns=['email', 'link'], index=[0])
    df.to_csv(path, mode='w', header=True)


    print('Collecting Google urls...')
    google_urls = url_list


    print('Searching for emails...')
    process = CrawlerProcess({'USER_AGENT': 'Mozilla/5.0'})
    process.start() 

    for i in small_site.Website.iteritems():
        print('Searching for emails...')
        process.crawl(MailSpider, start_urls=google_urls, path=path, reject=reject)
        ##process.start()

        print('Cleaning emails...')
        df = pd.read_csv(path, index_col=0)
        df.columns = ['email', 'link']
        df = df.drop_duplicates(subset='email')
        df = df.reset_index(drop=True)
        df.to_csv(path, mode='w', header=True)


    return df


url_list = small_site
path = 'email.csv'

df = get_info(url_list, path)

由于没有收到任何错误消息，我不确定我要去哪里。如果您需要其他信息，请询问。我一直在努力争取将近一个月的时间，现在我感觉好像只是在撞墙。

几周后，大部分代码在文章Web scraping to extract contact information— Part 1: Mailing Lists上找到。但是，我没有成功将其扩展到我的需要。在整合他们的Google搜索功能以获取基本URL的情况下，它一劳永逸。

在此先感谢您提供的任何帮助。

Answer 1

花了一段时间，但答案终于到了我这里。以下是最终的答案。与原始问题一样，这将适用于更改列表。

更改最终很小。我需要添加以下用户定义的函数。

def get_urls(io, sheet_name):
    data = pd.read_excel(io, sheet_name)
    urls = data['Website'].to_list()
    return urls

从那里开始，对get_info()用户定义的函数进行了简单的更改。我们需要将此函数中的google_urls设置为get_urls函数，并传递列表。此功能的完整代码如下。

def get_info(io, sheet_name, path, reject=[]):
    
    create_file(path)
    df = pd.DataFrame(columns=['email', 'link'], index=[0])
    df.to_csv(path, mode='w', header=True)
    
    print('Collecting Google urls...')
    google_urls = get_urls(io, sheet_name)
    
    print('Searching for emails...')
    process = CrawlerProcess({'USER_AGENT': 'Mozilla/5.0'})
    process.crawl(MailSpider, start_urls=google_urls, path=path, reject=reject)
    process.start()
    
    print('Cleaning emails...')
    df = pd.read_csv(path, index_col=0)
    df.columns = ['email', 'link']
    df = df.drop_duplicates(subset='email')
    df = df.reset_index(drop=True)
    df.to_csv(path, mode='w', header=True)
    
    return df

无需其他任何更改即可运行它。希望这会有所帮助。

Answer 2

我修改了一些脚本，这些脚本通过Shell运行了以下脚本，并且可以运行。也许它将为您提供一个起点。

我建议您使用外壳程序，因为它在抓取过程中总是会引发错误和其他消息

class MailSpider(scrapy.Spider):

    name = 'email'
    start_urls = [
        'http://10torr.com/',
        'https://www.10000drops.com/',
        'https://www.11wells.com/',
        'https://117westspirits.com/',
        'https://www.onpointdistillery.com/',
    ]

    def parse(self, response):
        self.log('A response from %s just arrived!' % response.url)
        links = LxmlLinkExtractor(allow=()).extract_links(response)
        links = [str(link.url) for link in links]
        links.append(str(response.url))

        for link in links:
            yield scrapy.Request(url=link, callback=self.parse_link) 

    def parse_link(self, response):

        html_text = str(response.text)
        mail_list = re.findall('\w+@\w+\.{1}\w+', html_text)

        dic = {'email': mail_list, 'link': str(response.url)}

        for key in dic.keys():
            yield {
                'email' : dic['email'],
                'link': dic['link'],
            }

通过 Anaconda shell scrapy crawl email -o test.jl

进行爬网时，将产生以下输出

{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/"}
{"email": ["8b4e078a51d04e0e9efdf470027f0ec1@sentry.wixpress", "bundle@3.2", "fetch@3.0", "bolt@2.3", "5oclock@11wells.com", "5oclock@11wells.com", "5oclock@11wells.com"], "link": "https://www.11wells.com"}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/shop?olsPage=search&keywords="}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/shop?olsPage=search&keywords="}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/shop"}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/shop?olsPage=cart"}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/home"}
{"email": ["8b4e078a51d04e0e9efdf470027f0ec1@sentry.wixpress", "bundle@3.2", "fetch@3.0", "bolt@2.3", "5oclock@11wells.com", "5oclock@11wells.com", "5oclock@11wells.com"], "link": "https://www.11wells.com"}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/home"}
{"email": ["info@ndiscovered.com"], "link": "https://117westspirits.com/117%C2%B0-west-spirits-1"}
...
...
...

请参阅Scrapy docs以获取更多信息