Question

我正在尝试提取TripAdvisor上每家餐厅的电子邮件地址。

我已经尝试过了，但是一直返回[]：

response.xpath('//*[@class= "restaurants-detail-overview-cards-LocationOverviewCard__detailLink--iyzJI restaurants-detail-overview-cards-LocationOverviewCard__contactItem--89flT6"]')

TripAdvisor页面上的代码段如下：

<div class="restaurants-detail-overview-cards-LocationOverviewCard__detailLink--iyzJI restaurants-detail-overview-cards-LocationOverviewCard__contactItem--1flT6"><span><a href="mailto:info@canopylounge.my?subject=?"><span class="ui_icon email restaurants-detail-overview-cards-LocationOverviewCard__detailLinkIcon--T_k32"></span><span class="restaurants-detail-overview-cards-LocationOverviewCard__detailLinkText--co3ei">Email</span><span class="ui_icon external-link-no-box restaurants-detail-overview-cards-LocationOverviewCard__upLinkIcon--1oVn1"></span></a></span></div>

Answer 1

首先：您在班级名称上有误。

第二个：它在<div>中是类，但是@href在<a>中。而且<a>不在<div>之后，因此您需要

'//*[@class="..."]//a/@href'

（我跳过类名，因为它太长了，无法显示）

您可以尝试使用

代替这么长的类名

'//a[contains(@href, "mailto")]/@href'

我使用xpath测试了lxml

text = '''<div class="restaurants-detail-overview-cards-LocationOverviewCard__detailLink--iyzJI restaurants-detail-overview-cards-LocationOverviewCard__contactItem--1flT6">
<span><a href="mailto:info@canopylounge.my?subject=?">
<span class="ui_icon email restaurants-detail-overview-cards-LocationOverviewCard__detailLinkIcon--T_k32"></span>
<span class="restaurants-detail-overview-cards-LocationOverviewCard__detailLinkText--co3ei">Email</span>
<span class="ui_icon external-link-no-box restaurants-detail-overview-cards-LocationOverviewCard__upLinkIcon--1oVn1"></span>
</a></span>
</div>'''

import lxml.html

soup = lxml.html.fromstring(text)

print(soup.xpath('//*[@class="restaurants-detail-overview-cards-LocationOverviewCard__detailLink--iyzJI restaurants-detail-overview-cards-LocationOverviewCard__contactItem--1flT6"]//a/@href'))
print(soup.xpath('//a[contains(@href, "mailto")]/@href'))

Answer 2

这是方法之一：

import requests
from scrapy import Selector

site_link = 'https://www.tripadvisor.com/Restaurant_Review-g60713-d11882449-Reviews-Coin_Op_Game_Room-San_Francisco_California.html'

res = requests.get(site_link)
sel = Selector(res)
email = sel.xpath("//*[contains(@class,'LocationOverviewCard__contactItem--')]//a[contains(@href,'mailto:')]/@href").get()
email = email.split("mailto:")[1].split("?")[0] if email else ""
print(email)

输出：

info@coinopsf.com

Answer 3

Selector还有一个.re()方法，用于使用正则表达式提取数据。

In [2]: response.xpath('//a[contains(@href, "mailto")]/@href')
Out[2]: [<Selector xpath='//a[contains(@href, "mailto")]/@href' data='mailto:info@coinopsf.com?subject=?'>]

In [3]: response.xpath('//a[contains(@href, "mailto")]/@href').get()
Out[3]: 'mailto:info@coinopsf.com?subject=?'

In [4]: response.xpath('//a[contains(@href, "mailto")]/@href').re('mailto:(.*)\?\w')
Out[4]: ['info@coinopsf.com']
In [5]: response.xpath('//a[contains(@href, "mailto")]/@href').re('mailto:([^?]*)')
Out[5]: ['info@coinopsf.com']

如何使用scrapy提取电子邮件地址？

3 个答案: