我正在尝试抓取以下HTML代码:
<ul class="results-list" id="search-results">
<li>
<h3 class="name">First John</h3>
<div class="details">
<a href="mailto:example@mail.com" class="email">email</a>
<span class="phone">999999999</span>
</div>
</li>
<li>
<h3 class="name">Second John</h3>
<div class="details">
<a href="mailto:example@mail.com" class="email">email</a>
<span class="phone">999999999</span>
</div>
</li>
</ul>
当我运行我的蜘蛛时,我得到2行,包含相同的信息。我有名字,电子邮件,电话列,例如在名称栏中,我会得到: 第一约翰,第二约翰。
我的Scrapy代码如下:
people= response.xpath('//ul[@class="results-list"]/li')
for person in people:
item = SpiderItem()
item['Name'] = person.xpath(
'//h3/text()').extract()
item['Email'] = person.xpath(
'//div[@class="details"]/a/@href').extract()
item['Phone'] = person.xpath(
'//div[@class="details"]/span[@class="phone"]/text()').extract()
yield item
但是,当我运行scrapy crawl MySpider -o output.csv
时,我会在所有行中获得相同的信息。
答案 0 :(得分:1)
您在xpath表达式上使用绝对路径,将它们更改为:
for person in people:
item = SpiderItem()
item['Name'] = person.xpath(
'.//h3/text()').extract_first()
item['Email'] = person.xpath(
'.//div[@class="details"]/a/@href').extract_first()
item['Phone'] = person.xpath(
'.//div[@class="details"]/span[@class="phone"]/text()').extract_first()
yield item