Python 使用带有 Scrapy 的 css 选择器获取 href
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Get href using css selector with Scrapy
提问by Marco Dinatsoli
I want to get the hrefvalue:
我想获得href价值：
<span class="title">
<a href="https://www.example.com"></a>
</span>
I tried this:
我试过这个：
Link = Link1.css('span[class=title] a::text').extract()[0]
But I just get the text inside the <a>. How can I get the link inside the href?
但我只是在<a>. 我怎样才能得到里面的链接href？
采纳答案by paul trmbrth
What you're looking for is:
你要找的是：
Link = Link1.css('span[class=title] a::attr(href)').extract()[0]
Since you're matching a span"class" attribute also, you can even write
由于您还匹配span“类”属性，您甚至可以编写
Link = Link1.css('span.title a::attr(href)').extract()[0]
Please note that ::textpseudo element and ::attr(attributename)functional pseudo element are NOTstandard CSS3 selectors. They're extensions to CSS selectors in Scrapy 0.20.
请注意，::text伪元素和::attr(attributename)功能伪元素不是标准的 CSS3 选择器。它们是 Scrapy 0.20 中 CSS 选择器的扩展。
Edit (2017-07-20): starting from Scrapy 1.0, you can use .extract_first()instead of .extract()[0]
编辑（2017-07-20）：从Scrapy 1.0开始，可以使用.extract_first()代替.extract()[0]
Link = Link1.css('span[class=title] a::attr(href)').extract_first()
Link = Link1.css('span.title a::attr(href)').extract_first()
回答by Eddy
Link = Link1.css('span.title a::attr(href)').extract_first()
回答by Jorgesys
This will do the job:
这将完成这项工作：
Link = Link1.css('span.title a::attr(href)').extract()
Linkwill have the value : https://www.example.com
Link将具有以下值：https: //www.example.com
