如何在Python Scrapy中的子div中使用特殊的src获取href
为了获取站点的所有图像,我编写了以下代码:如何在Python Scrapy中的子div中使用特殊的src获取href,python,xpath,css-selectors,scrapy,selector,Python,Xpath,Css Selectors,Scrapy,Selector,为了获取站点的所有图像,我编写了以下代码: content = Selector(text = html) all_images= content.css('img') i = 0 for image in all_images: src = image.css("::attr('src')").extract_first() 在获得图像的src之后,现在我想获得每个图像的href <a href="/rayons/images" onclick="ga('send', 'e
content = Selector(text = html)
all_images= content.css('img')
i = 0
for image in all_images:
src = image.css("::attr('src')").extract_first()
在获得图像的src之后,现在我想获得每个图像的href
<a href="/rayons/images" onclick="ga('send', 'event', 'computer HP', 'htwe', 'ope_xxl_s22Englos');">
<img src="/mySrc/" alt="something" class="ze-content">
</a>
当我知道Src时,如何获得href AFAIK,您不能使用CSS进行父级搜索。在本例中,XPath更适合。您可以这样做:
for image in all_images:
src = image.css("::attr('src')").extract_first()
href = image.xpath('parent::a/@href').extract_first()
或者,使用XPath:
href = image.xpath('../@href').extract_first()