Question

<a class="a-link-normal a-text-normal" href="/Art-Dutch-Republic-1585-Everyman/dp/0297833693/ref=sr_1_1?keywords=9780297833697&qid=1574351815&sr=8-1"> <span class="a-size-medium a-color-base a-text-normal">Art of the Dutch Republic 1585 - 1718 (Everyman Art Library)</span> </a>

如何使用CSS选择器或Xpath获取href的值？

Answer 1

这里是一个例子：

    def parse(self, response):
        # iterate over all href
        for href in response.xpath("//a[@class='class-name']/@href"):
            # extract href as a string
            url = href.extract()

Answer 2

CSS选择器示例：

links = response.css("a.a-link-normal.a-text-normal::attr(href)").extract()

Answer 3

尝试一下response.css('.a-link-normal ::attr(href)').extract()

Answer 4

您可以通过以下选择器来实现这一点

a.your_calss_name::attr(href)

如何通过使用CSS选择器的类名从链接中获取href值？

4 个答案: