Css 如何通过scrapy提取类似a href中的文本

我想按如下方式提取文本

子主题_选择器='。子主题：：文本'， 年份选择器='。年份：：文本' 但我不知道如何提取主题，你能帮我吗？ 主题选择器='//a[contains@href，/set/theme-]/@href'

<div class='tags floatleft'>
    <a href='/sets/10251-1/Brick-Bank'>10251-1</a> 
    <a href='/sets/theme-Creator-Expert'>Creator Expert</a> 
    <a class='subtheme' href='/sets/theme-Creator-Expert/subtheme-Modular-Buildings'>Modular Buildings</a> 
    <a class='year' href='/sets/theme-Creator-Expert/year-2016'>2016</a> 
</div>

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你说得对。您可以非常简单地对其进行测试，甚至无需实际删除站点：

import scrapy

TEXT = """
<div class='tags floatleft'>
    <a href='/sets/10251-1/Brick-Bank'>10251-1</a> 
    <a href='/sets/theme-Creator-Expert'>Creator Expert</a> 
    <a class='subtheme' href='/sets/theme-Creator-Expert/subtheme-Modular-Buildings'>Modular Buildings</a> 
    <a class='year' href='/sets/theme-Creator-Expert/year-2016'>2016</a> 
</div>
"""

s = scrapy.Selector(text=TEXT)
link = s.xpath('//a[contains(@href,"/sets/theme-")]/@href').extract_first()
text = s.xpath('//a[contains(@href,"/sets/theme-")]/text()').extract_first()
print(link)
print(text)

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你是什么意思？原因是什么？这是最后一个'a[href*=\/sets\/theme-]：：text'

/sets/theme-Creator-Expert
Creator Expert