如何通过scrapy提取类似href中的文本

Question

我想提取如下文本，

SUBTHEME_SELECTOR = '.subtheme::text',
YEAR_SELECTOR = '.year::text'
但我不知道如何提取主题，你能帮我吗？
THEME_SELECTOR = '//a[contains(@href, "/sets/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>

Answer 1

你没看错。即使没有真正抓取站点，您也可以非常简单地测试它：

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)

生产：

/sets/theme-Creator-Expert
Creator Expert

如何通过scrapy提取类似href中的文本

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