如何使用nokogiri从列表中获取“asin”标签？

Question

我正在尝试使用nokogiri从亚马逊html页面获取ASIN号码，但使用xpath没有成功。我已经用firepath尝试过了，但仍然一无所获。如果只获取URL，然后运行ruby REGEX来获取ASIN，会不会更好？如果是这样的话，正则表达式会是什么样子呢？

#!/usr/bin/env ruby -w
require 'nokogiri'
require 'open-uri'
url = "http://www.amazon.com/gp/new-releases/books/3839/ref=zg_bsnr_nav"
doc = Nokogiri::HTML(open(url))

puts doc.xpath('//zg_list').each do | node|
  p node['asin']
end

这就是当它打印出url时我得到的。

#!/usr/bin/env ruby -w
require 'nokogiri'
require 'open-uri'
url = "http://www.amazon.com/gp/new-releases/books/3839/ref=zg_bsnr_nav"
doc = Nokogiri::HTML(open(url))

l = doc.css('div.zg_image a').map { |link| 
  link['href'] 
  }
puts l # => /Introducing-ZBrush-4-Eric-Keller/dp/0470527641/ref=zg_bsnr_3839_20/183-0702383-0095048

Answer 1

对我来说，Nokogiri中的css方法比XPath更容易使用。给定您发布的URL处的HTML，以下代码应检索每个项目的“asin”属性：

doc.css("div.zg_item").map { |e| e["asin"] }

我认为正确的XPath应该是这样的：

doc.xpath("//div[contains(@class, 'zg_item') and @asin]")

Answer 2

您可以使用CSS accessors或XPath：

#!/usr/bin/env ruby -w

require 'nokogiri'
require 'open-uri'

url = "http://www.amazon.com/gp/new-releases/books/3839/ref=zg_bsnr_nav"

doc = Nokogiri::HTML(open(url))

# CSS
# puts doc.search('div[class="zg_item zg_sparseListItem"]').each { |n| p n['asin'] }

# XPath
puts doc.search('//div[@class="zg_item zg_sparseListItem"]').each { |n| p n['asin'] }

# >> "1934356549"
# >> "0596802471"
# >> "B004M8T01Q"
# >> "0596809158"
# >> "0470943327"
# >> "B004MMEJ36"
# >> "1935182641"
# >> "B004RDOPJI"
# >> "1449390501"
# >> "1449389716"
# >> "B004IWRH4I"
# >> "0470527641"
# >> "0735650926"
# >> "1430231475"
# >> "0321751043"
# >> "B004NBZ65G"
# >> "B004TMNSJK"
# >> "0132091518"
# >> "144030842X"
# >> "1430234040"
# >> 0