RUBY - 网页抓取 - (OpenURI :: HTTPError)

Question

我正在尝试用ruby编写一个简单的Web抓取代码.它工作到第29个网址,然后我收到此错误消息:

C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:346:in `open_http': 500 Internal Server Er
ror (OpenURI::HTTPError)
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:775:in `buffer_open'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:203:in `block in open_loop'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:201:in `catch'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:201:in `open_loop'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:146:in `open_uri'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:677:in `open'
        from C:/Ruby193/lib/ruby/1.9.1/open-uri.rb:33:in `open'
        from test.rb:24:in `block (2 levels) in <main>'
        from test.rb:18:in `each'
        from test.rb:18:in `block in <main>'
        from test.rb:14:in `each'
        from test.rb:14:in `<main>'

我的代码:

require 'rubygems'  
require 'nokogiri'  
require 'open-uri'  

aFile=File.new('data.txt', 'w')

ag = 0
  for i in 1..40 do
    agenzie = ag + 1

    #change url parameter 

    url = "http://www.infotrav.it/dettaglio.do?sort=*RICOVIAGGI*&codAgenzia=" + "#{ ag }"  
    doc = Nokogiri::HTML(open(url))
    aFile=File.open('data.txt', 'a')
    aFile.write(doc.at_css("table").text)
    aFile.close
  end

你有什么想法来解决它吗？谢谢!

如

Answer 1

在这里，让我为您清理一下：

File.open('data.txt', 'w') do |aFile|
  (1..40).each do |ag|
    url = "http://www.infotrav.it/dettaglio.do?sort=*RICOVIAGGI*&codAgenzia=#{ag}"
    response = open(url) rescue nil
    next unless response
    doc = Nokogiri::HTML(response)
    aFile << doc.at_css("table").text
  end
end

笔记：

• 使用块样式 File.open 意味着当块退出时文件将自行关闭
• 使用each来迭代而不是for循环