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Ruby用Nokogiri解析HTTPresponse

用Nokogiri解析HTTP响应

嗨,我在使用Nokogiri解析HTTPresponse对象时遇到问题.

我用这个函数在这里获取一个网站:

获取链接

def fetch(uri_str, limit = 10)


  # You should choose better exception.
  raise ArgumentError, 'HTTP redirect too deep' if limit == 0

  url = URI.parse(URI.encode(uri_str.strip))
  puts url

  #get path
  req = Net::HTTP::Get.new(url.path,headers)
  #start TCP/IP
  response = Net::HTTP.start(url.host,url.port) { |http|
        http.request(req)
  }
  case response
  when Net::HTTPSuccess
    then #print final redirect to a file
    puts "this is location" + uri_str
    puts "this is the host #{url.host}"
    puts "this is the path #{url.path}"

    return response
    # if you get a …
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ruby nokogiri

6
推荐指数
1
解决办法
6964
查看次数

绕过不可变的字符串python

好的,我开始编码:

lastentry = 'first'
campdata = {'a1'=0,
            'b2'=0}

class Someclass:
      def on_window1_destroy(self, widget, data=None):
           print campdata

      def func1(self)
          lastentry = 'b2'

      def func2(self)
          lastentry = 'a1'

      def func2(self)
          campdata[lastcall] +=1
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但后来我发现python字符串(和整数)是不可变的......

那么我该怎样绕过?

python

0
推荐指数
1
解决办法
161
查看次数

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