Question

我有一个控制器，它将一个url作为参数传递，我试图在该URL上刮取整个页面。但是当我尝试阅读网址时，我收到以下错误：No such file or directory @ rb_sysopen - www.google.com

控制器：

lass PageScraperController < ApplicationController
    require 'nokogiri'
    require 'open-uri'
    require 'diffy'
    require 'htmlentities'

    def scrape
    require 'open-uri'
        @url = watched_link_params.to_s
        @url = @url.slice(9..@url.length-3)
        puts "LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOG#{@url}"

        page = Nokogiri::HTML(open(@url))
        # coder = HTMLEntities.new
        # encodedHTML = coder.encode(page)
        puts page

     end

    def watched_link_params

        params.require(:default).permit(:url)

    end

end

Answer 1

试试这个：

def scrape
  @url = watched_link_params[:url]

  page = Nokogiri::HTML(open(@url))
  puts page
end

您需要传递整个网址，包括协议指示符;也就是说，您需要使用http://www.google.com代替www.google.com：

>> params = ActionController::Parameters.new(default: {url: 'http://www.google.com'})
>> watched_link_params = params.require(:default).permit(:url)
>> @url = watched_link_params[:url]
"http://www.google.com"
>> page = Nokogiri::HTML(open(@url))

Rails不会读取与nokogiri和open-uri的链接

1 个答案: