如果一个网站返回一个'503服务不可用‘的错误,那么open-uri就会抛出一个异常。例如:
require 'open-uri'
open('http://www.google.co.uk/sorry/?continue=http://www.google.co.uk/search%3Fq%3Dhello%26oq%3Dhello%26ie%3DUTF-8')
# OpenURI::HTTPError: 503 Service Unavailable
# ...但是,如果您随后在web浏览器中访问它,它实际上会显示一个带有验证码的页面,而不是一个错误。
如何确保open-uri不会抛出异常,而是实际处理响应并为我提供页面内容?
发布于 2013-09-05 19:38:41
OpenURI::HTTPError有一个io属性,您可以检查它以获得您想要的东西。io是一个StringIO对象,上面定义了几个单例方法(例如status):
require 'open-uri'
begin
open('http://www.google.co.uk/sorry/?continue=http://www.google.co.uk/search%3Fq%3Dhello%26oq%3Dhello%26ie%3DUTF-8')
rescue OpenURI::HTTPError => error
response = error.io
response.status
# => ["503", "Service Unavailable"]
response.string
# => <!DOCTYPE html PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">\n<html DIR=\"LTR\">\n<head><meta http-equiv=\"content-type\" content=\"text/html; charset=utf-8\"><meta name=\"viewport\" content=\"initial-scale=1\">...
end 但是,对于此任务,Net::HTTP模块可能是更好的替代方案:
require 'net/http'
response = Net::HTTP.get_response(URI.parse('http://www.google.co.uk/sorry/?continue=http://www.google.co.uk/search%3Fq%3Dhello%26oq%3Dhello%26ie%3DUTF-8'))
response.code
# => "503"
response.body
# => "<!DOCTYPE html PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">\n<html DIR=\"LTR\">\n<head><meta http-equiv=\"content-type\" content=\"text/html; ...https://stackoverflow.com/questions/18634684
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