使用Google App Engine在python中获取很多网址

问题描述

在我的RequestHandler子类中，我试图获取网址范围:

In my subclass of RequestHandler, I am trying to fetch range of urls:

class GetStats(webapp2.RequestHandler):
    def post(self):

    lastpage = 50
    for page in range(1, lastpage):
        tmpurl = url + str(page)
        response = urllib2.urlopen(tmpurl, timeout=5)
        html = response.read()
        # some parsing html
        heap.append(result_of_parsing)

    self.response.write(heap)

但是它可以处理大约30个网址(页面加载时间很长，但是可以).如果超过30个，我会报错:

But it works with ~ 30 urls (page is loading long but it is works).In case more than 30 I am getting an error:

错误:服务器错误

服务器遇到错误，无法完成您的请求.

请在30秒内重试.

有什么方法可以获取很多网址?可能更理想或更合适?多达几百页?

Is there any way to fetch a lot of urls? May be more optimal or smth?Up to several hundreds of pages?

更新:

我正在使用BeautifulSoup解析每个页面.我在gae日志中发现了这种追溯:

I am using BeautifulSoup to parse every single page. I found this traceback in gae logs:

Traceback (most recent call last):
  File "/base/data/home/runtimes/python27/python27_lib/versions/1/google/appengine/runtime/wsgi.py", line 267, in Handle
result = handler(dict(self._environ), self._StartResponse)
  File "/base/data/home/runtimes/python27/python27_lib/versions/third_party/webapp2-2.5.2/webapp2.py", line 1529, in __call__
rv = self.router.dispatch(request, response)
  File "/base/data/home/runtimes/python27/python27_lib/versions/third_party/webapp2-2.5.2/webapp2.py", line 1278, in default_dispatcher
return route.handler_adapter(request, response)
  File "/base/data/home/runtimes/python27/python27_lib/versions/third_party/webapp2-2.5.2/webapp2.py", line 1102, in __call__
return handler.dispatch()
  File "/base/data/home/runtimes/python27/python27_lib/versions/third_party/webapp2-2.5.2/webapp2.py", line 570, in dispatch
return method(*args, **kwargs)
  File "/base/data/home/apps/s~gae/1.379703839015039430/main.py", line 68, in post
heap = get_times(tmp_url, 160)
  File "/base/data/home/apps/s~gae/1.379703839015039430/main.py", line 106, in get_times
soup = BeautifulSoup(html)
  File "libs/bs4/__init__.py", line 168, in __init__
self._feed()
  File "libs/bs4/__init__.py", line 181, in _feed
self.builder.feed(self.markup)
  File "libs/bs4/builder/_htmlparser.py", line 56, in feed
super(HTMLParserTreeBuilder, self).feed(markup)
  File "/base/data/home/runtimes/python27/python27_dist/lib/python2.7/HTMLParser.py", line 114, in feed
self.goahead(0)
  File "/base/data/home/runtimes/python27/python27_dist/lib/python2.7/HTMLParser.py", line 155, in goahead
startswith = rawdata.startswith
 DeadlineExceededError

推荐答案

之所以失败，是因为您只有60秒的时间将响应返回给用户，而我猜想它花费的时间会更长.

It's failing because you only have 60 seconds to return a response to the user and I'm going to guess it's taking longer then that.

您将要使用以下内容: https://cloud.google.com/appengine/articles/deferred

You will want to use this: https://cloud.google.com/appengine/articles/deferred

创建一个10分钟超时的任务.然后，您可以立即返回给用户，他们可以稍后通过另一个处理程序(您创建的)提取"结果.如果收集所有URL花费的时间超过10分钟，则必须将其拆分为更多任务.

to create a task that has a 10 minute time out. Then you can return instantly to the user and they can "pick up" the results at a later time via another handler (that you create). If collecting all the URLs takes longer then 10 minutes you'll have to split them up into further tasks.

查看此内容: https://cloud.google.com/appengine/articles/deadlineexceedederrors

了解为什么不能再超过60秒.

to understand why you cannot go longer then 60 seconds.

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