python - 关于scrapy爬虫AJAX页面

Question

• 问题：
  爬取信息页面为：知乎话题广场

当点击加载的时候，用Chrome 开发者工具，可以看到Network中，实际请求的链接为：

FormData为：

urlencode：

• 然后我的代码为：

        ...
        data = response.css('.zh-general-list::attr(data-init)').extract()
        param = json.loads(data[0])
        topic_id = param['params']['topic_id']
        # hash_id = param['params']['hash_id']
        hash_id = ""
        for i in range(32):
            params = json.dumps({"topic_id": topic_id,"hash_id": hash_id, "offset":i*20})
            payload = {"method": "next", "params": params, "_xsrf":_xsrf}
            print payload
            yield scrapy.Request(
                    url="https://m.zhihu.com/node/TopicsPlazzaListV2?" + urlencode(payload),
                    headers=headers,
                    meta={
                        "proxy": proxy,
                        "cookiejar": response.meta["cookiejar"],
                    },
                    callback=self.get_topic_url,
            )

• 执行爬虫之后，返回的是：

{'_xsrf': u'161c70f5f7e324b92c2d1a6fd2e80198', 'params': '{"hash_id": "", "offset": 140, "topic_id": 253}', 'method': 'next'}
^C^C^C{'_xsrf': u'161c70f5f7e324b92c2d1a6fd2e80198', 'params': '{"hash_id": "", "offset": 160, "topic_id": 253}', 'method': 'next'}
2016-05-09 11:09:36 [scrapy] DEBUG: Retrying <GET https://m.zhihu.com/node/TopicsPlazzaListV2?_xsrf=161c70f5f7e324b92c2d1a6fd2e80198&params=%7B%22hash_id%22%3A+%22%22%2C+%22offset%22%3A+80%2C+%22topic_id%22%3A+253%7D&method=next> (failed 1 times): [<twisted.python.failure.Failure OpenSSL.SSL.Error: [('SSL routines', 'SSL3_READ_BYTES', 'ssl handshake failure')]>]
2016-05-09 11:09:36 [scrapy] DEBUG: Retrying <GET https://m.zhihu.com/node/TopicsPlazzaListV2?_xsrf=161c70f5f7e324b92c2d1a6fd2e80198&params=%7B%22hash_id%22%3A+%22%22%2C+%22offset%22%3A+40%2C+%22topic_id%22%3A+253%7D&method=next> (failed 1 times): [<twisted.python.failure.Failure OpenSSL.SSL.Error: [('SSL routines', 'SSL3_READ_BYTES', 'ssl handshake failure')]>]

• 是不是url哪里错了？请指点一二。

Answer 1

DEBUG: Retrying <GET 这么大个提示看不出来。。。你抓包时的请求是post。scrapy时为什么要用get。。。

Answer 2

写爬虫应该一步一步来，不能一步到位，不然哪里错了都不知道，一般是先获取到你要的数据，然后再去解析过滤。
你先发送请求看看能不能获取到你想要的数据，如果没有，可能是url不对或者被拦截了

Answer 3

#coding=utf-8

import requests

headers = {'Content-Type':'application/x-www-form-urlencoded; charset=UTF-8'}
url = 'https://www.zhihu.com/node/TopicsPlazzaListV2'
data = 'method=next&params=%7B%22topic_id%22%3A833%2C%22offset%22%3A0%2C%22hash_id%22%3A%22%22%7D'

r = requests.post(url, data, headers=headers)
print r.text

Answer 4

少年教你一个大招。