小编use*_*601的帖子

每当尝试在Python中使用oAuth2时,我似乎都遇到了SSL问题.我花了大部分时间试图调试它,但似乎无法搞清楚.

这是我的Python脚本(很简单):

import oauth2.oauth2 as oauth
import urlparse
import time

## If you're actually processing requests, you'll want this
# import simplejson


### GET A REQUEST TOKEN ###

consumer = oauth.Consumer(key="***KEYHERE***", secret="***KEYSECRETHERE***")

request_token_url = 'https://api.instagram.com/oauth/access_token'

client = oauth.Client(consumer)
resp, content = client.request(request_token_url, "GET")

request_token = dict(urlparse.parse_qsl(content))


token = oauth.Token(request_token['oauth_token'], request_token['oauth_token_secret'])

以及Python解释器中的这些错误:

Traceback (most recent call last):
  File "E:\Projects\oAuth2Test\test.py", line 16, in <module>
    resp, content = client.request(request_token_url, "GET")
  File "E:\Projects\oAuth2Test\oauth2\oauth2.py", line 682, in request
    connection_type=connection_type)
  File "E:\Projects\oAuth2Test\httplib2\httplib2.py", line 1445, in request …

仅使用Twitter作为示例而忽略了它们具有完全可用的API的事实,以下脚本从用户页面获取当前的第5条推文.

import urllib2
from lxml import etree

xpathselector = "/html/body/div/div[2]/div/div[5]/div[2]/div/ol/li[5]/div/div/p"
url =  "https://twitter.com/bmthofficial"
response = urllib2.urlopen(url)
htmlparser = etree.HTMLParser()
tree = etree.parse(response, htmlparser)
result = tree.xpath(xpathselector)

print result[0].text

在这篇文章的时候它打印:

从下午2点30分,赢取阅读节门票,并介绍

现在,它打印<p> </ p>的内容,我将如何获取P的类名？它的HTML看起来像这样.

<p class="js-tweet-text tweet-text">From 2.30pm, win tickets to Reading Festival, and introduce <a dir="ltr" class="twitter-atreply pretty-link" href="/bmthofficial"><s>@</s><b>bmthofficial</b></a> onstage!</p>

任何帮助表示赞赏!谢谢!

希望这是经验丰富的人的快速答案.我有一个包含URL的XML文件,我想从XML中获取URL,然后将其输入到我编写的下载脚本中.我唯一的问题是我似乎无法正确解析XML中的url.这就是它的样子:

<program new-version="1.1.1.1" name="ProgramName">
<download-url value="http://website.com/file.exe"/>
</program>

提前致谢!