我在网上找到了这个脚本:

import httplib, urllib
params = urllib.urlencode({'number': 12524, 'type': 'issue', 'action': 'show'})
headers = {"Content-type": "application/x-www-form-urlencoded",
            "Accept": "text/plain"}
conn = httplib.HTTPConnection("bugs.python.org")
conn.request("POST", "", params, headers)
response = conn.getresponse()
print response.status, response.reason
302 Found
data = response.read()
data
'Redirecting to <a href="http://bugs.python.org/issue12524">http://bugs.python.org/issue12524</a>'
conn.close()

但我不明白如何用PHP使用它，params变量里面的所有东西是什么，或者如何使用它。你能帮我把这个弄起来吗?

如果我这样做

url = "http://example.com?p=" + urllib.quote(query)

它不编码/到%2F(破坏OAuth规范化) 它不处理Unicode(它抛出异常)

有更好的图书馆吗?

我得到以下错误:

Exception in thread Thread-3:
Traceback (most recent call last):
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 810, in        __bootstrap_inner
self.run()
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/threading.py", line 763, in  run
self.__target(*self.__args, **self.__kwargs)
File "/Users/Matthew/Desktop/Skypebot 2.0/bot.py", line 271, in process
info = urllib2.urlopen(req).read()
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 154, in urlopen
return opener.open(url, data, timeout)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 431, in open
response = self._open(req, data)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 449, in _open
'_open', req)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 409, in _call_chain
result = func(*args)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1240, in https_open
context=self._context)
File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 1197, in do_open
raise URLError(err)
URLError: <urlopen error [SSL: CERTIFICATE_VERIFY_FAILED] certificate verify failed (_ssl.c:581)>

下面是导致这个错误的代码:

if input.startswith("!web"):
    input = input.replace("!web ", "")      
    url = "https://domainsearch.p.mashape.com/index.php?name=" + input
    req = urllib2.Request(url, headers={ 'X-Mashape-Key': 'XXXXXXXXXXXXXXXXXXXX' })
    info = urllib2.urlopen(req).read()
    Message.Chat.SendMessage ("" + info)

我正在使用的API要求我使用HTTPS。我怎样才能让它绕过验证呢?

我试图抓取一个网站，但它给了我一个错误。

我正在使用以下代码:

import urllib.request
from bs4 import BeautifulSoup

get = urllib.request.urlopen("https://www.website.com/")
html = get.read()

soup = BeautifulSoup(html)

print(soup)

我得到以下错误:

File "C:\Python34\lib\encodings\cp1252.py", line 19, in encode
    return codecs.charmap_encode(input,self.errors,encoding_table)[0]
UnicodeEncodeError: 'charmap' codec can't encode characters in position 70924-70950: character maps to <undefined>

我该怎么补救呢?

我试图在提交之前urlencode这个字符串。

queryString = 'eventName=' + evt.fields["eventName"] + '&' + 'eventDescription=' + evt.fields["eventDescription"]; 

在Python中，urllib, urllib2, urllib3和请求模块之间有什么区别?为什么有三个?他们似乎在做同样的事情……

我有一个小工具，我用来从一个网站上下载一个MP3文件，然后构建/更新一个播客XML文件，我已经添加到iTunes。

创建/更新XML文件的文本处理是用Python编写的。但是，我在Windows .bat文件中使用wget来下载实际的MP3文件。我更喜欢用Python编写整个实用程序。

我努力寻找一种用Python实际下载该文件的方法，因此我使用了wget。

那么，如何使用Python下载文件呢?