Python 是否有用于解析 HTTP 请求和响应的模块？
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Does Python have a module for parsing HTTP requests and responses?
提问by Jeff Hammerbacher
httplib (now http.client) and friends all have conn.getresponse() and an HTTPResponse class, but the server-side operations of conn.getrequest() and an HTTPRequest class seem to be lacking.
httplib（现在是http.client）和朋友们都有conn.getresponse() 和一个HTTPResponse 类，但是conn.getrequest() 和一个HTTPRequest 类的服务器端操作似乎缺乏。
I understand that BaseHTTPServer and BaseHTTPRequestHandler can perform this functionality, but they don't expose these methods for use outside of the module.
我知道 BaseHTTPServer 和 BaseHTTPRequestHandler 可以执行此功能，但它们不会公开这些方法以供在模块外部使用。
Essentially what I want is BaseHTTPRequestHandler#parse_request to be a static method that returns an HTTPRequest object rather than populating member variables.
基本上我想要的是 BaseHTTPRequestHandler#parse_request 是一个静态方法，它返回一个 HTTPRequest 对象而不是填充成员变量。
回答by Brandon Rhodes
Jeff, to enable parsing I create a small nine-line subclass of the base HTTP request handler:
Jeff，为了启用解析，我创建了基本 HTTP 请求处理程序的一个九行小子类：
from BaseHTTPServer import BaseHTTPRequestHandler
from StringIO import StringIO
class HTTPRequest(BaseHTTPRequestHandler):
def __init__(self, request_text):
self.rfile = StringIO(request_text)
self.raw_requestline = self.rfile.readline()
self.error_code = self.error_message = None
self.parse_request()
def send_error(self, code, message):
self.error_code = code
self.error_message = message
You can now take a string with the text of an HTTP request inside and parse it by instantiating this class:
您现在可以获取一个包含 HTTP 请求文本的字符串，并通过实例化此类来解析它：
# Simply instantiate this class with the request text
request = HTTPRequest(request_text)
print request.error_code # None (check this first)
print request.command # "GET"
print request.path # "/who/ken/trust.html"
print request.request_version # "HTTP/1.1"
print len(request.headers) # 3
print request.headers.keys() # ['accept-charset', 'host', 'accept']
print request.headers['host'] # "cm.bell-labs.com"
# Parsing can result in an error code and message
request = HTTPRequest('GET\r\nHeader: Value\r\n\r\n')
print request.error_code # 400
print request.error_message # "Bad request syntax ('GET')"
回答by S.Lott
For server-side processing you want to look at something like wsgiref.
对于服务器端处理，您需要查看类似wsgiref 的内容。
The WSGI standard parses the request into a simple dictionary with all of the relevant headers and elements.
WSGI 标准将请求解析为包含所有相关标头和元素的简单字典。
