解析原始HTTP头部

Question

我有一个原始HTTP字符串，我想表示对象中的字段。有没有办法从HTTP字符串中解析单个头？

'GET /search?sourceid=chrome&ie=UTF-8&q=ergterst HTTP/1.1\r\nHost: www.google.com\r\nConnection: keep-alive\r\nAccept: application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5\r\nUser-Agent: Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_6_6; en-US) AppleWebKit/534.13 (KHTML, like Gecko) Chrome/9.0.597.45 Safari/534.13\r\nAccept-Encoding: gzip,deflate,sdch\r\nAvail-Dictionary: GeNLY2f-\r\nAccept-Language: en-US,en;q=0.8\r\n
[...]'

Answer 1

Update: It’s 2019, so I have rewritten this answer for Python 3, following a confused comment from a programmer trying to use the code. The original Python 2 code is now down at the bottom of the answer.

标准库中有非常好的工具，既可以解析RFC 821头，也可以解析整个HTTP请求。下面是一个示例请求字符串（请注意，Python将其视为一个大字符串，即使我们为了可读性而将其分成几行），我们可以将其提供给我的示例：

request_text = (
    b'GET /who/ken/trust.html HTTP/1.1\r\n'
    b'Host: cm.bell-labs.com\r\n'
    b'Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3\r\n'
    b'Accept: text/html;q=0.9,text/plain\r\n'
    b'\r\n'
)

正如@trypy所指出的，您可以使用Python的email消息库来解析头-不过，我们应该添加的是，一旦您完成创建，生成的Message对象就像头字典一样：

from email.parser import BytesParser
request_line, headers_alone = request_text.split(b'\r\n', 1)
headers = BytesParser().parsebytes(headers_alone)

print(len(headers))     # -> "3"
print(headers.keys())   # -> ['Host', 'Accept-Charset', 'Accept']
print(headers['Host'])  # -> "cm.bell-labs.com"

当然，这会忽略请求行，或者让您自己解析它。结果发现有一个更好的解决方案。

如果您使用标准库的BaseHTTPRequestHandler，它将为您解析HTTP。尽管它的文档有点晦涩——标准库中的整个HTTP和URL工具套件都有问题——但是要使它解析字符串，您所要做的就是（a）将字符串包装在BytesIO()中，（b）阅读raw_requestline以便它随时可以被解析，（c）捕获解析过程中出现的任何错误代码，而不是让它尝试将它们写回客户端（因为我们没有这样的代码！）。

下面是我们对标准库类的专门化：

from http.server import BaseHTTPRequestHandler
from io import BytesIO

class HTTPRequest(BaseHTTPRequestHandler):
    def __init__(self, request_text):
        self.rfile = BytesIO(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

同样，我希望标准库的人已经意识到HTTP解析应该以一种不需要我们编写九行代码来正确调用它的方式进行，但是你能做什么呢？下面是如何使用这个简单的类：

# Using this new class is really easy!

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())   # ['Host', 'Accept-Charset', 'Accept']
print(request.headers['host'])  # "cm.bell-labs.com"

如果在解析过程中出现错误，error_code将不会是None：

# Parsing can result in an error code and message

request = HTTPRequest(b'GET\r\nHeader: Value\r\n\r\n')

print(request.error_code)     # 400
print(request.error_message)  # "Bad request syntax ('GET')"

我更喜欢使用这样的标准库，因为我怀疑他们已经遇到并解决了任何边缘情况，如果我自己尝试用正则表达式重新实现一个Internet规范，这些情况可能会让我感到不快。

旧Python 2代码

这是我第一次写这个答案时的原始代码：

request_text = (
    'GET /who/ken/trust.html HTTP/1.1\r\n'
    'Host: cm.bell-labs.com\r\n'
    'Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3\r\n'
    'Accept: text/html;q=0.9,text/plain\r\n'
    '\r\n'
    )

以及：

# Ignore the request line and parse only the headers

from mimetools import Message
from StringIO import StringIO
request_line, headers_alone = request_text.split('\r\n', 1)
headers = Message(StringIO(headers_alone))

print len(headers)     # -> "3"
print headers.keys()   # -> ['accept-charset', 'host', 'accept']
print headers['Host']  # -> "cm.bell-labs.com"

以及：

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

以及：

# Using this new class is really easy!

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"

以及：

# 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')"

Answer 2

mimetools自Python 2.3以来一直被弃用，并完全从Python 3中删除（link）。

以下是在Python 3中应如何操作：

import email
import io
import pprint

# […]

request_line, headers_alone = request_text.split('\r\n', 1)
message = email.message_from_file(io.StringIO(headers_alone))
headers = dict(message.items())
pprint.pprint(headers, width=160)

Answer 3

如果去掉GET行，这似乎可以正常工作：

import mimetools
from StringIO import StringIO

he = "Host: www.google.com\r\nConnection: keep-alive\r\nAccept: application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5\r\nUser-Agent: Mozilla/5.0 (Macintosh; U; Intel Mac OS X 10_6_6; en-US) AppleWebKit/534.13 (KHTML, like Gecko) Chrome/9.0.597.45 Safari/534.13\r\nAccept-Encoding: gzip,deflate,sdch\r\nAvail-Dictionary: GeNLY2f-\r\nAccept-Language: en-US,en;q=0.8\r\n"

m = mimetools.Message(StringIO(he))

print m.headers

解析示例并将信息从第一行添加到对象的方法是：

import mimetools
from StringIO import StringIO

he = 'GET /search?sourceid=chrome&ie=UTF-8&q=ergterst HTTP/1.1\r\nHost: www.google.com\r\nConnection: keep-alive\r\n'

# Pop the first line for further processing
request, he = he.split('\r\n', 1)    

# Get the headers
m = mimetools.Message(StringIO(he))

# Add request information
m.dict['method'], m.dict['path'], m.dict['http-version'] = request.split()    

print m['method'], m['path'], m['http-version']
print m['Connection']
print m.headers
print m.dict