hug*_*gie 170 python http python-requests
在使用requests
模块时,有没有办法打印原始HTTP请求?
我不想只是标题,我想要请求行,标题和内容打印输出.是否有可能看到最终由HTTP请求构造的内容?
小智 183
由于v1.2.3请求添加了PreparedRequest对象.根据文档"它包含将发送到服务器的确切字节".
可以使用它来打印请求,如下所示:
import requests
req = requests.Request('POST','http://stackoverflow.com',headers={'X-Custom':'Test'},data='a=1&b=2')
prepared = req.prepare()
def pretty_print_POST(req):
"""
At this point it is completely built and ready
to be fired; it is "prepared".
However pay attention at the formatting used in
this function because it is programmed to be pretty
printed and may differ from the actual request.
"""
print('{}\n{}\r\n{}\r\n\r\n{}'.format(
'-----------START-----------',
req.method + ' ' + req.url,
'\r\n'.join('{}: {}'.format(k, v) for k, v in req.headers.items()),
req.body,
))
pretty_print_POST(prepared)
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产生:
-----------START-----------
POST http://stackoverflow.com/
Content-Length: 7
X-Custom: Test
a=1&b=2
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然后你可以发送实际的请求:
s = requests.Session()
s.send(prepared)
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这些链接是可用的最新文档,因此它们可能会更改内容: 高级 - 准备请求和API - 低级别类
gon*_*opp 42
注意:这个答案已经过时了.更新版本的requests
支持直接获取请求内容,如AntonioHerraizS的回答文档.
由于它只处理更高级别的对象,例如标题和方法类型,因此无法获得请求的真实原始内容.使用发送请求,但还不能与原始数据处理-它使用.这是请求的代表性堆栈跟踪:requests
requests
urllib3
urllib3
httplib
-> r= requests.get("http://google.com")
/usr/local/lib/python2.7/dist-packages/requests/api.py(55)get()
-> return request('get', url, **kwargs)
/usr/local/lib/python2.7/dist-packages/requests/api.py(44)request()
-> return session.request(method=method, url=url, **kwargs)
/usr/local/lib/python2.7/dist-packages/requests/sessions.py(382)request()
-> resp = self.send(prep, **send_kwargs)
/usr/local/lib/python2.7/dist-packages/requests/sessions.py(485)send()
-> r = adapter.send(request, **kwargs)
/usr/local/lib/python2.7/dist-packages/requests/adapters.py(324)send()
-> timeout=timeout
/usr/local/lib/python2.7/dist-packages/requests/packages/urllib3/connectionpool.py(478)urlopen()
-> body=body, headers=headers)
/usr/local/lib/python2.7/dist-packages/requests/packages/urllib3/connectionpool.py(285)_make_request()
-> conn.request(method, url, **httplib_request_kw)
/usr/lib/python2.7/httplib.py(958)request()
-> self._send_request(method, url, body, headers)
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在httplib
机器内部,我们可以看到HTTPConnection._send_request
间接使用HTTPConnection._send_output
,最终创建原始请求和正文(如果存在),并用于HTTPConnection.send
单独发送它们.send
终于到达了套接字.
由于没有任何钩子可以做你想做的事情,作为最后的手段,你可以通过补丁httplib
来获取内容.这是一个脆弱的解决方案,如果httplib
发生变化,您可能需要对其进行调整.如果您打算使用此解决方案分发软件,您可能需要考虑打包httplib
而不是使用系统,这很容易,因为它是纯粹的python模块.
唉,不用多说,解决方案:
import requests
import httplib
def patch_send():
old_send= httplib.HTTPConnection.send
def new_send( self, data ):
print data
return old_send(self, data) #return is not necessary, but never hurts, in case the library is changed
httplib.HTTPConnection.send= new_send
patch_send()
requests.get("http://www.python.org")
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产生输出:
GET / HTTP/1.1
Host: www.python.org
Accept-Encoding: gzip, deflate, compress
Accept: */*
User-Agent: python-requests/2.1.0 CPython/2.7.3 Linux/3.2.0-23-generic-pae
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Emi*_*röm 36
更好的想法是使用requests_toolbelt库,它可以将请求和响应转储为字符串,供您打印到控制台.它使用上述解决方案无法处理的文件和编码来处理所有棘手的情况.
它就像这样简单:
import requests
from requests_toolbelt.utils import dump
resp = requests.get('https://httpbin.org/redirect/5')
data = dump.dump_all(resp)
print(data.decode('utf-8'))
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来源:https://toolbelt.readthedocs.org/en/latest/dumputils.html
您只需输入以下命令即可安装:
pip install requests_toolbelt
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Pay*_*man 33
import requests
response = requests.post('http://httpbin.org/post', data={'key1':'value1'})
print(response.request.body)
print(response.request.headers)
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我正在使用请求版本2.18.4和Python 3
requests
支持所谓的事件挂钩(从 2.23 开始,实际上只有response
挂钩)。该钩子可用于请求打印完整的请求-响应对的数据,包括有效的 URL、标题和正文,例如:
import textwrap
import requests
def print_roundtrip(response, *args, **kwargs):
format_headers = lambda d: '\n'.join(f'{k}: {v}' for k, v in d.items())
print(textwrap.dedent('''
---------------- request ----------------
{req.method} {req.url}
{reqhdrs}
{req.body}
---------------- response ----------------
{res.status_code} {res.reason} {res.url}
{reshdrs}
{res.text}
''').format(
req=response.request,
res=response,
reqhdrs=format_headers(response.request.headers),
reshdrs=format_headers(response.headers),
))
requests.get('https://httpbin.org/', hooks={'response': print_roundtrip})
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运行它打印:
---------------- request ----------------
GET https://httpbin.org/
User-Agent: python-requests/2.23.0
Accept-Encoding: gzip, deflate
Accept: */*
Connection: keep-alive
None
---------------- response ----------------
200 OK https://httpbin.org/
Date: Thu, 14 May 2020 17:16:13 GMT
Content-Type: text/html; charset=utf-8
Content-Length: 9593
Connection: keep-alive
Server: gunicorn/19.9.0
Access-Control-Allow-Origin: *
Access-Control-Allow-Credentials: true
<!DOCTYPE html>
<html lang="en">
...
</html>
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如果响应是二进制的,您可能想要更改res.text
为res.content
。
小智 6
这是一个代码,与上面相同,但是带有响应头:
import socket
def patch_requests():
old_readline = socket._fileobject.readline
if not hasattr(old_readline, 'patched'):
def new_readline(self, size=-1):
res = old_readline(self, size)
print res,
return res
new_readline.patched = True
socket._fileobject.readline = new_readline
patch_requests()
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我花了很多时间寻找这个,所以如果有人需要,我就把它留在这里。
@AntonioHerraizS 答案的一个分支(如评论中所述,HTTP 版本缺失)
使用以下代码获取表示原始 HTTP 数据包的字符串而不发送它:
import requests
def get_raw_request(request):
request = request.prepare() if isinstance(request, requests.Request) else request
headers = '\r\n'.join(f'{k}: {v}' for k, v in request.headers.items())
body = '' if request.body is None else request.body.decode() if isinstance(request.body, bytes) else request.body
return f'{request.method} {request.path_url} HTTP/1.1\r\n{headers}\r\n\r\n{body}'
headers = {'User-Agent': 'Test'}
request = requests.Request('POST', 'https://stackoverflow.com', headers=headers, json={"hello": "world"})
raw_request = get_raw_request(request)
print(raw_request)
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结果:
POST / HTTP/1.1
User-Agent: Test
Content-Length: 18
Content-Type: application/json
{"hello": "world"}
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还可以在响应对象中打印请求
Run Code Online (Sandbox Code Playgroud)r = requests.get('https://stackoverflow.com') raw_request = get_raw_request(r.request) print(raw_request)
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