使用urllib发出发布请求 [英] Making post request using urllib

问题描述

我正在尝试向API提供程序发出请求

I am trying to make request on API provider

curl "https://api.infermedica.com/dev/parse" \
  -X "POST" \
  -H "App_Id: 4c177c" -H "App_Key: 6852599182ba85d70066986ca2b3" \
  -H "Content-Type: application/json" \
  -d '{"text": "i feel smoach pain but no couoghing today"}'    

此curl请求给出响应。

This curl request gives response.

但是当我尝试编写代码时，同样的请求

But same request when I try to make in code

self.headers = { "App_Id": "4c177c", "App_Key": "6852599182ba85d70066986ca2b3", "Content-Type": "application/json", "User-Agent": "M$

self.url = "https://api.infermedica.com/dev/parse"

data = { "text": text }
json_data = json.dumps(data)
req = urllib2.Request(self.url, json_data.replace(r"\n", "").replace(r"\r", ""), self.headers)
response = urllib2.urlopen(req).read()

它给出

Traceback (most recent call last):
  File "symptoms_infermedia_api.py", line 68, in <module>
    SymptomsInfermedia().getResponse(raw_input("Enter comment"))
  File "symptoms_infermedia_api.py", line 39, in getResponse
    response = urllib2.urlopen(req).read()
  File "/usr/lib/python2.7/urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 410, in open
    response = meth(req, response)
  File "/usr/lib/python2.7/urllib2.py", line 523, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.7/urllib2.py", line 448, in error
    return self._call_chain(*args)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 531, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden

推荐答案

这将是使用python的等效请求请求库。

This would be the equivalent request using the python requests library.

url = "https://api.infermedica.com/dev/parse"
headers = {
    'App_Id': '4c177c',
    'App_Key': '6852599182ba85d70066986ca2b3',
    'Content-Type': 'application/json',
}
data = {'text': 'i feel stomach pain but no coughing today'}

r = requests.post(url, headers=headers, data=json.dumps(data))
print r.status_code
print r.json()

但是您真正的问题是您为其API使用了错误的标题键。它是 App_Id 和 App-key ，而不是 App_Id 和 App_key 。看起来像这样：

But your real problem is that you're using the wrong header keys for their api. It's App-Id and App-key, not App_Id and App_key. It would look like this:

headers = {
    'App-Id': 'xx', 
    'App-key': 'xxxx', 
    'Accept': 'application/json', 
    'Content-Type': 'application/json',
    'Dev-Mode': 'true'}

data = {'text': 'i feel stomach pain but no coughing today'}
r = requests.post(url, headers=headers, data=json.dumps(data))

同样值得注意的是，它们有一个 Python api 为您完成所有这些操作。

Also worth noting, they have a python api that does all this for you.

这篇关于使用urllib发出发布请求的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！