[英]Making post request using urllib
I am trying to make request on API provider 我正在尝试向API提供程序发出请求
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"}'
This curl request gives response. 该卷曲请求给出响应。
But same request when I try to make in code 但是当我尝试使用代码进行相同的请求时
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()
It gives 它给
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
This would be the equivalent request using the python requests
library. 这将是使用python
requests
库的等效请求。
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()
But your real problem is that you're using the wrong header keys for their api. 但是,您真正的问题是您为其API使用了错误的标题键。 It's
App-Id
and App-key
, not App_Id
and App_key
. 它是
App-Id
和App-key
,不是App_Id
和App_key
。 It would look like this: 它看起来像这样:
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))
Also worth noting, they have a python api that does all this for you. 同样值得注意的是,他们有一个python API为您完成所有这些工作。
json_data = json.dumps(data)
is not the correct way to prepare POST data. json_data = json.dumps(data)
不是准备POST数据的正确方法。
You should use urllib.urlencode()
to do the job: 您应该使用
urllib.urlencode()
来完成这项工作:
import urllib
data = { "text": text }
req = urllib2.Request(self.url, urllib.urlencode(data), self.headers)
response = urllib2.urlopen(req).read()
class urllib2.Request(url[, data][, headers][, origin_req_host][, unverifiable]) This class is an abstraction of a URL request.
类urllib2.Request(url [,data] [,headers] [,origin_req_host] [,unverifiable])此类是URL请求的抽象。
data may be a string specifying additional data to send to the server, or None if no such data is needed.
data可以是一个字符串,用于指定要发送到服务器的其他数据;如果不需要,则为None。 Currently HTTP requests are the only ones that use data;
当前,HTTP请求是唯一使用数据的请求。 the HTTP request will be a POST instead of a GET when the data parameter is provided.
提供data参数时,HTTP请求将是POST而不是GET。 data should be a buffer in the standard application/x-www-form-urlencoded format.
数据应为标准application / x-www-form-urlencoded格式的缓冲区。 The urllib.urlencode() function takes a mapping or sequence of 2-tuples and returns a string in this format.
urllib.urlencode()函数采用2元组的映射或序列,并以这种格式返回字符串。
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