Translate curl to python [duplicate] - python

This question already has answers here:
Using headers with the Python requests library's get method
(4 answers)
Closed 7 years ago.
I have a curl request that works:
curl "https://api.propublica.org/campaign-finance/v1/2016/candidates/search.json?query=Wilson"
-H "X-API-Key: PROPUBLICA_API_KEY"
How can I translate this into Python? I tried the following:
payload = {'X-API-Key': 'myapikey'}
r = requests.get("https://api.propublica.org/campaign-finance/v1/2016/candidates/search.json?query=Wilson", params = payload)
Then, I got:
>>> print(r.url)
https://api.propublica.org/campaign-finance/v1/2016/candidates/search.json?query=Wilson&X-API-Key=myapikey
>>> r.text
u'{"message": "Forbidden"}'

The simplest way to translate your curl work to python will be to use pycurl instead of requests.
Your Forbidden issue, however, does not depend on using requests or pycurl. It comes from sending the X-API-Key as query parameters instead of sending it as header (as you did in the curl call).

Try this out:
import urllib2
url = your_url
payload = {'X-API-Key': 'myapikey'}
req = Request(url, payload, {'Content-Type': 'application/json'})
open = urlopen(req)
after this you can use whatever way you want to print. Hope this works for you.

Related

Translating a curl post request into python-requests does not work [duplicate]

This question already has answers here:
How to POST JSON data with Python Requests?
(10 answers)
Closed 3 years ago.
curl -i -H "Content-Type: application/json" -X POST -d '{"message":
"test"}' http://localhost:5000/members
This is my old code which was working as a command (returns a 200). now i want to write a python skript for that using requests. I am trying this:
header = {"Content-type": "application/json"}
r = requests.post("http://127.0.0.1:5000/members", data={"message": "test"}, headers=header)
Which returns an 400. Why? How can i fix that? Its btw a flask rest api running on that port.
Your trying to send the data without actually combining it. You just need to surround the {} with '.
header = {"Content-type": "application/json"}
r = requests.post("http://127.0.0.1:5000/members", data='{"message": "test"}', headers=header)

Curl works but python requests doesn't

When I do curl, I get a response:
root#3d7044bac92f:/home/app/tmp# curl -H "Content-type: application/json" -X GET https://github.com/timeline.json -k
{"message":"Hello there, wayfaring stranger. If you\u2019re reading this then you probably didn\u2019t see our blog post a couple of years back announcing that this API would go away: http://git.io/17AROg Fear not, you should be able to get what you need from the shiny new Events API instead.","documentation_url":"https://developer.github.com/v3/activity/events/#list-public-events"}
However, when I do python requests to the same URL I get a status 410.
import requests
headers = {
'Content-type': 'application/json',
}
r = requests.get('https://github.com/timeline.json')
print r.json
root#3d7044bac92f:/home/app/tmp# python rest.py
<bound method Response.json of <Response [410]>>
What gives?
The host is a standard Ubuntu docker image and only installed Curl and some python modules. Python -V is 2.7
Note: I looked at this question but I can't telnet into above server so that solution doesn't apply to me:
Curl works but not Python requests
You've made at least two errors in your program.
1) You haven't specified the data= or headers parameters to the requests.get() call. Try this:
r = requests.get('https://github.com/timeline.json', data=data, headers=headers)
2) .json is a method, not a data attribute of the response object. As a method, it must be called in order to be effective. Try this:
print r.json()

How to POST a local file using urllib2 in Python?

I am a complete Python noob and am trying to run cURL equivalents using urllib2. What I want is a Python script that, when run, will do the exact same thing as the following cURL command in Terminal:
curl -k -F docfile=#myLocalFile.csv http://myWebsite.com/extension/extension/extension
I found the following template on a tutorial page:
import urllib
import urllib2
url = "https://uploadWebsiteHere.com"
data = "{From: 'sender#email.com', To: 'recipient#email.com', Subject: 'Postmark test', HtmlBody: 'Hello dear Postmark user.'}"
headers = { "Accept" : "application/json",
"Conthent-Type": "application/json",
"X-Postmark-Server-Token": "abcdef-1234-46cc-b2ab-38e3a208ab2b"}
req = urllib2.Request(url, data, headers)
response = urllib2.urlopen(req)
the_page = response.read()
but I am completely lost on the 'data' and 'headers' vars. The urllib2 documentation (https://docs.python.org/2/library/urllib2.html) defines the 'data' input as "a string specifying additional data to send to the server" and the 'headers' input as "a dictionary". I am totally out of my depth in trying to follow this documentation and do not see why a dictionary is necessary when I could accomplish this same task in terminal by only specifying the file and URL. Thoughts, please?
The data you are posting doesn't appear to be valid JSON. Assuming the server is expecting valid JSON, you should change that.
Your curl invocation does not pass any optional headers, so you shouldn't need to provide much in the request. If you want to verify the exact headers you could add -vi to the curl invocation and directly match them in the Python code. Alternatively, this works for me:
import urllib2
url = "http://localhost:8888/"
data = '{"From": "sender#email.com", "To": "recipient#email.com", "Subject": "Postmark test", "HtmlBody": "Hello dear Postmark user."}'
headers = {
"Content-Type": "application/json"
}
req = urllib2.Request(url, data, headers)
response = urllib2.urlopen(req)
the_page = response.read()
It probably is in your best interest to switch over to using requests, but for something this simple the standard library urllib2 can be made to work.
What I want is a Python script that, when run, will do the exact same thing as the following cURL command in Terminal:
$ curl -k -F docfile=#myLocalFile.csv https://myWebsite.com/extension...
curl -F sends the file using multipart/form-data content type. You could reproduce it easily using requests library:
import requests # $ pip install requests
with open('myLocalFile.csv','rb') as input_file:
r = requests.post('https://myWebsite.com/extension/...',
files={'docfile': input_file}, verify=False)
verify=False is to emulate curl -k.

Python requests - POST data from a file

I have used curl to send POST requests with data from files.
I am trying to achieve the same using python requests module. Here is my python script
import requests
payload=open('data','rb').read()
r = requests.post('https://IP_ADDRESS/rest/rest/2', auth=('userid', 'password'), data=payload , verify=False)
print r.text
Data file looks like below
'ID' : 'ISM03'
But my script is not POSTing the data from file. Am I missing something here.
In Curl , I used to have a command like below
Curl --data #filename -ik -X POST 'https://IP_ADDRESS/rest/rest/2'
You do not need to use .read() here, simply stream the object directly. You do need to set the Content-Type header explicitly; curl does this when using --data but requests doesn't:
with open('data','rb') as payload:
headers = {'content-type': 'application/x-www-form-urlencoded'}
r = requests.post('https://IP_ADDRESS/rest/rest/2', auth=('userid', 'password'),
data=payload, verify=False, headers=headers)
I've used the open file object as a context manager so that it is also auto-closed for you when the block exits (e.g. an exception occurs or requests.post() successfully returns).

Sending JSON request with Python

I'm new to web services and am trying to send the following JSON based request using a python script:
http://myserver/emoncms2/api/post?apikey=xxxxxxxxxxxxx&json={power:290.4,temperature:19.4}
If I paste the above into a browser, it works as expected. However, I am struggling to send the request from Python. The following is what I am trying:
import json
import urllib2
data = {'temperature':'24.3'}
data_json = json.dumps(data)
host = "http://myserver/emoncms2/api/post"
req = urllib2.Request(host, 'GET', data_json, {'content-type': 'application/json'})
response_stream = urllib2.urlopen(req)
json_response = response_stream.read()
How do I add the apikey data into the request?
Thank you!
Instead of using urllib2, you can use requests. This new python lib is really well written and it's easier and more intuitive to use.
To send your json data you can use something like the following code:
import json
import requests
data = {'temperature':'24.3'}
data_json = json.dumps(data)
payload = {'json_payload': data_json, 'apikey': 'YOUR_API_KEY_HERE'}
r = requests.get('http://myserver/emoncms2/api/post', data=payload)
You can then inspect r to obtain an http status code, content, etc
Even though this doesnt exactly answer OPs question, it should be mentioned here that requests module has a json option that can be used like this:
import requests
requests.post(
'http://myserver/emoncms2/api/post?apikey=xxxxxxxxxxxxx',
json={"temperature": "24.3"}
)
which would be equivalent to the curl:
curl 'http://myserver/emoncms2/api/post?apikey=xxxxxxxxxxxxx' \
-H 'Content-Type: application/json' \
--data-binary '{"temperature":"24.3"}'
Maybe the problem is that json.dumps puts " and in the json you put in the url there are no "s.
For example:
data = {'temperature':'24.3'}
print json.dumps(data)
prints:
{"temperature": "24.3"}
and not:
{temperature: 24.3}
like you put in your url.
One way of solving this (which is trouble prone) is to do:
json.dumps(data).replace('"', '')

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