how to construct the curl command from python requests module?

Question

Python requests is a good module to ease my web REST API access programming, I usually do like below

import json url = 'https://api.github.com/some/endpoint' payload = {'some': 'data'} headers = {'Content-type': 'application/json', 'Accept': 'application/json'}  r = requests.post(url, data=json.dumps(payload), headers=headers) 

And when there is error comes out, I want to see what happen behind it. Constructing the curl command to reproduce in command line is the common way, since this is the standard way which is most described in RESP API document

try:     r = requests.post(url, data=json.dumps(payload), headers=headers) except Exception as ex:     print "try to use curl command below to reproduce"     print curl_request(url,"POST",headers,payload) 

It will be nice I can generate curl command sample for this request, see good example in libcloud's debug, I can't find a simple way to construct, below are the method I want to create by myself.

# below code is just pseudo code, not correct  def curl_request(url,method,headers,payloads):     # construct curl sample from requests' structure     # $ curl -v -H "Accept: application/json" -H "Content-type: application/json"      # -d '{"some":"data"}'      # -X POST https://api.github.com/some/endpoint     request = "curl -v "     for header in headers:         print header         request = request + '-H "' + header + ": " + headers[header] + '" '     for payload in payloads:         request = request + '-d {} "' + payload + ": " + payloads[payload] + '" '              request = request + "-X %s %s" % (method,url)     return request 

It will also be nice if we have method in requests already

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Below are the final solution get the answer, works for me. Show it here for your reference

def curl_request(url,method,headers,payloads):     # construct the curl command from request     command = "curl -v -H {headers} {data} -X {method} {uri}"     data = ""      if payloads:         payload_list = ['"{0}":"{1}"'.format(k,v) for k,v in payloads.items()]         data = " -d '{" + ", ".join(payload_list) + "}'"     header_list = ['"{0}: {1}"'.format(k, v) for k, v in headers.items()]     header = " -H ".join(header_list)     print command.format(method=method, headers=header, data=data, uri=url)     

Answer 1

You can also use curlify to do this.

$ pip install curlify ... import curlify print(curlify.to_curl(r.request))  # r is the response object from the requests library. 

Answer 2

This method existed in requests once upon a time but it is far from being remotely relevant to the module. You could create a function that takes a response and inspects its request attribute.

The request attribute is a PreparedRequest object so it has headers, and body attributes. The body is what you pass to curl with -d and the headers can be generated as you did above. Finally you'll want to pluck off the url attribute from the request object and send that. The hooks don't matter to you unless you're doing something with a custom authentication handler.

req = response.request  command = "curl -X {method} -H {headers} -d '{data}' '{uri}'" method = req.method uri = req.url data = req.body headers = ['"{0}: {1}"'.format(k, v) for k, v in req.headers.items()] headers = " -H ".join(headers) return command.format(method=method, headers=headers, data=data, uri=uri) 

That should work. Your data will be properly formatted whether it is as multipart/form-data or anything else.