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…

Accepted Answer

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.

how to construct the curl command from python requests module?

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