如何从python请求模块构造curl命令? [英] how to construct the curl command from python requests module?

问题描述

Python requests 是一个很好的模块，可以简化我的 Web REST API 访问编程，我通常会像下面这样

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)

当出现错误时，我想看看它背后发生了什么.构建curl命令在命令行中重现是常用的方式，因为这是RESP API文档中描述最多的标准方式

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)

我可以为此请求生成 curl 命令示例会很好，请参阅 中的好示例libcloud的debug，找不到简单的构造方法，下面是我想自己创建的方法.

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

如果我们已经在 requests 中有方法也很好

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

以下是得到答案的最终解决方案，对我有用.在这里展示以供您参考

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)    

推荐答案

这个方法曾经存在于请求中，但它远非与模块远程相关.您可以创建一个接受响应并检查其 request 属性的函数.

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.

request 属性是一个 PreparedRequest 对象，因此它具有 headers 和 body 属性.正文是您使用 -d 传递给 curl 的内容，并且可以像上面一样生成标题.最后，您需要从 request 对象中取出 url 属性并发送它.除非您使用自定义身份验证处理程序执行某些操作，否则钩子对您无关紧要.

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)

那应该有效.您的数据将被正确格式化，无论是 multipart/form-data 还是其他任何形式.

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

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