如何构建来自python请求模块的curl命令？

Question

Python requests是一个很好的模块，以减轻我的网络REST API访问编程，我通常不喜欢下面

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) 

当有错误出现，我想看看它后面发生什么。构建curl命令在命令行再现是常见的方式，因为这是

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) 

这将是很好我可以生成此请求curl命令样品，其最在RESP API文献中描述的标准方法，参见在libcloud's debug中很好的例子，我找不到一个简单的方法来构建，下面是我想自己创建的方法。

# 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 

这也将是很好，如果我们在requests有方法已经

---

下面是最终的解决方案得到了答案，我的作品。在此处显示以供参考

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

此方法存在于请求中，但与远程模块远没有关系。您可以创建一个函数来获取响应并检查其属性request。

request属性是PreparedRequest对象，所以它具有headers和body属性。正文是你通过与-d卷曲并且可以像上面那样生成标题。最后，您需要从request对象中摘取url属性并发送该属性。除非你使用自定义身份验证处理程序进行操作，否则钩子对你无关紧要。

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) 

这应该工作。您的数据将被格式化为multipart/form-data或其他任何内容。