Question

Python requests是一个很好的模块来简化我的Web REST API访问编程，我通常会在下面这样做

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。构造curl命令以在命令行中重现是常用方法，因为这是RESP API文档中最常描述的标准方法

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，请参阅libcloud's debug中的好示例，我找不到一种简单的构造方法，下面是我要创建的方法我自己。

# 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

，那也很好

以下是最终解决方案得到答案，适合我。在此处显示供您参考

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

这种方法曾经存在于请求中，但远不是与模块远程相关。您可以创建一个接收响应并检查其request属性的函数。

request属性是PreparedRequest对象，因此它具有headers和body属性。身体是你传递给-d卷曲的身体，你可以像上面那样生成标题。最后，您需要从url对象中取出request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，否则挂钩对您无关紧要。

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)

应该工作。无论是multipart/form-data还是其他任何内容，您的数据都将被正确格式化。

Answer 2

您也可以使用curlify进行此操作。

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

如何从python请求模块构造curl命令？

2 个答案: