将POST请求卷入pycurl代码

Question

我正在尝试将以下curl请求转换为pycurl:

curl -v
-H Accept:application/json \
-H Content-Type:application/json \
-d "{
    name: 'abc',
    path: 'def',
    target: [ 'ghi' ]
}" \
-X POST http://some-url

我有以下python代码:

import pycurl, json

c = pycurl.Curl()
c.setopt(pycurl.URL, 'http://some-url')
c.setopt(pycurl.HTTPHEADER, ['Accept: application/json'])
data = json.dumps({"name": "abc", "path": "def", "target": "ghi"})
c.setopt(pycurl.POST, 1)
c.setopt(pycurl.POSTFIELDS, data)
c.setopt(pycurl.VERBOSE, 1)
c.perform()
print curl_agent.getinfo(pycurl.RESPONSE_CODE)
c.close()

执行此操作我有一个错误415:不支持的媒体类型,所以我改变了:

c.setopt(pycurl.HTTPHEADER, ['Accept: application/json'])

成:

c.setopt(pycurl.HTTPHEADER, [ 'Content-Type: application/json' , 'Accept: application/json'])

这次我有400:不好的请求.但是使用curl的bash代码可行.你知道我应该在python代码中修复什么吗？

Answer 1

PycURL 是用 C 语言编写的 libcurl 库的包装器，因此它的 Python API 可能有点令人费解。由于有些人提倡使用 python 请求，我只想指出它不是一个完美的替代品。对我来说，它缺乏 DNS 解析超时是一个交易破坏者。我也发现它在我的 Raspberry Pi 上慢得多。这个比较可能是相关的： Python Requests vs PyCurl Performance

所以这里有一些不能回避 OP 的问题：

import pycurl
import json
from cStringIO import StringIO

curl = pycurl.Curl()
curl.setopt(pycurl.URL, 'http://some-url')
curl.setopt(pycurl.HTTPHEADER, ['Accept: application/json',
                                'Content-Type: application/json'])
curl.setopt(pycurl.POST, 1)

# If you want to set a total timeout, say, 3 seconds
curl.setopt(pycurl.TIMEOUT_MS, 3000)

## depending on whether you want to print details on stdout, uncomment either
# curl.setopt(pycurl.VERBOSE, 1) # to print entire request flow
## or
# curl.setopt(pycurl.WRITEFUNCTION, lambda x: None) # to keep stdout clean

# preparing body the way pycurl.READDATA wants it
# NOTE: you may reuse curl object setup at this point
#  if sending POST repeatedly to the url. It will reuse
#  the connection.
body_as_dict = {"name": "abc", "path": "def", "target": "ghi"}
body_as_json_string = json.dumps(body_as_dict) # dict to json
body_as_file_object = StringIO(body_as_json_string)

# prepare and send. See also: pycurl.READFUNCTION to pass function instead
curl.setopt(pycurl.READDATA, body_as_file_object) 
curl.setopt(pycurl.POSTFIELDSIZE, len(body_as_json_string))
curl.perform()

# you may want to check HTTP response code, e.g.
status_code = curl.getinfo(pycurl.RESPONSE_CODE)
if status_code != 200:
    print "Aww Snap :( Server returned HTTP status code {}".format(status_code)

# don't forget to release connection when finished
curl.close()

在libcurl curleasy setopts 文档中有一些更有趣的功能值得查看

Answer 2

在您的bash示例中，属性target是一个数组，在Python 示例中，属性是一个字符串。

尝试这个：

data = json.dumps({"name": "abc", "path": "def", "target": ["ghi"]})

我也强烈建议您检查requests具有更好API 的库：

import requests
data = {"name": "abc", "path": "def", "target": ["ghi"]}
response = requests.post('http://some-url', json=data)
print response.status_code