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()

执行此操作时出现错误415：不支持的媒体类型，所以我更改了：

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

成：

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

这次我有400：请求不好。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

在bash示例中，属性target是一个数组，在Python示例中它是一个字符串。

试试这个：

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

我还强烈建议您查看具有更好API的requests库：

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

Answer 2

我知道现在已超过一年了，但请尝试删除标头值中的空白。

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

我也更喜欢使用请求模块，因为API /方法干净且易于使用。

Answer 3

PycURL是用C语言编写的libcurl库的包装，因此其Python API可能让人有些困惑。正如某些人提倡使用python请求一样，我只想指出这不是一个完美的替代方法。对我而言，其缺乏DNS解析超时功能会破坏交易。我也发现Raspberry Pi的速度慢得多。这种比较可能是相关的： Python Requests vs PyCurl Performance

因此，有些事情无法回避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()

还有一些更有趣的功能值得在libcurl curleasy setopts documentation中查看

Answer 4

我有类似的问题，我使用了您的代码示例，但对httpheader部分进行了如下更新：

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

Answer 5

使用请求库更简单。（http://docs.python-requests.org/en/latest）

我为原始curl自定义标题添加了python代码。

import json
import requests

url = 'http://some-url'
headers = {'Content-Type': "application/json; charset=xxxe", 'Accept': "application/json"}
data = {"name": "abc", "path": "def", "target":  ["ghi"]}
res = requests.post(url, json=data, headers=headers)
print (res.status_code)
print (res.raise_for_status())

将POST请求卷入pycurl代码

5 个答案: