如何发送POST请求?

use*_*569 243 python urllib httplib

我在网上找到了这个脚本:

import httplib, urllib
params = urllib.urlencode({'number': 12524, 'type': 'issue', 'action': 'show'})
headers = {"Content-type": "application/x-www-form-urlencoded",
            "Accept": "text/plain"}
conn = httplib.HTTPConnection("bugs.python.org")
conn.request("POST", "", params, headers)
response = conn.getresponse()
print response.status, response.reason
302 Found
data = response.read()
data
'Redirecting to <a href="http://bugs.python.org/issue12524">http://bugs.python.org/issue12524</a>'
conn.close()
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但我不明白如何使用它与PHP或params变量内的所有内容或如何使用它.试着让它发挥作用,我可以请一点帮助吗?

小智 359

如果你真的想用Python处理HTTP,我强烈推荐Requests:HTTP for Humans.适用于您的问题的POST快速入门是:

>>> import requests
>>> r = requests.post("http://bugs.python.org", data={'number': 12524, 'type': 'issue', 'action': 'show'})
>>> print(r.status_code, r.reason)
200 OK
>>> print(r.text[:300] + '...')

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>
Issue 12524: change httplib docs POST example - Python tracker

</title>
<link rel="shortcut i...
>>> 
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  • 如果你需要发送一个JSON对象,你应该这样做:`json = {'number':12524 ...`而不是`data = ...` (7认同)
  • 为什么答案说“如果你真的想使用Python处理HTTP”?处理 HTTP 请求是一个坏主意吗?如果是这样,为什么?谁能解释一下吗? (5认同)
  • 请更改data = {'number':12524,以读取数据= {'数字':'12524',.我自己会改变它,但编辑需要超过6个字符.谢谢 (2认同)
  • 如何获取json结果? (2认同)

sti*_*til 139

如果你需要你的脚本是可移植的,而你宁愿没有任何第三方依赖,那么这就是你纯粹在Python 3中发送POST请求的方式.

from urllib.parse import urlencode
from urllib.request import Request, urlopen

url = 'https://httpbin.org/post' # Set destination URL here
post_fields = {'foo': 'bar'}     # Set POST fields here

request = Request(url, urlencode(post_fields).encode())
json = urlopen(request).read().decode()
print(json)
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样本输出:

{
  "args": {}, 
  "data": "", 
  "files": {}, 
  "form": {
    "foo": "bar"
  }, 
  "headers": {
    "Accept-Encoding": "identity", 
    "Content-Length": "7", 
    "Content-Type": "application/x-www-form-urlencoded", 
    "Host": "httpbin.org", 
    "User-Agent": "Python-urllib/3.3"
  }, 
  "json": null, 
  "origin": "127.0.0.1", 
  "url": "https://httpbin.org/post"
}
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  • 这段代码只能在Python 3中使用,就像我在答案中所说的那样. (6认同)

d1j*_*i1b 35

您无法使用urllib(仅限GET)实现POST请求,而是尝试使用requests模块,例如:

例1.0:

import requests

base_url="www.server.com"
final_url="/{0}/friendly/{1}/url".format(base_url,any_value_here)

payload = {'number': 2, 'value': 1}
response = requests.post(final_url, data=payload)

print(response.text) #TEXT/HTML
print(response.status_code, response.reason) #HTTP
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例1.2:

>>> import requests

>>> payload = {'key1': 'value1', 'key2': 'value2'}

>>> r = requests.post("http://httpbin.org/post", data=payload)
>>> print(r.text)
{
  ...
  "form": {
    "key2": "value2",
    "key1": "value1"
  },
  ...
}
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例1.3:

>>> import json

>>> url = 'https://api.github.com/some/endpoint'
>>> payload = {'some': 'data'}

>>> r = requests.post(url, data=json.dumps(payload))
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  • 谢谢.data = json.dumps(payload)是我的用例的关键 (4认同)

Pra*_*ell 9

requests通过点击REST API端点,使用库进行GET,POST,PUT或DELETE。在中传递其余api端点url,在中传递url有效载荷(dict),data并在中传递头/元数据headers

import requests, json

url = "bugs.python.org"

payload = {"number": 12524, 
           "type": "issue", 
           "action": "show"}

header = {"Content-type": "application/x-www-form-urlencoded",
          "Accept": "text/plain"} 

response_decoded_json = requests.post(url, data=payload, headers=header)
response_json = response_decoded_json.json()

print response_json
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  • 此代码在缩进和标题参数名称方面存在问题。 (2认同)
  • “ headers”参数是错误的,而且这里没有json。我们应该使用json.dumps(pauload) (2认同)

小智 6

您的数据字典包含表单输入字段的名称,您只需保持正确的值即可找到结果。 表单视图 标题配置浏览器以检索您声明的数据类型。使用 requests 库可以轻松发送 POST:

import requests

url = "https://bugs.python.org"
data = {'@number': 12524, '@type': 'issue', '@action': 'show'}
headers = {"Content-type": "application/x-www-form-urlencoded", "Accept":"text/plain"}
response = requests.post(url, data=data, headers=headers)

print(response.text)
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更多关于请求对象:https : //requests.readthedocs.io/en/master/api/