python - Python 3.5 / Pastebin“错误的 API 请求，无效的 api_option”

Question

我正在开发一个 twitch irc 机器人，我想要的组件之一是机器人能够在关闭时将引号保存到 pastebin 粘贴，然后在启动时检索相同的引号。

我从保存部分开始，遇到了一个我似乎无法获得有效帖子的路障，我想不出方法。

#!/usr/bin/env python3

import urllib.parse
import urllib.request

# --------------------------------------------- Pastebin Requisites --------------------------------------------------

pastebin_key = 'my pastebin key'  # developer api key, required. GET: http://pastebin.com/api
pastebin_password = 'password'  # password for pastebin_username
pastebin_postexp = 'N'  # N = never expire
pastebin_private = 0  # 0 = Public 1 = unlisted 2 = Private
pastebin_url = 'http://pastebin.com/api/api_post.php'
pastebin_username = 'username'  # user corresponding with key


# --------------------------------------------- Value clean up --------------------------------------------------

pastebin_password = urllib.parse.quote(pastebin_password, safe='/')
pastebin_username = urllib.parse.quote(pastebin_username, safe='/')

# --------------------------------------------- Pastebin Functions --------------------------------------------------

def post(title, content):  # used for posting a new paste
    pastebin_vars = {'api_option': 'paste', 'api_user_key':     pastebin_username, 'api_paste_private': pastebin_private,
                 'api_paste_name': title, 'api_paste_expire_date': pastebin_postexp,  'api_dev_key': pastebin_key,
                 'api_user_password': pastebin_password, 'api_paste_code': content}
    try:
        str_to_paste = ', '.join("{!s}={!r}".format(key, val) for (key, val) in pastebin_vars.items())  # dict to str :D
        str_to_paste = str_to_paste.replace(":", "")  # remove :
        str_to_paste = str_to_paste.replace("'", "")  # remove '
        str_to_paste = str_to_paste.replace(")", "")  # remove )
        str_to_paste = str_to_paste.replace(", ", "&")  # replace dividers with &
        urllib.request.urlopen(pastebin_url, urllib.parse.urlencode(pastebin_vars)).read()
        print('did that work?')
    except:
        print("post submit failed :(")
        print(pastebin_url + "?" + str_to_paste)  # print the output for test

 post("test", "stuff")

我愿意导入更多的库和东西，在连续两天工作后不太确定我做错了什么：S

score 1 · Accepted Answer

import urllib.parse
import urllib.request

PASTEBIN_KEY = 'xxx'
PASTEBIN_URL = 'https://pastebin.com/api/api_post.php'
PASTEBIN_LOGIN_URL = 'https://pastebin.com/api/api_login.php'
PASTEBIN_LOGIN = 'my_login_name'
PASTEBIN_PWD = 'yyy'

def pastebin_post(title, content):
    login_params = dict(
        api_dev_key=PASTEBIN_KEY,
        api_user_name=PASTEBIN_LOGIN,
        api_user_password=PASTEBIN_PWD
    )

    data = urllib.parse.urlencode(login_params).encode("utf-8")
    req = urllib.request.Request(PASTEBIN_LOGIN_URL, data)

    with urllib.request.urlopen(req) as response:
        pastebin_vars = dict(
            api_option='paste',
            api_dev_key=PASTEBIN_KEY,
            api_user_key=response.read(),
            api_paste_name=title,
            api_paste_code=content,
            api_paste_private=2,
        )
        return urllib.request.urlopen(PASTEBIN_URL, urllib.parse.urlencode(pastebin_vars).encode('utf8')).read()

rv = pastebin_post("This is my title", "These are the contents I'm posting")
print(rv)

结合上面两个不同的答案给了我这个可行的解决方案。

score 0 · Accepted Answer

首先，您的try/except块正在丢弃实际错误。您几乎不应该在except不捕获或重新引发原始异常的情况下使用“裸”子句。有关完整说明，请参阅本文。

删除try/except后，您将看到潜在的错误：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "paste.py", line 42, in post
    urllib.request.urlopen(pastebin_url, urllib.parse.urlencode(pastebin_vars)).read()
  File "/usr/lib/python3.4/urllib/request.py", line 161, in urlopen
    return opener.open(url, data, timeout)
  File "/usr/lib/python3.4/urllib/request.py", line 461, in open
    req = meth(req)
  File "/usr/lib/python3.4/urllib/request.py", line 1112, in do_request_
    raise TypeError(msg)
TypeError: POST data should be bytes or an iterable of bytes. It cannot be of type str.

这意味着您正在尝试将 unicode 字符串传递给需要字节的函数。当您执行 I/O（例如在磁盘上读取/写入文件，或通过 HTTP 发送/接收数据）时，您通常需要将任何 unicode 字符串编码为字节。有关unicode 与字节以及何时需要编码和解码的详细说明，请参阅此演示文稿。

接下来，这一行：

urllib.request.urlopen(pastebin_url, urllib.parse.urlencode(pastebin_vars)).read()

正在丢弃响应，因此您无法知道 API 调用的结果。将此分配给一个变量或从您的函数中返回它，以便您可以检查该值。它可以是粘贴的 URL，也可以是来自 API 的错误消息。

接下来，我认为您的代码向 API 发送了许多不必要的参数，并且您的str_to_paste语句不是必需的。

我可以使用以下更简单的代码进行粘贴：

import urllib.parse
import urllib.request

PASTEBIN_KEY = 'my-api-key' # developer api key, required. GET: http://pastebin.com/api
PASTEBIN_URL = 'http://pastebin.com/api/api_post.php'

def post(title, content):  # used for posting a new paste
    pastebin_vars = dict(
        api_option='paste',
        api_dev_key=PASTEBIN_KEY,
        api_paste_name=title,
        api_paste_code=content,
    )
    return urllib.request.urlopen(PASTEBIN_URL, urllib.parse.urlencode(pastebin_vars).encode('utf8')).read()

在这里它正在使用中：

>>> post("test", "hello\nworld.")
b'http://pastebin.com/v8jCkHDB'

score 0 · Accepted Answer

直到现在我才知道 pastebin。我阅读了他们的 api 并第一次尝试了它，它工作得非常好。

这是我所做的：

我登录以获取api_user_key.
将其与api_dev_key.
看了下官网，帖子还在。

这是代码：

import urllib.parse
import urllib.request


def post(url, params):
    data = urllib.parse.urlencode(login_params).encode("utf-8")
    req = urllib.request.Request(login_url, data)
    with urllib.request.urlopen(req) as response:
       return response.read()


# Logging in to fetch api_user_key
login_url = "http://pastebin.com/api/api_login.php"
login_params = {"api_dev_key": "<the dev key they gave you",
                "api_user_name": "<username goes here>",
                "api_user_password": "<password goes here>"}
api_user_key = post(login_url, login_params)


# Posting some random text
post_url = "http://pastebin.com/api/api_post.php"
post_params = {"api_dev_key": "<the dev key they gave you",
               "api_option": "paste",
               "api_paste_code": "<head>Testing</head>",
               "api_paste_private": "0",
               "api_paste_name": "testing.html",
               "api_paste_expire_date": "10M",
               "api_paste_format": "html5",
               "api_user_key": api_user_key}
response = post(post_url, post_params)

发布内容只需要前三个参数，其余参数是可选的。

score 0 · Accepted Answer

0

在撰写本文时，API 似乎不接受 http 请求，因此请确保网址格式为https://pas...

于 2021-03-04T02:38:24.313 回答

python - Python 3.5 / Pastebin“错误的 API 请求，无效的 api_option”

4 回答 4

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