Question

我正在尝试为我的小组构建一个slackbot，我尝试了示例代码和其他一些东西，但它没有向该组发送消息。

首先我尝试通过终端

export SLACK_API_TOKEN="my_token_id"

然后

from slackclient import SlackClient
import os

slack_token = os.environ["SLACK_API_TOKEN"]
sc = SlackClient(slack_token)

sc.api_call(
  "chat.postMessage",
  channel="#random",
  text="Hello from Python! :tada:",
  thread_ts="283.5127(dummy_id)",
  reply_broadcast=False
)

print(sc)

#<slackclient.client.SlackClient object at 0x109b77ba8>

但松弛组中没有消息。

我尝试使用此代码：

from slackclient import SlackClient
import os
slack_token = os.environ['SLACK_API_TOKEN']
sc = SlackClient(slack_token)
print(sc.api_call("channels.list"))

重新调整：

{'error': 'invalid_auth', 'ok': False}

我没有得到我做错了，访问令牌是正确的，我想通过机器人发布一些消息，所以我如何创建一个机器人松弛和使用该机器人我可以通过python发送消息？

Answer 1

当我用php＆amp ;;实现一个松弛的机器人时，我遇到了类似的问题。 symfony的。正确地创建和配置松弛的app，bot和OAuth权限并不是那么简单。

如果您需要，我在此博客文章中解释了所有这些配置：pilcrow

此外，我在PHP中的代码非常类似于解析Slack请求并发布到其API所需的代码。

摘要，TL; DR：

转到https://blog.eleven-labs.com/en/en/replace-erp-by-slack-bot-with-dialogflow-and-symfony/并点击“创建新应用”。
在此应用配置中，转到左侧菜单＆＃39; Bot Users＆＃39;或者来自＆＃39;基本信息＆＃39; ＆GT; ＆＃39;添加功能和＃39; ＆GT; ＆＃39;机器人＆＃39;
仍然在此应用配置中，转到菜单＆＃39; OAuth＆amp;权限＆＃39;并允许范围＆＃39;聊天：写：bot＆＃39;并复制＆＃39; OAuth访问令牌＆＃39;
从您的代码中，拨打＆＃39; chat.postMessage＆＃39;使用＆＃39;授权＆＃39;的API方法标头使用以前的标记值。

Answer 2

通过网络上的一些示例构建了这个：SELECT_TO_DATE 和 liza daly - brobot : github.com

当然不是最好的实现，但它可以作为（我认为）

的适当答案

import random
import time
import re
from slackclient import SlackClient
bot_id = None
slack_token = 'xoxb-no.more.mister.nice.gui'
sc = SlackClient(slack_token)

# constants

RTM_READ_DELAY = 1 # 1 second delay between reading from RTM
DEFAULT_RESPONSE = "greetings: 'hello', 'hi', 'greetings', 'sup', 'what's     up' / commands: 'do'"
DEFAULT_COMMAND = "do"
MENTION_REGEX = "^<@(|[WU].+?)>(.*)"

def parse_bot_commands(slack_events):
    """
    parses a list of events coming from the slack rtm api to find bot commands
    :param slack_events:
    :return:
"""
for event in slack_events:
    if event["type"] == "message" and not "subtype" in event:
        user_id, message = parse_direct_mention(event["text"])
        if user_id == bot_id:
            return message, event["channel"]
return None, None
def parse_direct_mention(message_text):
"""
finds direct message and returns user id
:param message_text:
:return:
"""
matches = re.search(MENTION_REGEX, message_text)
# the first group contains the user name, the second group contains
# the remaining message

return (matches.group(1), matches.group(2).strip()) if matches else (None, None)
def handle_command(command, channel):
"""
    executes bot command if the command is known
:param command:
:param channel:
:return:
"""
GREETING_KEYWORDS = ("hello", "hi", "greetings", "sup", "what's up",)
GREETING_RESPONSES = ["'sup brah", "hey", "*headnod*", "didjageddathingahsencha?"]

# default response is help text for the user
default_response = "Not sure what you mean. Try *{}*.".format(DEFAULT_RESPONSE)

# finds and executes the given command, filling the response
response = None

#implement more commands below this line
if command in GREETING_KEYWORDS:
    response = random.choice(GREETING_RESPONSES)
else:
    if command.startswith(DEFAULT_COMMAND):
        response = "Sure...write some more code and I'll do that"

# Sends the response back to the channel
sc.api_call(
    "chat.postMessage",
    channel="#the_danger_room",
    as_user="true:",
    text=response or default_response)

if __name__ == "__main__":
if sc.rtm_connect(with_team_state=False):
    print("Connected and running!")

    #call web api method auth.test to get bot usre id
    bot_id = sc.api_call("auth.test")["user_id"]
    while True:
        command, channel = parse_bot_commands(sc.rtm_read())
        if command:
            handle_command(command, channel)
        time.sleep(RTM_READ_DELAY)
    else:
        print("Connection failed. Exception traceback printed above.")

如何在python中使用slackbot？

2 个答案: