简体   繁体   English

Telegram Bot响应Python列表中的特定命令

[英]Telegram Bot respond to specific command in Python list

I am making a Telegram bot that can can access database to reply users' query. 我正在制作一个可以访问数据库以回复用户查询的Telegram机器人。 The bot need to respond to specific request of certain data in database. 机器人需要响应数据库中某些数据的特定请求。 I was able to solve for when users request for all data but I am stuck with individual data. 当用户请求所有数据时,我能够解决问题,但是我只能使用单个数据。 I am using telegram.ext from telegram package in python. 我使用telegram.exttelegram在Python包。 Here is what I have done so far. 到目前为止,这是我所做的。

from telegram.ext import Updater, CommandHandler, MessageHandler, Filters
import MySQLdb

currr = [] # global list var ~don't bash me for using global in python please, I'm a newbie

# request for all data in database
def request2(bot, update):
    db = MySQLdb.connect(host = "local", user = "root", passwd = "pwd", db = "mydb")
    cur = db.cursor()
    cur.execute("select ID from table")
    ID = cur.fetchall()

    cur.execute("SELECT ID, temp FROM table2 order by indexs desc")
    each_rows = cur.fetchall()
    for IDs in ID:
        for each_row in each_rows:
            if str(each_row[0])[0:4]==str(ID)[2:6]:
                update.message.reply_text('reply all related data here')
                break

# request for single data
def individualreq(bot, update):
    db = pymysql.connect(host = "localhost", user = "root", passwd = "pwd", db = "mydb")
    update.message.reply_text('reply individual data to users here')

def main():
    updater = Updater("TOKEN")
    dp = updater.dispatcher

    global currr
    # get all ID form database
    db = MySQLdb.connect(host = "localhost", user = "root", passwd = "pwd", db = "mydb")
    cur = db.cursor()
    cur.execute("select ID from table")
    curr_ID = cur.fetchall()

    # example ID = 'F01', 'F02', 'F03'
    for curr_IDs in curr_ID:
        currr.append(curr_IDs[0])

    # request all data
    dp.add_handler(CommandHandler("all", request2))

    # request individual data
    dp.add_handler(CommandHandler(currr, individualreq)) # list command in currr[]

if __name__ == '__main__':
    main()

I am looking for a way to pass the current command which is also the ID in database that user request in the currr[] list to the individualreq(bot, update) function so that only data of the called ID is being replied. 我正在寻找一种方法,将当前命令(也是用户在currr[]列表中请求的数据库中的ID)传递给currr[] individualreq(bot, update)函数,以便仅对被调用ID的数据进行复制。 Users will select from a list of ID in telegram and the command handler can pass the selected ID to the function. 用户将从电报中的ID列表中进行选择,命令处理程序可以将所选的ID传递给函数。 I have not found a way to pass the ID to the function. 我还没有找到将ID传递给函数的方法。 Could someone help me to solve this please. 有人可以帮我解决这个问题。 Thanks 谢谢

I find out a solution for my question from the answer provided by Oluwafemi Sule. 我从Oluwafemi Sule提供的答案中找到了我的问题的解决方案。 CommandHandler can pass the arguments of the command to the function by adding pass_args=True in the CommandHandler . CommandHandler可以通过添加传递命令到函数的参数pass_args=TrueCommandHandler

dp.add_handler(CommandHandler(currr, individualreq, pass_args=True))

To print out the args in the function, the function need to receive the args. 要打印出函数中的参数,该函数需要接收参数。

def individualreq(bot, update, args):
    # id store the args value
    id = update.message.text
    print(id[1:]) # [1:] is to get rid of the / in id

You can outright make individualreq a closure. 您可以直接将individualreq封闭。

CommandHandler takes a command or list of command to listen to and a list other options. CommandHandler接受一个命令或命令列表以进行侦听,并列出其他选项。 There is a pass_user_data option that allows for user data to be passed to the callback. 有一个pass_user_data选项 ,该选项允许将用户数据传递给回调。

dp.add_handler(CommandHandler(currr, individualreq, pass_user_data=True))

The signature for individualreq callback will be updated to take the user_data 对于签名individualreq回调将被更新采取user_data

def individualreq(bot, update, user_data=None):
     #user_data is a dict
     print(user_data)

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM