Hi, I would like your help with an issue that I'm facing. I want to hold multiple telegram bots, but using a single source code, just changing the bot id. To do so, I'm using the following structure:
My wsgi loads a py file, as expected:
import sys
project_home = '/home/xxxxxx/bots/telegramBot1'
if project_home not in sys.path:
sys.path = [project_home] + sys.path
# import flask app but need to call it "application" for WSGI to work
from flask_app import app as application # noqa
And inside of the flask_app, I call another py, which I'm using as a generic library, I have a folder called DEV, and another called PROD, and inside of this folder, I have a bot_source.py:
import json
import sys
import os
from flask import Flask
from flask import request
f = open(os.path.dirname(os.path.abspath(__file__)) + '/customer.json',)
customerData = json.load(f)
f.close()
project_home = '/home/xxxxx/telegram_bot_source/' + customerData['version'] + '/'
if project_home not in sys.path:
sys.path = [project_home] + sys.path
import bot_source as bot_source
bot_source.setVariables(customerData)
bot_source.setWebhook()
app = Flask(__name__)
@app.route('/', methods=['POST'])
def doPost():
return bot_source.doPost(request)
And inside of the bot_source, it's where I do my thing for the bot. My issue, it's that when a function inside of the bot_source.py has an issue, I don't see the error inside the com.error.log, but I can see prints inside com.server.log
For example, I have this setWebhook function, inside of the bot_source:
def setWebhook():
print('setWebhook')
response = rq.get(telegramUrl + "/deleteWebhook")
print('deleteWebhook ' + json.dumps(response) )
response = rq.get(telegramUrl + "/setWebhook?url=" + webAppUrl)
print('setWebhook ' + json.dumps(response) )
I can see the "setWebhook" print, inside the com.server.log, but the next print (which is the one throwing me an error) I cannot see it, and there's no error inside the com.error.log
What am I doing wrong ? Can anybody, please, shed me some light ?
Thank you