There is a script that sends the file to all users from the database. But recipients see all other recipients. How to hide them?

from flask import Flask, render_template, make_response from flask_celery import make_celery from flask_sqlalchemy import SQLAlchemy from flask_mail import Mail, Message import os app = Flask(__name__) app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///subscribe.db' app.config['MAIL_SERVER'] = 'mail.post.com' app.config['MAIL_PORT'] = 25 app.config['MAIL_USERNAME'] = 'UserName' app.config['MAIL_PASSWORD'] = 'Secret' db = SQLAlchemy(app) celery = make_celery(app) mail = Mail(app) class Subscribe(db.Model): id = db.Column('id', db.Integer, primary_key=True) email = db.Column('email', db.String(50), unique=True, nullable=False) company = db.Column(db.String(50), nullable=False) def __repr__(self): return f"Subscribe('{self.company}', '{self.email}')" @app.route('/sendmail', methods=['GET', 'POST']) def index(): with app.app_context(): theuser = Subscribe.query.filter(Subscribe.id).all() msg = Message('Hello', sender='flask@post.com', recipients=[theuser.email for theuser in Subscribe.query.all()]) msg.body = "testing" msg.html = "<b>testing</b>" with app.open_resource(r"C:\dir\worry_cat.jpg") as fp: msg.attach("image.png", "image/png", fp.read()) for theuser in Subscribe.query.all(): mail.send(msg) return 'Message sent!'

    1 answer 1

    Bypass users in a loop and send an email to each separately:

     with app.open_resource(r"C:\dir\worry_cat.jpg") as fp: img = fp.read() for theuser in Subscribe.query.all(): msg = Message('Hello', sender='flask@post.com', recipients=[theuser.email]) msg.body = "testing" msg.html = "<b>testing</b>" msg.attach("image.png", "image/png", img) mail.send(msg)
    • Thanks, that is necessary. Your answer, as always, helped ignoramus! - Vadim