import pymysql
import sys
import json
import random
import threading
import logging
from sqlalchemy import func
from concurrent.futures import ThreadPoolExecutor, as_completed
from datetime import datetime
from typing import List, Dict, Union, Any
from stripe_payment_processor import StripePaymentProcessor
from config import Config
from app import create_app, db
from models import Payments, PaymentBatch, SinglePayments, PaymentPlans
from splynx import Splynx, SPLYNX_URL, SPLYNX_KEY, SPLYNX_SECRET
from services import (
    log_script_start, log_script_completion, log_batch_created, 
    log_payment_intent_followup
)



if __name__ == "__main__":
    app = create_app()
    
    with app.app_context():
        # Find customer_ids that appear more than once
        duplicate_customer_ids = db.session.query(
            Payments.Stripe_Customer_ID
        ).group_by(
            Payments.Stripe_Customer_ID
        ).having(
            func.count(Payments.Stripe_Customer_ID) > 1
        ).all()

        # Get the actual duplicate records
        duplicate_records = db.session.query(Payments).filter(
            Payments.Stripe_Customer_ID.in_([row[0] for row in duplicate_customer_ids])
        ).order_by(Payments.Stripe_Customer_ID).all()
        i = 0
        has_charge = 0
        for a in duplicate_records:
            i += 1
            #print(a.Stripe_Customer_ID)
            if a.Stripe_Charge_ID != None:
                has_charge += 1

        print(i, i/2, has_charge, has_charge/2)


        ranked_duplicates = db.session.query(
            Payments.id,
            func.row_number().over(
                partition_by=Payments.Stripe_Customer_ID,
                order_by=Payments.id
            ).label('rn')
        ).filter(
            Payments.Stripe_Customer_ID.in_([row[0] for row in duplicate_customer_ids])
        ).subquery()

        # Get only the first record (rn = 1) for each customer_id
        first_duplicates = db.session.query(Payments).join(
            ranked_duplicates, Payments.id == ranked_duplicates.c.id
        ).filter(ranked_duplicates.c.rn == 1).order_by(Payments.Stripe_Customer_ID).all()

        i = 0
        has_charge = 0
        for a in first_duplicates:
            i += 1
            #print(a.id, a.Splynx_ID)
            print(a.Stripe_Charge_ID)
            if a.Stripe_Charge_ID != None:
                has_charge += 1

        print(i, has_charge)