如何在 mysql 中优化我的排名查询？

Question

Multiple records are being saved for each date against each subscription id. There is coins column, All subscriber gets few coins. It is basically a rewarding system. I want to get the rank of current subscriber. There are 2 tables that are being used.

1. user_rewards (which saves all the records with coins)
2. coins_history (if coins are used, then i add the used coins in coins history table)

For example if user has earned total of 1000 coins, it remains 1000 in user reward table, and if he has used 200 coins somewhere, i add an entry in coins history table. So currently he has 800 coins left. I will have to find rank with current coins.

For example, I have 5 records exist in my user reward against 3 subscriber ids. I want to get the rank of one subscriber against his subscription_id.

+-----------------+--------------+-------+
| subscription_id | date_awarded | coins |
+-----------------+--------------+-------+
| 525252 | 2020-09-11 | 10 |
| 454545 | 2020-09-11 | 20 |
| 989898 | 2020-09-11 | 10 |
| 525252 | 2020-09-10 | 50 |
| 454545 | 2020-09-10 | 30 |
+-----------------+--------------+-------+

The subscription ids 525252 has 60 coins, 989898 has 10 coins and 454545 has 50 coins. But subscription id 525252 has used 40 of his coins, 40 coins are added in coins history table. So 454545 ranks first with 50 coins, 525252 ranks 2nd with 20 coins and 989898 ranks 3rd with 10 coins.

So for getting rank of user. I get sum of coins from user reward table, then minus it with coins history table against that subscription id. Then find the rank against his current unused coins. Below is my query, Query took 34.0794 sec.

SELECT COUNT(*)+1 AS aboveRank 
FROM 
(
   SELECT subscription_id  
   FROM user_rewards 
   GROUP BY subscription_id 
   HAVING SUM(coins) > 
    (Select (SELECT IFNULL(sum(coins), 0)  
             FROM user_rewards 
             WHERE subscription_id = 525252) - 
            (SELECT IFNULL(sum(coins), 0) as coins 
             FROM `coins_history` 
             where subscription_id = 525252) as z
    ) 
) t

Note: I have around 800000 records in one month, it will keep on increasing daily, So need to optimize my current query, I am using MySQL and PHP.

Answer 1

You can calculate the total coins per subscriber using union all and group by . Then use window functions for the ranking:

select c.*
from (select subscription_id, sum(coins) as coins,
             rank() over (order by coins desc) as seqnum
      from ((select subscription_id, - sum(coins) as coins
             from user_rewards ur
             group by subscription_id
            ) union all
            (select subscription_id, sum(coins)
             from coins_history ch
             group by subscription_id
            ) 
           ) c
       group by subscription_id
      ) c
where subscription_id = ?;