Calculate a running total in MySQL

Perhaps a simpler solution for you and prevents the database having to do a ton of queries. This executes just one query then does a little math on the results in a single pass.

SET @runtot:=0;
SELECT
   q1.d,
   q1.c,
   (@runtot := @runtot + q1.c) AS rt
FROM
   (SELECT
       DAYOFYEAR(`date`) AS d,
       COUNT(*) AS c
    FROM  `orders`
    WHERE  `hasPaid` > 0
    GROUP  BY d
    ORDER  BY d) AS q1

This will give you an additional RT (running total) column. Don't miss the SET statement at the top to initialize the running total variable first or you will just get a column of NULL values.

SELECT 
   DAYOFYEAR(O.`date`)  AS d, 
   COUNT(*),
   (select count(*) from `orders` 
       where  DAYOFYEAR(`date`) <= d and   `hasPaid` > 0)
FROM  
  `orders` as O
WHERE  
  O.`hasPaid` > 0
GROUP  BY d
ORDER  BY d

This will require some syntactical tuning (I don't have MySQL to test it), but it shows you the idea. THe subquery just has to go back and add up everything fresh that you already included in the outer query, and it has to do that for every row.

Take a look at this question for how to use joins to accomplish the same.

To address concerns about performance degradation with growing data: Since there are max. 366 days in a year, and I assume that you are not running this query against multiple years, the subquery will get evaluated up to 366 times. With proper indices on the date and the hasPaid flag, you'll be ok.

Starting with MySQL 8, you will be using window functions for this kind of query:

SELECT dayofyear(`date`) AS d, count(*), sum(count(*)) OVER (ORDER BY dayofyear(`date`))
FROM `orders`
WHERE `hasPaid` > 0
GROUP BY d
ORDER BY d

In the above query, the aggregate function count(*) is nested inside of the window function sum(..) OVER (..), which is possible because of the logical order of operations in SQL. If that's too confusing, you can easily resort to using a derived table or a WITH clause to better structure your query:

WITH daily (d, c) AS (
  SELECT dayofyear(`date`) AS d, count(*)
  FROM `orders`
  WHERE `hasPaid` > 0
  GROUP BY d
)
SELECT d, c, sum(c) OVER (ORDER BY d)
ORDER BY d