从两个不同的表中获取两列的总和

Question

I have two MySQL tables, with the following structure (I have removed irrelevant columns).

mysql> DESCRIBE `edinners_details`;
+------------------+------------------+------+-----+---------+----------------+
| Field            | Type             | Null | Key | Default | Extra          |
+------------------+------------------+------+-----+---------+----------------+
| details_id       | int(11) unsigned | NO   | PRI | NULL    | auto_increment |
| details_pupil_id | int(11) unsigned | NO   |     | NULL    |                |
| details_cost     | double unsigned  | NO   |     | NULL    |                |
+------------------+------------------+------+-----+---------+----------------+

mysql> DESCRIBE `edinners_payments`;
+------------------+------------------+------+-----+---------+----------------+
| Field            | Type             | Null | Key | Default | Extra          |
+------------------+------------------+------+-----+---------+----------------+
| payment_id       | int(11) unsigned | NO   | PRI | NULL    | auto_increment |
| payment_pupil_id | int(11) unsigned | NO   |     | NULL    |                |
| payment_amount   | float unsigned   | NO   |     | NULL    |                |
+------------------+------------------+------+-----+---------+----------------+

The way the system works is that you order a meal and each meal has a cost, each of these orders are stored in edinners_details . An example row would be as follows:

mysql> SELECT * FROM `edinners_details` LIMIT 1;
+------------+------------------+--------------+
| details_id | details_pupil_id | details_cost |
+------------+------------------+--------------+
|          1 |            18343 |           25 |
+------------+------------------+--------------+

Typically people will pay for these meals in bulk - if they have £40 worth of meals over the course of 20 days, they'll pay that off at the end of the month. Every time they pay, a new row goes into the edinners_payments table, an example row for this would be as follows:

mysql> SELECT * FROM `edinners_payments` LIMIT 1;
+------------+------------------+----------------+
| payment_id | payment_pupil_id | payment_amount |
+------------+------------------+----------------+
|          1 |            18343 |             20 |
+------------+------------------+----------------+

So from these two rows we can see that this person is current £5 in debt - they've had a £25 meal and only paid £20. Over time there will be many rows for each of the users of the system, I can easily work out how much worth of food they've had by doing a simple query such as

SELECT SUM(`details_cost`) AS `meal_total` 
FROM `edinners_details` 
WHERE `details_pupil_id` = '18343';

Then to get the amount of money they have paid, I simply do this query:

SELECT SUM(`payment_amount`) AS `payment_total` 
FROM `edinners_payments` 
WHERE `payment_pupil_id` = '18343';

My ultimate goal is to be able to see who owes the most money, but to loop every user of my users table and run these two queries for them, I believe it would be quite slow, so ideally what I would like to do is combine the two above queries into one, and perhaps an additional column which would be ( meal_total - payment_total ) which would give me the amount owed. I've tried a few methods to make this work, including joins and sub-queries but they all seem to repeat every relevant line from the edinners_details for each of the edinners_payments line - so if there were 3 details and 4 payments, you would have 12 rows pulled out which means doing a SUM() on the columns gives me a value which far exceeds what it should be. A good way to demonstrate this would be to run this query:

SELECT * FROM (
    SELECT `details_cost` AS `cost` 
    FROM `edinners_details` 
    WHERE `details_pupil_id` = '18343'
    GROUP BY `details_id`
) AS `details`, (
    SELECT `payment_amount` AS `amount` 
    FROM `edinners_payments` 
    WHERE `payment_pupil_id` = '18343'
    GROUP BY `payment_id`
) AS `payment`;

which gives me the following result:

+------+--------+
| cost | amount |
+------+--------+
|  2.5 |     20 |
|  2.5 |      6 |
|  2.5 |      3 |
|  2.5 |   1200 |
|  2.5 |     20 |
|  2.5 |      6 |
|  2.5 |      3 |
|  2.5 |   1200 |
|  2.5 |     20 |
|  2.5 |      6 |
|  2.5 |      3 |
|  2.5 |   1200 |
|  2.5 |     20 |
|  2.5 |      6 |
|  2.5 |      3 |
|  2.5 |   1200 |
|  2.5 |     20 |
|  2.5 |      6 |
|  2.5 |      3 |
|  2.5 |   1200 |
+------+--------+

Adding the SUM into this, as such:

SELECT SUM(`details`.`cost`) AS `details_cost`, SUM(`payment`.`amount`) AS `payment_total` FROM (
    SELECT `details_cost` AS `cost` 
    FROM `edinners_details` 
    WHERE `details_pupil_id` = '18343'
    GROUP BY `details_id`
) AS `details`, (
    SELECT `payment_amount` AS `amount` 
    FROM `edinners_payments` 
    WHERE `payment_pupil_id` = '18343'
    GROUP BY `payment_id`
) AS `payment`;

Gives me the following result:

+--------------+---------------+
| details_cost | payment_total |
+--------------+---------------+
|           50 |          6145 |
+--------------+---------------+

If this was working, details_cost would be 12.5 and payment_total would be 1229, but that is not the case. You can clearly see the repetition in the above result, I apologise that all the costs are 2.5, it makes it a little less obvious, but they are 5 seperate meal orders with 4 payments having been made. Does anyone know how I would go about getting the SUM() of the meal order costs and the SUM() of the payments at the same time?

Thanks

Answer 1

I have only PostgreSQL available at hand and this is what I came up with:

SELECT coalesce(costs.pupil_id, amounts.pupil_id) as pupil_id,
       coalesce(amount_sum, 0) as amount_sum,
       coalesce(cost_sum, 0) as cost_sum,
       coalesce(amount_sum, 0) - coalesce(cost_sum, 0) as debit
FROM (
       SELECT details_pupil_id AS pupil_id,
              sum(details_cost) AS cost_sum
       FROM edinners_details
       GROUP BY details_pupil_id
     ) costs
     FULL OUTER JOIN 
     (
       SELECT payment_pupil_id AS pupil_id,
              sum(payment_amount) AS amount_sum
       FROM edinners_payments
       GROUP BY payment_pupil_id
     ) amounts ON costs.pupil_id = amounts.pupil_id;

It's grouping records in each table by pupil_id to properly count the sum and then joins them to get the difference. There is full outer join to handle cases when someone doesn't have any payments (but has dinners) and doesn't have any dinners (but has payments).

From what I've read MySQL doesn't support FULL OUTER JOIN (bump...) so you must emulate it with UNION:

SELECT coalesce(costs.pupil_id, amounts.pupil_id) AS pupil_id,
       coalesce(amount_sum, 0) as amount_sum,
       coalesce(cost_sum, 0) as cost_sum,
       coalesce(amount_sum, 0) - coalesce(cost_sum, 0) AS debit
FROM (
       SELECT details_pupil_id AS pupil_id,
              sum(details_cost) AS cost_sum
       FROM edinners_details
       GROUP BY details_pupil_id
     ) costs
     LEFT OUTER JOIN 
     (
       SELECT payment_pupil_id AS pupil_id,
              sum(payment_amount) AS amount_sum
       FROM edinners_payments
       GROUP BY payment_pupil_id
     ) amounts ON costs.pupil_id = amounts.pupil_id
UNION
SELECT coalesce(costs.pupil_id, amounts.pupil_id) AS pupil_id,
       coalesce(amount_sum, 0) as amount_sum,
       coalesce(cost_sum, 0) as cost_sum,
       coalesce(amount_sum, 0) - coalesce(cost_sum, 0) AS debit
FROM (
       SELECT details_pupil_id AS pupil_id,
              sum(details_cost) AS cost_sum
       FROM edinners_details
       GROUP BY details_pupil_id
     ) costs
     RIGHT OUTER JOIN 
     (
       SELECT payment_pupil_id AS pupil_id,
              sum(payment_amount) AS amount_sum
       FROM edinners_payments
       GROUP BY payment_pupil_id
     ) amounts ON costs.pupil_id = amounts.pupil_id;

Answer 2

Currently, your query is performing a CROSS JOIN , which joins every row from the first table to every row in the second, and so is returning a lot of redundant results. Both tables have a pupil_id , however, so we can use this to join the correct records from each table.

SELECT
  d.detail_pupil_id AS pupil_id,
  SUM(d.details_cost) AS cost,
  SUM(p.payment_amount) AS amount
FROM `edinners_details` d
INNER JOIN `edinners_payments` p ON d.detail_pupil_id = p.payment_pupil_id
GROUP BY pupil_id;

You could take this further by performing a join to your users table and returning all the data you need in a single query.

SELECT
  users.id,
  users.name,
  payment.cost,
  payment.amount
FROM `users`
INNER JOIN (
  SELECT
    d.detail_pupil_id AS pupil_id,
    SUM(d.details_cost) AS cost,
    SUM(p.payment_amount) AS amount
  FROM `edinners_details` d
  INNER JOIN `edinners_payments` p ON d.detail_pupil_id = p.payment_pupil_id
  GROUP BY pupil_id
) payment ON payment.pupil_id = users.id
ORDER BY users.id ASC;

Answer 3

The following works for me, although it looks ugly. In a MySQL DB:

SELECT
    t1.p_id, t1.cost, t2.amount
FROM
    (SELECT
        details_pupil_id AS p_id, SUM(details_cost) AS cost
     FROM
        edinners_details
     GROUP BY
        details_pupil_id) t1,
    (SELECT
        payment_pupil_id AS p_id, SUM(payment_amount) AS amount
     FROM
        edinners_payments
     GROUP BY
        payments_pupil_id) t2
WHERE
    t1.p_id = t2.p_id

/* Getting pupils with dinners but no payment */
UNION
    SELECT
        details_pupil_id, SUM(details_cost) cost, 0
    FROM
        edinners_details
    WHERE
        details_pupil_id NOT IN (SELECT DISTINCT payment_pupil_id FROM edinners_payments)
    GROUP BY
        details_pupil_id

/* Getting pupils with payment but no dinners */
UNION
    SELECT
        payment_pupil_id, 0, SUM(payment_amount)
    FROM
        edinners_payments
    WHERE
        payment_pupil_id NOT IN (SELECT DISTINCT details_pupil_id FROM edinners_details)
    GROUP BY
        payment_pupil_id

Answer 4

You could try something like the following which should return you the amount of debit and id of the pupil with the greatest debt (if a pupil has overpaid his debt will be negative):

select t.id, max(t.debit) from (select details_pupil_id as id, (sum(details_cost) - (select sum(payment_amount) from edinners_payments where payment_pupil_id = details_pupil_id)) as debit from edinners_details group by details_pupil_id) as t;

so if you have the following situation:

mysql> select * from edinners_details;
+------------+------------------+--------------+
| details_id | details_pupil_id | details_cost |
+------------+------------------+--------------+
|          1 |            18343 |           25 |
|          2 |            18344 |           17 |
|          3 |            18343 |           11 |
|          4 |            18344 |            2 |
|          5 |            18344 |            7 |
|          6 |            18343 |           12 |
|          7 |            18343 |           12 |
|          8 |            18343 |           35 |
|          9 |            18344 |           30 |
+------------+------------------+--------------+

mysql> select * from edinners_payments;
+------------+------------------+----------------+
| payment_id | payment_pupil_id | payment_amount |
+------------+------------------+----------------+
|          1 |            18343 |             20 |
|          2 |            18344 |             25 |
|          3 |            18343 |             12 |
|          4 |            18344 |             25 |
|          5 |            18343 |             22 |
|          6 |            18344 |             11 |
|          7 |            18343 |              8 |
|          8 |            18344 |              2 |
+------------+------------------+----------------+

when running the above query you should get:

+-------+--------------+
| id    | max(t.debit) |
+-------+--------------+
| 18343 |           33 |
+-------+--------------+

if you like a list of debits for each pupil you can run:

select details_pupil_id as id, (sum(details_cost) - (select sum(payment_amount) from edinners_payments where payment_pupil_id = details_pupil_id)) as debit from edinners_details group by details_pupil_id;

which should get you this result:

+-------+-------+
| id    | debit |
+-------+-------+
| 18343 |    33 |
| 18344 |    -7 |
+-------+-------+

I hope this helps.