Unknown column in mysql subquery

Question

I am trying to get the avg of an item so I am using a subquery.

Update: I should have been clearer initially, but i want the avg to be for the last 5 items only

First I started with

SELECT 
y.id
FROM (
    SELECT *
        FROM (
                SELECT *
                FROM products
                WHERE itemid=1
        ) x  
    ORDER BY id DESC
    LIMIT 15 
) y;

Which runs but is fairly useless as it just shows me the ids.

I then added in the below

SELECT
y.id,
(SELECT AVG(deposit) FROM (SELECT deposit FROM products WHERE id < y.id ORDER BY id DESC LIMIT 5)z) AVGDEPOSIT
FROM (
    SELECT *
        FROM (
                SELECT *
                FROM products
                WHERE itemid=1
        ) x  
    ORDER BY id DESC
    LIMIT 15 
) y;

When I do this I get the error Unknown column 'y.id' in 'where clause', upon further reading here I believe this is because when the queries go down to the next level they need to be joined?

So I tried the below ** removed un needed suquery

SELECT
y.id,
(SELECT AVG(deposit) FROM (
    SELECT deposit 
    FROM products
    INNER JOIN y as yy ON products.id = yy.id       
    WHERE id < yy.id 
    ORDER BY id DESC 
    LIMIT 5)z
    ) AVGDEPOSIT
FROM (
    SELECT *
    FROM products
    WHERE itemid=1
    ORDER BY id DESC
    LIMIT 15 
) y;

But I get Table 'test.y' doesn't exist. Am I on the right track here? What do I need to change to get what I am after here?

The example can be found here in sqlfiddle.

CREATE TABLE products
    (`id` int, `itemid` int, `deposit` int);

    INSERT INTO products
    (`id`, `itemid`, `deposit`)
VALUES
(1, 1, 50),
(2, 1, 75),
(3, 1, 90),
(4, 1, 80),
(5, 1, 100),
(6, 1, 75),
(7, 1, 75),
(8, 1, 90),
(9, 1, 90),
(10, 1, 100);

Given my data in this example, my expected result is below, where there is a column next to each ID that has the avg of the previous 5 deposits.

id | AVGDEPOSIT
10 | 86 (deposit value of (id9+id8+id7+id6+id5)/5) to get the AVG
 9 | 84
 8 | 84
 7 | 84
 6 | 79
 5 | 73.75

Answer 1

I'm not an MySQL expert (in MS SQL it could be done easier), and your question looks a bit unclear for me, but it looks like you're trying to get average of previous 5 items.

If you have Id without gaps, it's easy:

select
    p.id,
    (
        select avg(t.deposit)
        from products as t
        where t.itemid = 1 and t.id >= p.id - 5 and t.id < p.id
    ) as avgdeposit
from products as p
where p.itemid = 1
order by p.id desc
limit 15

If not, then I've tri tried to do this query like this

select
    p.id,
    (
        select avg(t.deposit)
        from (
            select tt.deposit
            from products as tt
            where tt.itemid = 1 and tt.id < p.id
            order by tt.id desc
            limit 5
        ) as t
    ) as avgdeposit
from products as p
where p.itemid = 1
order by p.id desc
limit 15

But I've got exception Unknown column 'p.id' in 'where clause'. Looks like MySQL cannot handle 2 levels of nesting of subqueries. But you can get 5 previous items with offset, like this:

select
    p.id,
    (
        select avg(t.deposit)
        from products as t
        where t.itemid = 1 and t.id > coalesce(p.prev_id, -1) and t.id < p.id
    ) as avgdeposit
from 
(
    select
        p.id,
        (
            select tt.id
            from products as tt
            where tt.itemid = 1 and tt.id <= p.id
            order by tt.id desc
            limit 1 offset 6
        ) as prev_id
    from products as p
    where p.itemid = 1
    order by p.id desc
    limit 15
) as p

sql fiddle demo

Answer 2

This is my solution. It is easy to understand how it works, but at the same time it can't be optimized much since I'm using some string functions, and it's far from standard SQL. If you only need to return a few records, it could be still fine.

This query will return, for every ID, a comma separated list of previous ID, ordered in ascending order:

SELECT p1.id, p1.itemid, GROUP_CONCAT(p2.id ORDER BY p2.id DESC) previous_ids
FROM
  products p1 LEFT JOIN products p2
  ON p1.itemid=p2.itemid AND p1.id>p2.id
GROUP BY
  p1.id, p1.itemid
ORDER BY
  p1.itemid ASC, p1.id DESC

and it will return something like this:

| ID | ITEMID |      PREVIOUS_IDS |
|----|--------|-------------------|
| 10 |      1 | 9,8,7,6,5,4,3,2,1 |
|  9 |      1 |   8,7,6,5,4,3,2,1 |
|  8 |      1 |     7,6,5,4,3,2,1 |
|  7 |      1 |       6,5,4,3,2,1 |
|  6 |      1 |         5,4,3,2,1 |
|  5 |      1 |           4,3,2,1 |
|  4 |      1 |             3,2,1 |
|  3 |      1 |               2,1 |
|  2 |      1 |                 1 |
|  1 |      1 |            (null) |

then we can join the result of this query with the products table itself, and on the join condition we can use FIND_IN_SET(src, csvalues) that return the position of the src string inside the comma separated values:

ON FIND_IN_SET(id, previous_ids) BETWEEN 1 AND 5

and the final query looks like this:

SELECT
  list_previous.id,
  AVG(products.deposit)
FROM (
  SELECT p1.id, p1.itemid, GROUP_CONCAT(p2.id ORDER BY p2.id DESC) previous_ids
  FROM
    products p1 INNER JOIN products p2
    ON p1.itemid=p2.itemid AND p1.id>p2.id
  GROUP BY
    p1.id, p1.itemid
  ) list_previous LEFT JOIN products
  ON list_previous.itemid=products.itemid
     AND FIND_IN_SET(products.id, previous_ids) BETWEEN 1 AND 5
GROUP BY
  list_previous.id
ORDER BY
  id DESC

Please see fiddle here. I won't recommend using this trick for big tables, but for small sets of data it is fine.