Given 2 tables, I want to generate top 3 highest amount from [Purchase] table.
Additional criteria is [Crocs] must be included in top 3 of the records.
I have following SQL, but it cannot generates the result as I wanted (Result A), please guide me on how to pull out the result in Result B. Thank you.
Table (Purchase):
Purchase_ID | StoreID | Amount
------------|---------|--------
1 | 21 | 22
2 | 23 | 13
3 | 25 | 6
4 | 26 | 23
5 | 28 | 18
Table (Store):
Store_ID | StoreName
---------|----------
21 | Adidas
22 | Nike
23 | Puma
24 | New Balance
25 | Crocs
26 | Converse
SQL:
SELECT IF(SUM(amount) IS NULL, 0, SUM(amount)) as totalAmount
FROM (
SELECT a.amount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
GROUP BY a.amount
HAVING b.StoreName = 'Crocs'
ORDER BY a.amount DESC
LIMIT 3
) t
Result A: $6
Explanation A: Amount of Crocs is $6
Result B: $51
Explanation B: Total Amount of top 3 = $22 (Adidas) + 23 (Puma) + $6 (Crocs)
The answer from scaisEdge is almost right, but the first query could also return a row with crocs and the sorting is wrong (order by max(a.amount) limit 2 means that the lowest 2 results will be shown). Additionally you could wrap the query in another select query to sort the results
SELECT * FROM (
SELECT b.storename, max(a.amount) as maxAmount
FROM purchase a
INNER JOIN store b ON a.store_id = b.storeid
WHERE b.storename != 'crocks'
GROUP BY a.storename
ORDER BY max(a.amount) DESC
LIMIT 2
UNION
SELECT b.storename, a.amount as maxAmount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
WHERE b.storename='crocks'
ORDER BY a.amount DESC
LIMIT 1
) ORDER BY maxAmount DESC
You could use an union
SELECT b.storename, max(a.amount)
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
GROUP BY a.storename
order by max(a.amount) limit 2
union
SELECT b.storename, a.amount
FROM purchase a
INNER JOIN store b
ON a.store_id = b.storeid
where b.storename='crocks'
try this one:
SELECT sum(amount)as sum_amount,a.store_id,storename,category from
(select amount,store_id from tbl_purchase) as a
inner JOIN
(select store_id,storename,category from tbl_store)as b on a.store_id = b.store_id where b.category = 'supermarket' GROUP BY category
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
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
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.
This is maybe not the simplest solution, but it does do the job and is an interesting variation and in my opinion transparent. I simulate the analytical functions that I know from Oracle.
As we do not assume the id to be consecutive the counting of the rows is simulated by increasing #rn each row. Next products table including the rownum is joint with itself and only the rows 2-6 are used to build the average.
select p2id, avg(deposit), group_concat(p1id order by p1id desc), group_concat(deposit order by p1id desc)
from ( select p2.id p2id, p1.rn p1rn, p1.deposit, p2.rn p2rn, p1.id p1id
from (select p.*,#rn1:=#rn1+1 as rn from products p,(select #rn1 := 0) r) p1
, (select p.*,#rn2:=#rn2+1 as rn from products p,(select #rn2 := 0) r) p2 ) r
where p2rn-p1rn between 1 and 5
group by p2id
order by p2id desc
;
Result:
+------+--------------+---------------------------------------+------------------------------------------+
| p2id | avg(deposit) | group_concat(p1id order by p1id desc) | group_concat(deposit order by p1id desc) |
+------+--------------+---------------------------------------+------------------------------------------+
| 10 | 86.0000 | 9,8,7,6,5 | 90,90,75,75,100 |
| 9 | 84.0000 | 8,7,6,5,4 | 90,75,75,100,80 |
| 8 | 84.0000 | 7,6,5,4,3 | 75,75,100,80,90 |
| 7 | 84.0000 | 6,5,4,3,2 | 75,100,80,90,75 |
| 6 | 79.0000 | 5,4,3,2,1 | 100,80,90,75,50 |
| 5 | 73.7500 | 4,3,2,1 | 80,90,75,50 |
| 4 | 71.6667 | 3,2,1 | 90,75,50 |
| 3 | 62.5000 | 2,1 | 75,50 |
| 2 | 50.0000 | 1 | 50 |
+------+--------------+---------------------------------------+------------------------------------------+
SQL Fiddle Demo: http://sqlfiddle.com/#!2/c13bc/129
I want to thank this answer on how to simulate analytical functions in mysql: MySQL get row position in ORDER BY
It looks like you just want:
SELECT
id,
(SELECT AVG(deposit)
FROM (
SELECT deposit
FROM products
ORDER BY id DESC
LIMIT 5) last5
) avgdeposit
FROM products
The inner query gets the last 5 rows added to product, the query that wraps that gets the average for their deposits.
I'm going to simplify your query a bit so I can explain it.
SELECT
y.id,
(
SELECT AVG(deposit) FROM
(
SELECT deposit
FROM products
LIMIT 5
) z
) AVGDEPOSIT
FROM
(
SELECT *
FROM
(
SELECT *
FROM products
) x
LIMIT 15
) y;
My guess would be that you just need to insert some AS keywords in there. I'm sure someone else will come up with something more elegant, but for now you can try it out.
SELECT
y.id,
(
SELECT AVG(deposit) FROM
(
SELECT deposit
FROM products
LIMIT 5
) z
) AS AVGDEPOSIT
FROM
(
SELECT *
FROM
(
SELECT *
FROM products
) AS x
LIMIT 15
) y;
Here's one way to do it in MySQL:
SELECT p.id
, ( SELECT AVG(deposit)
FROM ( SELECT #rownum:=#rownum+1 rn, deposit, id
FROM ( SELECT #rownum:=0 ) r
, products
ORDER BY id ) t
WHERE rn BETWEEN p.rn-5 AND p.rn-1 ) avgdeposit
FROM ( SELECT #rownum1:=#rownum1+1 rn, id
FROM ( SELECT #rownum1:=0 ) r
, products
ORDER BY id ) p
WHERE p.rn >= 5
ORDER BY p.rn DESC;
It's a shame MySQL doesn't support the WITH clause or windowing functions. Having both would greatly simplify the query to the following:
WITH tbl AS (
SELECT id, deposit, ROW_NUMBER() OVER(ORDER BY id) rn
FROM products
)
SELECT id
, ( SELECT AVG(deposit)
FROM tbl
WHERE rn BETWEEN t.rn-5 AND t.rn-1 )
FROM tbl t
WHERE rn >= 5
ORDER BY rn DESC;
The latter query runs fine in Postgres.
2 possible solutions here
Firstly using user variables to add a sequence number. Do this twice, and join the second set to the first where the sequence number is between the id - 1 and the id - 5. Then just use AVG. No correlated sub queries.
SELECT Sub3.id, Sub3.itemid, Sub3.deposit, AVG(Sub4.deposit)
FROM
(
SELECT Sub1.id, Sub1.itemid, Sub1.deposit, #Seq:=#Seq+1 AS Sequence
FROM
(
SELECT id, itemid, deposit
FROM products
ORDER BY id DESC
) Sub1
CROSS JOIN
(
SELECT #Seq:=0
) Sub2
) Sub3
LEFT OUTER JOIN
(
SELECT Sub1.id, Sub1.itemid, Sub1.deposit, #Seq1:=#Seq1+1 AS Sequence
FROM
(
SELECT id, itemid, deposit
FROM products
ORDER BY id DESC
) Sub1
CROSS JOIN
(
SELECT #Seq1:=0
) Sub2
) Sub4
ON Sub4.Sequence BETWEEN Sub3.Sequence + 1 AND Sub3.Sequence + 5
GROUP BY Sub3.id, Sub3.itemid, Sub3.deposit
ORDER BY Sub3.id DESC
Second one is cruder, and uses a correlated sub query (which is likely to perform poorly as the amount of data increases). Does a normal select but for the last column it has a sub query that refers to the id in the main select.
SELECT id, itemid, deposit, (SELECT AVG(P2.deposit) FROM products P2 WHERE P2.id BETWEEN P1.id - 5 AND p1.id - 1 ORDER BY id DESC LIMIT 5)
FROM products P1
ORDER BY id DESC
Is this what you are after?
SELECT m.id
, AVG(d.deposit)
FROM products m
, products d
WHERE d.id < m.id
AND d.id >= m.id - 5
GROUP BY m.id
ORDER BY m.id DESC
;
But can't be that simple. Firstly, the table cannot just contain one itemid (hence your WHERE clause); Second, the id cannot be sequential/without gaps within an itemid. Thirdly, you probably want to produce something that runs across itemid and not one itemid at a time. So here it is.
SELECT itemid
, m_id as id
, AVG(d.deposit) as deposit
FROM (
SELECT itemid
, m_id
, d_id
, d.deposit
, #seq := (CASE WHEN m_id = d_id THEN 0 ELSE #seq + 1 END) seq
FROM (
SELECT m.itemid
, m.id m_id
, d.id d_id
, d.deposit
FROM products m
, products d
WHERE m.itemid = d.itemid
AND d.id <= m.id
ORDER BY m.id DESC
, d.id DESC) d
, (SELECT #seq := 0) s
) d
WHERE seq BETWEEN 1 AND 5
GROUP BY itemid
, m_id
ORDER BY itemid
, m_id DESC
;
I've got a budget table:
user_id product_id budget created
-----------------------------------------------------------------
1 1 300 2011-12-01
2 1 400 2011-12-01
1 1 500 2011-12-03
2 2 400 2011-12-04
I've also got a manager_user table, joining a manager with the user
user_id manager_id product_id
------------------------------------
1 5 1
1 9 2
2 5 1
2 5 2
3 5 1
What I'd like to do is grab each of the user that's assigned to Manager #5, and also get their 'budgets'... but only the most recent one.
Right now my statement looks like this:
SELECT * FROM manager_user mu
LEFT JOIN budget b
ON b.user_id = mu.user_id AND b.product_id = mu.product_id
WHERE mu.manager_id = 5
GROUP BY mu.user_id, mu.product_id
ORDER BY b.created DESC;
The problem is it doesn't pull the most recent budget. Any suggestions? Thanks!
To accomplish your task you can do as follows:
select b1.user_id,
b1.budget
from budget b1 inner join (
select b.user_id,
b.product_id,
max(created) lastdate
from budget b
group by b.user_id, b.product_id ) q
on b1.user_id=q.user_id and
b1.product_id=q.product_id and
b1.created=q.lastdate
where b1.user_id in
(select user_id from manager_user where manager_id = 5);
I'm assuming here that your (user_id, product_id, created) combination is unique.
For what it's worth, here's the code that returned what I was looking for:
SELECT DISTINCT(b1.id),mu.user_id,mu.product_id,b1.budget,b1.created
FROM budget b1
INNER JOIN (
SELECT b.user_id, b.product_id, MAX(created) lastdate
FROM budget b
GROUP BY b.user_id, b.product_id) q
ON b1.user_id=q.user_id AND
b1.product_id=q.product_id AND
b1.created=q.lastdate
RIGHT JOIN manager_user mu
ON mu.user_id = b1.user_id AND
mu.product_id = b1.product_id
WHERE mu.manager_id = 5;
Thanks for the help Andrea!
I have a table call production
factory_id | factory_name | product_id
1 | A | 1
1 | A | 2
1 | A | 3
2 | B | 3
3 | C | 1
3 | C | 2
3 | C | 3
3 | C | 4
3 | C | 5
I'm trying to develop a query that will return two factory name pair such that every product of factory1 is produced by factory2, result looked like:
factory_name_1 | factory_name_2
A | C
B | A
B | C
I have some nested self join and renames, but I can't wrap my head around how I can apply EXISTS or IN for this scenario that does "for each product produced by factory X do condition". Thanks to any help in advanced.
Update:
Sorry that I forgot to paste my query:
select t0.fname0, t1.fname1
from (
select factory_id as fid0, factory_name as fname0, product_id as pid0, count(distinct factory_id, product_id) as pnum0
from production
group by factory_id
) t0
join
(
select factory_id as fid1, factory_name as fname1, product_id as pid1, count(distinct factory_id, product_id) as pnum1
from production
group by factory_id
) t1
where t0.fid0 <> t1.fid1
and t0.pnum0 < t1.pnum1
and t0.pid0 = t1.pid1;
Update 2: production is the only table. Expected output factory1 and factory2 are just the rename of factory_name attribute.
You need to JOIN the table for each factory pairing to make sure they "join" on the same product_ids, otherwise you might end up with similar counts for DISTINCT product_ids but these will not necessarily refer to the same product_ids.
This is my take on it:
SELECT bfna,afna, pcnt FROM (
SELECT a.factory_name afna, b.factory_name bfna, COUNT(DISTINCT b.product_id) commoncnt
FROM tbl a LEFT JOIN tbl b ON b.factory_name!=a.factory_name AND b.product_id=a.product_id
GROUP BY a.factory_name, b.factory_name
) c
INNER JOIN (
SELECT factory_name fna, COUNT(DISTINCT product_id) pcnt
FROM TBL GROUP BY factory_name
) d ON fna=bfna AND commoncnt=pcnt
ORDER BY bfna,afna
You can find a demo here: https://rextester.com/JJGCK84904
It produces:
bfna afna commoncnt
A C 3
B A 1
B C 1
For simplicity I left out the column factory_id as it does not add any information here.
Fun fact: as I am using only "bare-bone" SQL expressions, the above code will run on SQL-Server too without any changes.
You can do it this way:
select A as factory_name_1 , B as factory_name_2
from
(
select A, B, count(*) as Count_
from
(
select a.factory_name as A, b.factory_name as B
from yourtable a
inner join yourtable b
on a.product_id = b.product_id and a.factory_id <> b.factory_id
)a group by A, B
)a
inner join
(select factory_name, count(*) as Count_ from yourtable group by factory_name) b
on a.A = b.factory_name and a.Count_ = b.Count_
Order by 1
Output:
factory_name_1 factory_name_2
A C
B A
B C
The other solutions just seem more complicated than necessary. This is basically a self-join with aggregation:
with t as (
select t.*, count(*) over (partition by factory_id) as cnt
from tbl t
)
select t1.factory_id, t2.factory_id, t1.factory_name, t2.factory_name, count(*)
from t t1 join
t t2
on t1.product_id = t2.product_id and t1.factory_id <> t2.factory_id
group by t1.factory_id, t2.factory_id, t1.factory_name, t2.factory_name, t1.cnt
having count(*) = max(t1.cnt);
Here is a db<>fiddle.