So I have these 2 tables:
jobs:
-------------------------------------------------
id business_id other_columns
-------------------------------------------------
1 223 xxxxxx
-------------------------------------------------
1 12 xxxxxx
-------------------------------------------------
businesses_ratings:
--------------------------------------------------------------------------------------
id business_id professional communication safety respectful dependability
--------------------------------------------------------------------------------------
1 223 4 2 5 4 3
--------------------------------------------------------------------------------------
2 223 3 5 2 4 5
--------------------------------------------------------------------------------------
3 223 1 2 5 4 4
--------------------------------------------------------------------------------------
I want to select the jobs of a particular business_id, and append to each job the overall rating of that business_id, computed as AVG((AVG(professional), AVG(communication), AVG(safety), AVG(respectful), AVG(dependability))
Can I achieve this in one query?
LE: I'll append here the query I've tried ( containing also the WHERE clause, maybe it'll help explaining better what I need to achieve. Also the thrown error:
SELECT * FROM jobs
CROSS JOIN (
SELECT count(*) totalJobs FROM jobs
WHERE (
( JSON_CONTAINS(skills, '{"id":1,"val":"Carpenter"}') )
AND NOT JSON_CONTAINS(workers, '{"id":6,"fullname":"Cip"}')
AND NOT JSON_CONTAINS(applicants, '{"id":6,"fullname":"Cip"}')
)
) ttl
CROSS JOIN (
SELECT AVG(
(SELECT AVG(professional) FROM businesses_ratings WHERE business_id=jobs.business_id) +
(SELECT AVG(communication) FROM businesses_ratings WHERE business_id=jobs.business_id) +
(SELECT AVG(safety) FROM businesses_ratings WHERE business_id=jobs.business_id) +
(SELECT AVG(respectful) FROM businesses_ratings WHERE business_id=jobs.business_id) +
(SELECT AVG(dependability) FROM businesses_ratings WHERE business_id=jobs.business_id)
) business_rating FROM businesses_ratings WHERE business_id=jobs.business_id
) avg
WHERE (
( JSON_CONTAINS(skills, '{"id":1,"val":"Carpenter"}') )
AND NOT JSON_CONTAINS(workers, '{"id":6,"fullname":"Cip"}')
AND NOT JSON_CONTAINS(applicants, '{"id":6,"fullname":"Cip"}')
)
ORDER BY start_date LIMIT 3
and the error:
Unknown column 'jobs.business_id' in 'where clause'
I think you want aggregation and window functions:
select
business_id,
rank() over(
order by avg(professional) + avg(communication) + avg(safety)
) as rn
from businesses_ratings
group by business_id
You can expand the order by clause of the window function with additional columns as needed.
I am quite skeptical about the interest of ranking by the average of the 3 averages - but the above query seems like a reasonable interpretation of what you ask for.
In earlier versions, one option uses user-defined variables to compute the rank:
select t.*, #rn := #rn + 1 rn
from (
select
business_id,
avg(professional) + avg(communication) + avg(safety) sum_avg
from businesses_ratings
group by business_id
order by sum_avg
) t
cross join (select #rn := 0) x
Your query seems way more complicated than necessary. I think you want:
select br.business_id,
avg(professional),
avg(communication),
avg(safety),
avg(respectful),
avg(dependability),
(avg(professional) + avg(communication) + avg(safety) + avg(respectful) + avg(dependability)) / 5 as overall_avg
from businesses_ratings br
group by br.business_id;
I've got a query returning the following:
ID | Price
---------------
1 | 20
1 | 30
1 | 15
2 | 10
2 | 12
2 | 20
3 | 1
3 | 0
3 | 0
4 | 0
4 | 0
4 | 7
I'm wondering if there's a way I can get the sum of the lowest value for each ID. So in this case it would return 25.
15+10+0+0
You can use a subquery selecting the min price for each id, then sum those values:
select sum(minprice) as overallprice
from (
select min(price) minprice
from yourtable
group by id) t
You can create a sub-query that finds the lowest price per id and take the results from that and sum them together. In pseudo-code:
select
sum(lowest_price)
from (select id, min(price) as lowest_price from prices group by id) lowest_prices
You can do a query like below
Select sum (a) from
(
Select min (price) as a from yourtable
Group by id
) t
Another approach using partition without using group by statement
select sum(price.min_price) from
(select distinct id,min(price) over(partition by id) as min_price from prices) price
Some other approaches would be to use MySQL user variables or a self left join..
MySQL user variable solution
Query
SELECT
SUM(prices.Price)
FROM (
SELECT
prices.Price
, CASE
WHEN #id != prices.id
THEN 1
ELSE 0
END AS isMinGroupValue
, (#id := prices.id)
FROM
prices
CROSS JOIN (
SELECT
#id := 0
) AS init_user_params
ORDER BY
prices.ID ASC
, prices.price ASC
) AS prices
WHERE
prices.isMinGroupValue = 1
see demo https://www.db-fiddle.com/f/nzWqMQAxd7mvq589R7WuZ8/0
Self left join solution
Query
SELECT
SUM(prices1.Price)
FROM
prices prices1
LEFT JOIN
prices prices2
ON
prices1.ID = prices2.ID
AND
prices1.price > prices2.price
WHERE
prices2.ID IS NULL
see demo https://www.db-fiddle.com/f/nzWqMQAxd7mvq589R7WuZ8/1
I would use correlation subquery :
select sum(t.price) as overallprice
from table t
where price = (select min(price) from table t1 where t1.id = t.id);
Let's say I have the following table:
user_id | category_id | points
-------------------------------
1 | 1 | 4
2 | 1 | 2
2 | 1 | 5
1 | 2 | 3
2 | 2 | 2
1 | 3 | 1
2 | 3 | 4
1 | 3 | 8
Could someone please help me to construct a query to return user's rank per category - something like this:
user_id | category_id | total_points | rank
-------------------------------------------
1 | 1 | 4 | 2
1 | 2 | 3 | 1
1 | 3 | 9 | 1
2 | 1 | 7 | 1
2 | 2 | 2 | 2
2 | 3 | 4 | 2
First, you need to get the total points per category. Then you need to enumerate them. In MySQL this is most easily done with variables:
SELECT user_id, category_id, points,
(#rn := if(#cat = category_id, #rn + 1,
if(#cat := category_id, 1, 1)
)
) as rank
FROM (SELECT u.user_id, u.category_id, SUM(u.points) as points
FROM users u
GROUP BY u.user_id, u.category_id
) g cross join
(SELEct #user := -1, #cat := -1, #rn := 0) vars
ORDER BY category_id, points desc;
You want to get the SUM of points for each unique category_id:
SELECT u.user_id, u.category_id, SUM(u.points)
FROM users AS u
GROUP BY uc.category_id
MySQL doesn't have analytic functions like other databases (Oracle, SQL Server) which would be very convenient for returning a result like this.
The first three columns are straightforward, just GROUP BY user_id, category_id and a SUM(points).
Getting the rank column returned is a bit more of a problem. Aside from doing that on the client, if you need to do that in the SQL statement, you could make use of MySQL user-defined variables.
SELECT #rank := IF(#prev_category = r.category_id, #rank+1, 1) AS rank
, #prev_category := r.category_id AS category_id
, r.user_id
, r.total_points
FROM (SELECT #prev_category := NULL, #rank := 1) i
CROSS
JOIN ( SELECT s.category_id, s.user_id, SUM(s.points) AS total_points
FROM users s
GROUP BY s.category_id, s.user_id
ORDER BY s.category_id, total_points DESC
) r
ORDER BY r.category_id, r.total_points DESC, r.user_id DESC
The purpose of the inline view aliased as i is to initialize user defined variables. The inline view aliased as r returns the total_points for each (user_id, category_id).
The "trick" is to compare the category_id value of the previous row with the value of the current row; if they match, we increment the rank by 1. If it's a "new" category, we reset the rank to 1. Note this only works if the rows are ordered by category, and then by total_points descending, so we need the ORDER BY clause. Also note that the order of the expressions in the SELECT list is important; we need to do the comparison of the previous value BEFORE it's overwritten with the current value, so the assignment to #prev_category must follow the conditional test.
Also note that if two users have the same total_points in a category, they will get distinct values for rank... the query above doesn't give the same rank for a tie. (The query could be modified to do that as well, but we'd also need to preserve total_points from the previous row, so we can compare to the current row.
Also note that this syntax is specific to MySQL, and that this is behavior is not guaranteed.
If you need the columns in the particular sequence and/or the rows in a particular order (to get the exact resultset specified), we'd need to wrap the query above as an inline view.
SELECT t.user_id
, t.category_id
, t.total_points
, t.rank
FROM (
SELECT #rank := IF(#prev_category = r.category_id, #rank+1, 1) AS rank
, #prev_category := r.category_id AS category_id
, r.user_id
, r.total_points
FROM (SELECT #prev_categor := NULL, #rank := 1) i
CROSS
JOIN ( SELECT s.category_id, s.user_id, SUM(s.points) AS total_points
FROM users s
GROUP BY s.category_id, s.user_id
ORDER BY s.category_id, total_points DESC
) r
ORDER BY r.category_id, r.total_points DESC, r.user_id DESC
) t
ORDER BY t.user_id, t.category_id
NOTE: I've not setup a SQL Fiddle demonstration. I've given an example query which has only been desk checked.
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
;