mysql query lowest option sum - mysql

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);

Related

MySQL select not equal to limit in subquery

MySQL #1235 - This version of MySQL doesn't yet support 'LIMIT & IN/ALL/ANY/SOME subquery'
Given 1 table as following
Item | Name | Price
----- ------------ --------
1 | Adidas | 310.00
2 | Nike Run | 30.00
3 | Puma | 150.00
4 | Nike Women | 20.00
5 | NB | 20.00
Would like to select records and return the sum amount. Do not sum up the 2 highest prices' record.
SELECT SUM(Price) as total_amount
FROM `test`
WHERE Item NOT IN (
SELECT Price
FROM `test`
ORDER BY Price DESC
LIMIT 2)
Expected Result:
total_amount
------------
70.00
How to use JOIN or alternative LIMIT in Subquery in this query?
Thank you.
You need a temp table:
SELECT SUM(Price) FROM test WHERE Item NOT IN (
SELECT * FROM (
SELECT Item FROM test ORDER BY Price DESC LIMIT 2
) AS tmp
)
Here's one option using a subquery with limit / offset:
select sum(price)
from (
select *
from test
order by price desc
limit 999999999
offset 2
) t
SQL Fiddle Demo
Just make sure the limit value is greater than the number of potential rows (which evidently is 18446744073709551615)...
Or you could use user-defined variables:
select sum(price)
from (
select *, #rn:=#rn + 1 rn
from test cross join (select #rn:= 0) t
) t
where rn > 2
If you looking to exclude the 2 highest prices which could be more than 2 records, this will also work with user defined variables:
select sum(price)
from (
select *, #rn:=if(#prevPrice=price, #rn,
if(#prevPrice:=price, #rn + 1, #rn + 1)) rn
from test cross join (select #rn:= 0, #prevPrice:= null) t
) t
where rn > 2
More Fiddle

MySQL select upto first occurrence of condition matching

Id | Price
----------------
1 | 10
2 | 20
3 | 40
4 | 10
I need to select ids where first occurrence of summation of price is greater than or equal 55 matching from the bottom. At this case --
I will have 4,3,2 ids selected.
Well, this is kinda tricky for MySQL since it doesn't support any window fuctions and becuase you want to include the first occurrence as well. You can try this:
SELECT * FROM (
SELECT t.id,
(SELECT sum(s.price) FROM YourTable s
WHERE s.id <= t.id) as cuml_sum
FROM YourTable t) ss
WHERE ss.cuml_sum < 55
--Will select all the record will the sum < 55
UNION ALL
SELECT * FROM (
SELECT t.id,
(SELECT sum(s.price) FROM YourTable s
WHERE s.id <= t.id) as cuml_sum
FROM YourTable t) tt
WHERE tt.cuml_sum >= 55
ORDER BY tt.cuml_sum
LIMIT 1
--Will select the first record that have sum >= 55

Divide with the rollup value

I have this code where it sums up the hours of the employee and uses rollup to get the total of the hours:
SELECT IFNULL(users, 'Total') AS Employee,
SUM(actual) AS Amount
FROM table1
WHERE name = "ProjectName"
GROUP BY users
WITH ROLLUP
Employee | Amount
A | 15
B | 10
C | 10
Total | 35
What I would like to do for my third column (Percent) is to divide the sum(actual) to the value of the total to get the percentage.
But for that Percent column I don't need to get the Total Percent.
The total value is not constant to just 35.
Employee | Amount | Percent
A | 15 | 42.85
B | 10 | 28.57
C | 10 | 28.57
Total | 35 |
How can I do that?
Here's the sqlfiddle: http://sqlfiddle.com/#!2/4543b/5
This works as desired:
SET #project_name = 'ProjectName';
SELECT IFNULL(users, 'Total') AS Employee, SUM(actual) AS Amount,
IF(ISNULL(users), '', TRUNCATE(SUM(actual) / sum_table.amount_sum * 100, 2)
) AS Percent
FROM Table1
INNER JOIN (
SELECT SUM(actual) AS amount_sum
FROM Table1
WHERE name = #project_name
) AS sum_table
WHERE name = #project_name
GROUP BY users
WITH ROLLUP;
DEMO # SQL Fiddle
Perhaps a job best left to the logic tier of your application, but if you absolutely must do it in the data tier then you merely need to join your query with another that finds the overall total:
SELECT IFNULL(users, 'Total') AS Employee,
SUM(actual) AS Amount,
SUM(actual)/t.Total AS Percent
FROM Table1, (
SELECT SUM(actual) AS Total
FROM Table1
WHERE name = 'ProjectName'
) t
WHERE name = 'ProjectName'
GROUP BY users WITH ROLLUP
SELECT if(users is NULL,'Total',users) as Employee, sum(actual) as Amount,
(CASE
WHEN users is not null THEN CAST(sum(actual)/sum.sumAmt * 100 as DECIMAL(10,2))
END) as Percent
FROM Table1, (SELECT sum(actual) as sumAmt FROM Table1
WHERE name = 'ProjectName') sum
WHERE name = "ProjectName"
GROUP BY users
WITH ROLLUP
DEMO

Unknown column in mysql subquery

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
;

SQL get all rows sorted without duplicates

I have a table that looks something like this:
________________________
|id|value|date|approved|
-----------------------
What I need to be able to do is get each row where approved = 1. That part is obvious. For each occurrence of value, I only want the most recent row (sorted by date).
Meaning that with a table like this:
________________________
|id|value|date|approved|
-----------------------
|1 |Foo | 5 | 1 |
|2 |Bar | 6 | 1 |
|3 |Foo | 8 | 1 |
-----------------------
I only want the rows with id 2 and 3.
I assume I need to use DISTINCT somehow, but I'm not sure how. Could anyone help me out here?
SELECT m.*
FROM (
SELECT DISTINCT value
FROM mytable
) md
JOIN mytable m
ON m.id =
(
SELECT id
FROM mytable mi
WHERE mi.value = md.value
AND mi.approved = 1
ORDER BY
mi.value DESC, mi.date DESC, mi.id DESC
LIMIT 1
)
Create an index on (value, date, id) for this to work fast.
You need:
select id, value, date, approved where (value, date) in (
select value, max(date)
from your_table
group by value
);
Actually using GROUP BY will yeld better results. try something like this:
SELECT id, value, date, approved FROM table WHERE approved = 1 GROUP BY value ORDER BY date;
select id, value, date, approved
from mytable a
where approved = 1
and date =
(select max(b.date)
from mytable b
where b.approved = 1
and b.value = a.value)
select
id,
value,
date
from
( select
value, max( date ) as LastInstance
from
YourTable
where
approved = 1
group by
value ) PreQuery
join YourTable
on PreQuery.value = YourTable.value
and PreQuery.LastInstance = YourTable.LastInstance
and YourTable.approved = 1
order by
date
Try with this:
SELECT * DISTINCT FROM TABLA WHERE APPROVED = 1 ORDER BY DATE DESC
Hope this helps you.