I need to write a query that searches for a maximum value and if there more items having the same maximum value, it would return them both.
For example, if I have in my database
Item Price
Coffee 2.50
Tea 2.50
Cola 1.50
it would give Coffee and Tea as an answer.
Thank you in advance
This is typically solved using a window function:
select item,
price
from (
select item,
price,
dense_rank() over (order by price desc) as rnk
from the_table
) t
where rnk = 1;
You didn't specify your DBMS, therefore the above is standard (ANSI) SQL.
Alternatively ...
SELECT
Item,
Price
FROM the_table
WHERE Price = (SELECT MAX(Price) FROM the_table)
Or ...
SELECT
Item,
Price
FROM the_table
WHERE NOT EXISTS
(
SELECT Price
FROM the_table st
WHERE st.Price > the_table.Price
)
Or one more ...
SELECT
Item,
Price
FROM
the_table
INNER JOIN
(
SELECT MAX(Price) MaxPrice
FROM the_table
) Match ON
the_table.Price = match.MaxPrice
Just in case windowed functions aren't available.
Related
I am using SQL Server and I have a table "a"
month segment_id price
-----------------------------
1 1 100
1 2 200
2 3 50
2 4 80
3 5 10
I want to make a query which presents the original columns where the price will be the max per month
The result should be:
month segment_id price
----------------------------
1 2 200
2 4 80
3 5 10
I tried to write SQL code:
Select
month, segment_id, max(price) as MaxPrice
from
a
but I got an error:
Column segment_id is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause
I tried to fix it in many ways but didn't find how to fix it
Because you need a group by clause without segment_id
Select month, max(price) as MaxPrice
from a
Group By month
as you want results per each month, and segment_id is non-aggregated in your original select statement.
If you want to have segment_id with maximum price repeating per each month for each row, you need to use max() function as window analytic function without Group by clause
Select month, segment_id,
max(price) over ( partition by month order by segment_id ) as MaxPrice
from a
Edit (due to your lastly edited desired results) : you need one more window analytic function row_number() as #Gordon already mentioned:
Select month, segment_id, price From
(
Select a.*,
row_number() over ( partition by month order by price desc ) as Rn
from a
) q
Where rn = 1
I would recommend a correlated subquery:
select t.*
from t
where t.price = (select max(t2.price) from t t2 where t2.month = t.month);
The "canonical" solution is to use row_number():
select t.*
from (select t.*,
row_number() over (partition by month order by price desc) as seqnum
from t
) t
where seqnum = 1;
With the right indexes, the correlated subquery often performs better.
Only because it was not mentioned.
Yet another option is the WITH TIES clause.
To be clear, the approach by Gordon and Barbaros would be a nudge more performant, but this technique does not require or generate an extra column.
Select Top 1 with ties *
From YourTable
Order By row_number() over (partition by month order by price desc)
With not exists:
select t.*
from tablename t
where not exists (
select 1 from tablename
where month = t.month and price > t.price
)
or:
select t.*
from tablename inner join (
select month, max(price) as price
from tablename
group By month
) g on g.month = t.month and g.price = t.price
I want to select the ID of the Table Products with the lowest Price Grouped By Product.
ID Product Price
1 123 10
2 123 11
3 234 20
4 234 21
Which by logic would look like this:
SELECT
ID,
Min(Price)
FROM
Products
GROUP BY
Product
But I don't want to select the Price itself, just the ID.
Resulting in
1
3
EDIT: The DBMSes used are Firebird and Filemaker
You didn't specify your DBMS, so this is ANSI standard SQL:
select id
from (
select id,
row_number() over (partition by product order by price) as rn
from orders
) t
where rn = 1
order by id;
If your DBMS doesn't support window functions, you can do that with joining against a derived table:
select o.id
from orders o
join (
select product,
min(price) as min_price
from orders
group by product
) t on t.product = o.product and t.min_price = o.price;
Note that this will return a slightly different result then the first solution: if the minimum price for a product occurs more then once, all those IDs will be returned. The first solution will only return one of them. If you don't want that, you need to group again in the outer query:
select min(o.id)
from orders o
join (
select product,
min(price) as min_price
from orders
group by product
) t on t.product = o.product and t.min_price = o.price
group by o.product;
SELECT ID
FROM Products as A
where price = ( select Min(Price)
from Products as B
where B.Product = A.Product )
GROUP BY id
This will show the ID, which in this case is 3.
I'm trying to get the most expensive and cheapest items from two different tables.
The output should be one row with the values for MostExpensiveItem, MostExpensivePrice, CheapestItem, CheapestPrice
I was able to get the price of the most expensive and cheapest items in the two tables with following query:
SELECT
MAX(ExtrasPrice) as MostExpensivePrice, MIN(ExtrasPrice) as CheapestPrice
FROM
(
SELECT ExtrasPrice FROM Extras
UNION ALL
SELECT ItemPrice FROM Items
) foo
How can I add the names of the items (ItemName, ExtrasName) to my output? Again, there should only be one row as the output.
Try this:
SELECT TOP 1 FIRST_VALUE(Price) OVER (ORDER BY Price) AS MinPrice,
FIRST_VALUE(Name) OVER (ORDER BY Price) AS MinName,
LAST_VALUE(Price) OVER (ORDER BY Price DESC) AS MaxPrice,
LAST_VALUE(Name) OVER (ORDER BY Price DESC) AS MaxName
FROM (
SELECT ExtrasName AS Name, ExtrasPrice AS Price FROM Extras
UNION ALL
SELECT ItemName As Name, ItemPrice AS Price FROM Items) u
SQL Fiddle Demo
TOP 1 with order by clause should work for you. Try this
SELECT *
FROM (SELECT TOP 1 ExtrasPrice,ExtrasName
FROM Extras ORDER BY ExtrasPrice Asc),
(SELECT TOP 1 ItemPrice,ItemName
FROM Items ORDER BY ItemPrice Desc)
Note: Comma can be replaced with CROSS JOIN
You can use row_number() for this. If you are satisfied with two rows:
SELECT item, price
FROM (SELECT foo.*, row_number() over (order by price) as seqnum_asc,
row_number() over (order by price) as seqnum_desc
FROM (SELECT item, ExtrasPrice as price FROM Extras
UNION ALL
SELECT item, ItemPrice FROM Items
) foo
) t
WHERE seqnum_asc = 1 or seqnum_desc = 1;
EDIT:
If you have an index on "price" in both tables, then the cheapest method is probably:
with exp as (
(select top 1 item, ExtrasPrice as price
from Extras e
order by price desc
) union all
(select top 1 i.item, ItemPrice
from Items i
order by price desc
)
),
cheap as (
(select top 1 item, ExtrasPrice as price
from Extras e
order by price asc
) union all
(select top 1 i.item, ItemPrice
from Items i
order by price asc
)
)
select top 1 *
from exp
order by price desc
union all
select top 1 *
from cheap
order by price asc;
If you want this in one row, you can replace the final query with:
select e.*, c.*
from (select top 1 *
from exp
order by price desc
) e cross join
(select top 1 *
from cheap
order by price asc
) c
I'd like to select each pair of two columns in a database, but only select the entry with the lowest price. As a result, I want to output the id and the price column.
But it does not work:
My table:
id | category | type | name | price
1;"car";"pkw";"honda";1000.00
2;"car";"pkw";"bmw";2000.00
SQL:
select min(price) price, id
from cartable
group by category, type
Result:
Column "cartable.id" must be present in GROUP-BY clause or used in an aggregate function.
If you want the entry with the lowest price, then calculate the lowest price and join the information back in:
select ct.*
from cartable ct join
(select category, type, min(price) as price
from cartable
group by category, type
) ctp
on ct.category = ctp.category and ct.type = ctp.type and ct.price = ctp.price;
You can achieve this with EXISTS clause:
SELECT *
FROM cartable ct
WHERE
NOT EXISTS (
SELECT *
FROM cartable
WHERE ct.type = type and ct.category = categoery and ct.price < price)
For speed caparison can you try this:
SELECT DISTINCT ON (type, category), id, price
FROM cartable
ORDER BY price DESC
SELECT id, price
from cartable C
inner join
(
select min(price) as price , category, type
from cartable
group by category, type
)T
on T.category = C.category
and T.type = C.type
Most of the time you can't do much else than resolve to use Select - Over
select price, id
from(
select price, id, [rnk] = ROW_NUMBER() over( partition by category, type order by price)
from cartable
) as a
where [rnk]=1
Create index appropriately and performance are good.
In your example something like this:
CREATE NONCLUSTERED INDEX [foo]
ON [dbo].[cartable] ([category],[type])
INCLUDE ([price])
Maybe you can try:
select id, price from cartable
where price = (select min(price) from cartable);
I have a table as follows:
BRAND_ID PRODUCT_ID PRODUCT_DESC PRODUCT_TYPE
100 1000 Tools A
100 1500 Tools A
200 2000 Burgers B
300 3000 Clothing C
300 4000 Makeup D
300 5000 Clothing C
So a Brand can have multiple products, all of the same type or mixed types. If a brands products are all of the same type I need them first in the result, sorted by product type, followed by brands that have different product types. I can do this programatically but I wanted to see if there is a way to do it in the query.
I don't have access to Oracle, but I believe something along these lines should work...
WITH
ranked_data
AS
(
SELECT
COUNT(DISTINCT product_type) OVER (PARTITION BY brand_id) AS brand_rank,
MIN(product_type) OVER (PARTITION BY brand_id) AS first_product_type,
*
FROM
yourTable
)
SELECT
*
FROM
ranked_data
ORDER BY
brand_rank,
first_product_type,
brand_id,
product_type,
product_description
An alternative is to JOIN on to a sub-query to calculate the two sorting fields.
SELECT
yourTable.*
FROM
yourTable
INNER JOIN
(
SELECT
brand_id,
COUNT(DISTINCT product_type) AS brand_rank,
MIN(product_type) AS first_product_type,
FROM
yourTable
GROUP BY
brand_id
)
AS brand_summary
ON yourTable.brand_id = brand_summary.brand_id
ORDER BY
brand_summary.brand_rank,
brand_summary.first_product_type,
yourTable.brand_id,
yourTable.product_type,
yourTable.product_description
How about selecting from a sub-select that figures out number of distinct brands and then sorting by the count?
select t.BRAND_ID,
t.PRODUCT_ID,
t.PRODUCT_DESC,
t.PRODUCT_TYPE
from (select t2.BRAND_ID,
t2.PRODUCT_ID,
count(distinct t2.PRODUCT_TYPE) cnt
from YOURTABLE t2
group by t2.BRAND_ID, t2.PRODUCT_ID) data
join YOURTABLE t on t.BRAND_ID = data.BRAND_ID and t.PRODUCT_ID = data.PRODUCT_ID
order by data.cnt, BRAND_ID, PRODUCT_ID, PRODUCT_TYPE