I have a SQL problem (MS SQL Server 2012), where I only want one result per set, but have different items in some rows, so a group by doesn't work.
Here is the statement:
Select Deliverer, ItemNumber, min(Price)
From MyTable
Group By Deliverer, ItemNumber
So I want the deliverer with the lowest price for one item.
With this query I get the lowest price for each deliverer.
So a result like:
DelA 12345 1,25
DelB 11111 2,31
And not like
DelA 12345 1,25
DelB 12345 1,35
DelB 11111 2,31
DelC 11111 2,35
I know it is probably a stupid question with an easy solution, but I tried for about three hours now and just can't find a solution. Needles to say, I'm not very experienced with SQL.
Just Add an aggregate function to your deliverer field also, as appropriate (Either min or max). From your data, I guess you need min(deliverer) and hence use the below query to get your desired result.
Select mIN(Deliverer), ItemNumber, min(Price)
From MyTable
Group By ItemNumber;
EDIT:
Below query should help you get the deliverer with the lowest price item-wise:
SELECT TABA.ITEMNUMBER, TABA.MINPRICE, TABB.DELIVERER
FROM
(
SELECT ITEMNUMBER, MIN(PRICE) MINPRICE
FROM MYTABLE GROUP BY
ITEMNUMBER
) TABA JOIN
MYTABLE TABB
ON TABA.ITEMNUMBER=TABB.ITEMNUMBER AND
TABA.MINPRICE = TABB.PRICE
You should be able to do this with the RANK() (or DENSE_RANK()) functions, and a bit of partitioning, so something like:
; With rankings as (
SELECT Deliverer,
rankings.ItemNumber,
rankings.Price
RANK() OVER (PARTITION BY ItemNumber ORDER BY Price ASC) AS Ranking
FROM MyTable (Deliverer, ItemNumber, Price)
)
SELECT rankings.Deliverer,
rankings.ItemNumber,
rankings.Price
FROM rankings
WHERE ranking = 1
Related
This is my code:
SELECT ROW_NUMBER()OVER(order by date) as rowNo, id, amount
INTO #temptbl
FROM sales
WHERE code = 1000
I'm trying to group this temporary table and sum the amount by its id so the id will be unique.
*The row column is mandatory because it will be used later.
I've tried few ways like this,
SELECT ROW_NUMBER()OVER(order by date) as rowNo, id, amount
INTO #temptbl
FROM sales
WHERE code = 1000
GROUP BY id
and even tried subquey nad nesting but it wont work. I know the solution must be simple its just i cant see it yet. Thank you
You can use SUM with Partition BY like this:
SELECT ROW_NUMBER()OVER(order by date) as rowNo, id,
Sum(amount) over (partition by id) sumAmount
FROM sales
WHERE code = 1000
As you know, since you want the Row number then sum amount will be repeated for same ids
It is known that GROUP BY produces one row per group. I want to produce multiple rows per group. The particular use case is, for example, selecting two cheapest offerings for each item.
It is trivial for two or three elements in the group:
select type, variety, price
from fruits
where price = (select min(price) from fruits as f where f.type = fruits.type)
or price = (select min(price) from fruits as f where f.type = fruits.type
and price > (select min(price) from fruits as f2 where f2.type = fruits.type));
(Select n rows per group in mysql)
But I am looking for a query that can show n rows per group, where n is arbitrarily large. In other words, a query that displays 5 rows per group should be convertible to a query that displays 7 rows per group by just replacing some constants in it.
I am not constrained to any DBMS, so I am interested in any solution that runs on any DBMS. It is fine if it uses some non-standard syntax.
For any database that supports analytic functions\ window functions, this is relatively easy
select *
from (select type,
variety,
price,
rank() over ([partition by something]
order by price) rnk
from fruits) rank_subquery
where rnk <= 3
If you omit the [partition by something], you'll get the top three overall rows. If you want the top three for each type, you'd partition by type in your rank() function.
Depending on how you want to handle ties, you may want to use dense_rank() or row_number() rather than rank(). If two rows tie for first, using rank, the next row would have a rnk of 3 while it would have a rnk of 2 with dense_rank. In both cases, both tied rows would have a rnk of 1. row_number would arbitrarily give one of the two tied rows a rnk of 1 and the other a rnk of 2.
To save anyone looking some time, at the time of this writing, apparently this won't work because https://dev.mysql.com/doc/refman/5.7/en/subquery-restrictions.html.
I've never been a fan of correlated subqueries as most uses I saw for them could usually be written more simply, but I think this has changed by mind... a little. (This is for MySQL.)
SELECT `type`, `variety`, `price`
FROM `fruits` AS f2
WHERE `price` IN (
SELECT DISTINCT `price`
FROM `fruits` AS f1
WHERE f1.type = f2.type
ORDER BY `price` ASC
LIMIT X
)
;
Where X is the "arbitrary" value you wanted.
If you know how you want to limit further in cases of duplicate prices, and the data permits such limiting ...
SELECT `type`, `variety`, `price`
FROM `fruits` AS f2
WHERE (`price`, `other_identifying_criteria`) IN (
SELECT DISTINCT `price`, `other_identifying_criteria`
FROM `fruits` AS f1
WHERE f1.type = f2.type
ORDER BY `price` ASC, `other_identifying_criteria` [ASC|DESC]
LIMIT X
)
;
"greatest N per group problems" can easily be solved using window functions:
select type, variety, price
from (
select type, variety, price,
dense_rank() over (partition by type) order by price as rnk
from fruits
) t
where rnk <= 5;
Windows functions only work on SQL Server 2012 and above. Try this out:
SQL Server 2005 and Above Solution
DECLARE #yourTable TABLE(Category VARCHAR(50), SubCategory VARCHAR(50), price INT)
INSERT INTO #yourTable
VALUES ('Meat','Steak',1),
('Meat','Chicken Wings',3),
('Meat','Lamb Chops',5);
DECLARE #n INT = 2;
SELECT DISTINCT Category,CA.SubCategory,CA.price
FROM #yourTable A
CROSS APPLY
(
SELECT TOP (#n) SubCategory,price
FROM #yourTable B
WHERE A.Category = B.Category
ORDER BY price DESC
) CA
Results in two highest priced subCategories per Category:
Category SubCategory price
------------------------- ------------------------- -----------
Meat Chicken Wings 3
Meat Lamb Chops 5
I have the following table:
ItemID Price
1 10
2 20
3 12
4 10
5 11
I need to find the second lowest price. So far, I have a query that works, but i am not sure it is the most efficient query:
select min(price)
from table
where itemid not in
(select itemid
from table
where price=
(select min(price)
from table));
What if I have to find third OR fourth minimum price? I am not even mentioning other attributes and conditions... Is there any more efficient way to do this?
PS: note that minimum is not a unique value. For example, items 1 and 4 are both minimums. Simple ordering won't do.
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
select price from table where price in (
select
distinct price
from
(select t.price,rownumber() over () as rownum from table t) as x
where x.rownum = 2 --or 3, 4, 5, etc
)
Not sure if this would be the fastest, but it would make it easier to select the second, third, etc... Just change the TOP value.
UPDATED
SELECT MIN(price)
FROM table
WHERE price NOT IN (SELECT DISTINCT TOP 1 price FROM table ORDER BY price)
To find out second minimum salary of an employee, you can use following:
select min(salary)
from table
where salary > (select min(salary) from table);
This is a good answer:
SELECT MIN( price )
FROM table
WHERE price > ( SELECT MIN( price )
FROM table )
Make sure when you do this that there is only 1 row in the subquery! (the part in brackets at the end).
For example if you want to use GROUP BY you will have to define even further using:
SELECT MIN( price )
FROM table te1
WHERE price > ( SELECT MIN( price )
FROM table te2 WHERE te1.brand = te2.brand)
GROUP BY brand
Because GROUP BY will give you multiple rows, otherwise you will get the error:
SQL Error [21000]: ERROR: more than one row returned by a subquery used as an expression
I guess a simplest way to do is using offset-fetch filter from standard sql, distinct is not necessary if you don't have repeat values in your column.
select distinct(price) from table
order by price
offset 1 row fetch first 1 row only;
no need to write complex subqueries....
In amazon redshift use limit-fetch instead for ex...
Select distinct(price) from table
order by price
limit 1
offset 1;
You can either use one of the following:-
select min(your_field) from your_table where your_field NOT IN (select distinct TOP 1 your_field from your_table ORDER BY your_field DESC)
OR
select top 1 ColumnName from TableName where ColumnName not in (select top 1 ColumnName from TableName order by ColumnName asc)
I think you can find the second minimum using LIMIT and ORDER BY
select max(price) as minimum from (select distinct(price) from tableName order by price asc limit 2 ) --or 3, 4, 5, etc
if you want to find third or fourth minimum and so on... you can find out by changing minimum number in limit. you can find using this statement.
You can use RANK functions,
it may seem complex query but similar results like other answers can be achieved with the same,
WITH Temp_table AS (SELECT ITEM_ID,PRICE,RANK() OVER (ORDER BY PRICE) AS
Rnk
FROM YOUR_TABLE_NAME)
SELECT ITEM_ID FROM Temp_table
WHERE Rnk=2;
Maybe u can check the min value first and then place a not or greater than the operator. This will eliminate the usage of a subquery but will require a two-step process
select min(price)
from table
where min(price) <> -- "the min price you previously got"
I am almost a novie in database queries.
However,I do understand why and how correlated subqueries are expensive and best avoided.
Given the following simple example - could someone help replacing with a join to help understand how it scores better:
SQL> select
2 book_key,
3 store_key,
4 quantity
5 from
6 sales s
7 where
8 quantity < (select max(quantity)
9 from sales
10 where book_key = s.book_key);
Apart from join,what other option do we have to avoid the subquery.
In this case, it ought to be better to use a windowed-function on a single access to the table - like so:
with s as
(select book_key,
store_key,
quantity,
max(quantity) over (partition by book_key) mq
from sales)
select book_key, store_key, quantity
from s
where quantity < s.mq
Using Common Table Expressions (CTE) will allow you to execute a single primary SELECT statement and store the result in a temporary result set. The data can then be self-referenced and accessed multiple times without requiring the initial SELECT statement to be executed again and won't require possibly expensive JOINs. This solution also uses ROW_NUMBER() and the OVER clause to number the matching BOOK_KEYs in descending order based off of the quantity. You will then only include the records that have a quantity that is less than the max quantity for each BOOK_KEY.
with CTE as
(
select
book_key,
store_key,
quantity,
row_number() over(partition by book_key order by quantity desc) rn
from sales
)
select
book_key,
store_key,
quantity
from CTE where rn > 1;
Working Demo: http://sqlfiddle.com/#!3/f0051/1
Apart from join,what other option do we have to avoid the subquery.
You use something like this:
SELECT select max(quantity)
INTO #myvar
from sales
where book_key = s.book_key
select book_key,store_key,quantity
from sales s
where quantity < #myvar
I have a table which contains:
-an ID for a financial instrument
-the price
-the date the price was recorded
-the actual time the price was recorded
-the source of the price
I want to get the index ID, the latest price, price source and the date of this latest price, for each instrument, where the source is either "L" or "R". I prefer source "L" to "R", but the latest price is more important (so if the latest price date only has a source of "R"- take this, but if for the latest date we have both, take "L").
This is the SQL I have:
SELECT tab1.IndexID, tab1.QuoteDate, tab2.Source, tab2.ActualTime FROM
(SELECT IndexID, Max(QuoteDate) as QuoteDate FROM PricesTable GROUP BY IndexID) tab1
JOIN
(SELECT IndexID, Min(Source) AS Source, Max(UpdatedTime) AS ActualTime, QuoteDate FROM PricesTable WHERE Source IN ('L','R') GROUP BY IndexID, QuoteDate) tab2
ON tab1.IndexID = tab2.IndexID AND tab1.QuoteDate = tab2.QuoteDate
However, I also want to extract the price field but cannot get this due to the GROUP BY clause. I cannot extract the price without including price in either the GROUP BY, or an aggregate function.
Instead, I have had to join the above SQL code to another piece of SQL which just gets the prices and index IDs and joins on the index ID.
Is there a faster way of performing this query?
EDIT: thanks for the replies so far. Would it be possible to have some advice on which are more efficient in terms of performance?
Thanks
Use ROW_NUMBER within a subquery or CTE to order the rows how you're interested in them, then just select the rows that come at the top of that ordering. (Use PARITION so that row numbers are reaassigned starting at 1 for each IndexId):
;WITH OrderedValues as (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY IndexID ORDER BY QuoteDate desc,Source asc) as rn
FROM
PricesTable
)
SELECT * from OrderedValues where rn=1
Try:
select * from
(select p.*,
row_number() over (partition by IndexID
order by QuoteDate desc, Source) rn
from PricesTable p
where Source IN ('L','R')
) sq
where rn = 1
(This syntax should work in relatively recent versions of Oracle, SQLServer or PostgreSQL, but won't work in MySQL.)