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"
Related
I want to find the max value in a column
ID CName Tot_Val PName
--------------------------------
1 1 100 P1
2 1 10 P2
3 2 50 P2
4 2 80 P1
Above is my table structure. I just want to find the max total value only from the table. In that four row ID 1 and 2 have same value in CName but total val and PName has different values. What I am expecting is have to find the max value in ID 1 and 2
Expected result:
ID CName Tot_Val PName
--------------------------------
1 1 100 P1
4 2 80 P1
I need result same as like mention above
select Max(Tot_Val), CName
from table1
where PName in ('P1', 'P2')
group by CName
This is query I have tried but my problem is that I am not able to bring PName in this table. If I add PName in the select list means it will showing the rows doubled e.g. Result is 100 rows but when I add PName in selected list and group by list it showing 600 rows. That is the problem.
Can someone please help me to resolve this.
One possible option is to use a subquery. Give each row a number within each CName group ordered by Tot_Val. Then select the rows with a row number equal to one.
select x.*
from ( select mt.ID,
mt.CName,
mt.Tot_Val,
mt.PName,
row_number() over(partition by mt.CName order by mt.Tot_Val desc) as No
from MyTable mt ) x
where x.No = 1;
An alternative would be to use a common table expression (CTE) instead of a subquery to isolate the first result set.
with x as
(
select mt.ID,
mt.CName,
mt.Tot_Val,
mt.PName,
row_number() over(partition by mt.CName order by mt.Tot_Val desc) as No
from MyTable mt
)
select x.*
from x
where x.No = 1;
See both solutions in action in this fiddle.
You can search top-n-per-group for this kind of a query.
There are two common ways to do it. The most efficient method depends on your indexes and data distribution and whether you already have another table with the list of all CName values.
Using ROW_NUMBER
WITH
CTE
AS
(
SELECT
ID, CName, Tot_Val, PName,
ROW_NUMBER() OVER (PARTITION BY CName ORDER BY Tot_Val DESC) AS rn
FROM table1
)
SELECT
ID, CName, Tot_Val, PName
FROM CTE
WHERE rn=1
;
Using CROSS APPLY
WITH
CTE
AS
(
SELECT CName
FROM table1
GROUP BY CName
)
SELECT
A.ID
,A.CName
,A.Tot_Val
,A.PName
FROM
CTE
CROSS APPLY
(
SELECT TOP(1)
table1.ID
,table1.CName
,table1.Tot_Val
,table1.PName
FROM table1
WHERE
table1.CName = CTE.CName
ORDER BY
table1.Tot_Val DESC
) AS A
;
See a very detailed answer on dba.se Retrieving n rows per group
, or here Get top 1 row of each group
.
CROSS APPLY might be as fast as a correlated subquery, but this often has very good performance (and better than ROW_NUMBER():
select t.*
from t
where t.tot_val = (select max(t2.tot_val)
from t t2
where t2.cname = t.cname
);
Note: The performance depends on having an index on (cname, tot_val).
I want to use a query, showing the top two best Quantity. If the table is like the picture, how can the desired result be produced
You can use DENSE_RANK(). For example:
select
id, name, quantity
from (
select
id, name, quantity,
dense_rank() over(order by quantity desc) as rk
from t
) x
where rk <= 2
DENSE_RANK() computes a number for each row according to an ordering of your choosing. Identical values get the same number, and no numbers are skipped. See SQL Fiddle.
You can use the TOP/LIMIT functions in query. Which would allow you to select a specific number of rows.
using number as 5 you can get the desired result from:
SELECT columnname FROM tablename WHERE condition LIMIT number;
or
SELECT TOP (number)/(percent) columnname FROM tablename WHERE condition;
Currently i am trying to output the top row for 2 condition. One is max and one is min.
Current code
Select *
from (MY SELECT STATEMENT order by A desc)
where ROWNUM <= 1
UPDATE
I am now able to do for both condition. But i need the A to be the highest, if same then check for the B lowest.
E.g Lets say there is 2 rows, Both A is 100 and B is 50 for one and 60 for other.
In this case the 100:50 shld be choose because A is same then B is lowest.
E.g
Lets say there is 2 rows, A is 100 for one and 90 for other, since one is higher no need to check for B.
I tried using max and min but this method seems to work better, any suggestions
Well, after your clarification, you are looking for one record. With Max A. And the smallest B, in case there is more than one record with MAX A. This is simply:
Select *
from (MY SELECT STATEMENT order by A desc, B)
where ROWNUM = 1;
This sorts by A descending first, so you get all maximal A records first. Then it sorts by B, so inside each A group you get the least B first. This gives you the desired A record first, no matter if the found A is unique or not.
or avoid the vagaries of rownun and go for row_number() instead:
SELECT
*
FROM (
SELECT
*
, ROW_NUMBER (ORDER BY A DESC) adesc
, ROW_NUMBER (ORDER BY B ASC) basc
FROM SomeQuery
)
WHERE adesc = 1
OR basc = 1
footnote: select * is a convenience only, please replace with the actual columns required along with table names etc.
Try this if that works
Select *
from (MY SELECT STATEMENT order by A desc)
where ROWNUM <= 1
union
Select *
from (MY SELECT STATEMENT order by A asc)
where ROWNUM <= 1
SELECT * FROM
(Select foo.*, 0 as union_order
from (MY SELECT STATEMENT order by A desc) foo
where ROWNUM <= 1
UNION
Select foo.*, 1
from (MY SELECT STATEMENT order by B asc) foo
where ROWNUM <= 1)
ORDER BY
union_order
I need to get the latest price of an item (as part of a larger select statement) and I can't quite figure it out.
Table:
ITEMID DATE SALEPRICE
1 1/1/2014 10
1 2/2/2014 20
2 3/3/2014 15
2 4/4/2014 13
I need the output of the select to be '20' when looking for item 1 and '13' when looking for item 2 as per the above example.
I am using Oracle SQL
The most readable/understandable SQL (in my opinion) would be this:
select salesprice from `table` t
where t.date =
(
select max(date) from `table` t2 where t2.itemid = t.itemid
)
and t.itemid = 1 -- change item id here;
assuming your table's name is table and you only have one price per day and item (else the where condition would match more than one row per item). Alternatively, the subselect could be written as a self-join (should not make a difference in performance).
I'm not sure about the OVER/PARTITION used by the other answers. Maybe they could be optimized to better performance depending on the DBMS.
Maybe something like this:
Test data
DECLARE #tbl TABLE(ITEMID int,DATE DATETIME,SALEPRICE INT)
INSERT INTO #tbl
VALUES
(1,'1/1/2014',10),
(1,'2/2/2014',20),
(2,'3/3/2014',15),
(2,'4/4/2014',13)
Query
;WITH CTE
AS
(
SELECT
ROW_NUMBER() OVER(PARTITION BY ITEMID ORDER BY [DATE] DESC) AS rowNbr,
tbl.*
FROM
#tbl AS tbl
)
SELECT
*
FROM
CTE
WHERE CTE.rowNbr=1
Try this!
In sql-server may also work in Oracle sql
select * from
(
select *,rn=row_number()over(partition by ITEMID order by DATE desc) from table
)x
where x.rn=1
You need Row_number() to allocate a number to all records which is partition by ITEMID so each group will get a RN,then as you are ordering by date desc to get Latest record
SEE DEMO
You have three fields ID, Date and Total. Your table contains multiple rows for the same day which is valid data however for reporting purpose you need to show only one row per day. The row with the highest ID per day should be returned the rest should be hidden from users (not returned).
To better picture the question below is sample data and sample output:
ID, Date, Total
1, 2011-12-22, 50
2, 2011-12-22, 150
The correct result is:
2, 2012-12-22, 150
The correct output is single row for 2011-12-22 date and this row was chosen because it has the highest ID (2>1)
Assuming that you have a database that supports window functions, and that the date column is indeed just date (and not datetime), then something like:
SELECT
* --TODO - Pick columns
FROM
(
SELECT ID,[Date],Total,ROW_NUMBER() OVER (PARTITION BY [Date] ORDER BY ID desc) rn
FROM [Table]
) t
WHERE
rn = 1
Should produce one row per day - and the selected row for any given day is that with the highest ID value.
SELECT *
FROM table
WHERE ID IN ( SELECT MAX(ID)
FROM table
GROUP BY Date )
This will work.
SELECT *
FROM tableName a
INNER JOIN
(
SELECT `DATE`, MAX(ID) maxID
FROM tableName
GROUP BY `DATE`
) b ON a.id = b.MaxID AND
a.`date` = b.`date`
SQLFiddle Demo
Probably
SELECT * FROM your_table ORDER BY ID DESC LIMIT 1
Select MAX(ID),Data,Total from foo
for MySQL
Another simple way is
SELECT TOP 1 * FROM YourTable ORDER BY ID DESC
And, I think this is the most simple way!
SELECT * FROM TABLE_SUM S WHERE S.ID =
(
SELECT MAX(ID) FROM TABLE_SUM
WHERE CDATE = GG.CDATE
GROUP BY CDATE
)