I am trying to find the n largest numbers in a particular column in SQL Server.
We can find the largest value in a column and the 2nd largest value easily.
But how do I find say, 5 largest values in a column ?
You tagged this both for MySQL and SQL Server. In SQL Server you can use TOP:
SELECT TOP 5 yourColumn
FROM yourTable
ORDER BY someColumn DESC;
TOP limits the number of rows returned. To get the data with the largest/smallest values you will want to include an ORDER BY.
In MySQL you will use LIMIT
Another way to do this in SQL Server is using row_number():
select id
from
(
select id, row_number() over(order by id desc) rn
from yourtable
) x
where rn <= 5
See SQL Fiddle With Demo
In MySql you can use [LIMIT {[offset,] row_count }] to do this like so:
...
ORDER BY SomeField DESC
LIMIT #n;
For SQL Server you can use the TOP(n) to get the top n:
SELECT TOP(#n) SomeFieldName
FROM TABLE
ORDER BY SomeField DESC
For example:
SELECT TOP 5 items_sold
FROM tbl_PRODUCT
ORDER BY items_sold dESC
Update: If you have another table families with a foreign key family_ID to products table, and you want to find all products with the top n family id's. Then you can dot this:
SELECT *
FROM Products WHERE family_ID IN
(
SELECT TOP 5 family_ID
FROM families
ORDER BY family_ID DESC
)
Update 2: The topmost product in each family:
;WITH cte
AS
(
SELECT *,
ROW_NUMBER() OVER(PARTITION BY family_ID ORDER BY items_sold DESC) row_num
FROM #Products
)
SELECT * FROM cte
where row_num = 1
Order by family_ID
Here is alive demo
sql server
select min(val)
from your_table
where val in (select top 5 val from your_table order by val desc)
mysql
select min(val)
from your_table
where val in (select val from your_table order by val desc limit 5)
Related
I need to retrieve the last few entries from a table. I can retrieve them using:
SELECT TOP n *
FROM table
ORDER BY id DESC
That I looked everywhere and that's the only answer I could find, But that way I get them in reverse order. I need them in the same order as they are in the table because it's for a messaging interface.
Use a derived table:
select id, ...
from
(
select top n id, ...
from t
order by id desc
) dt
order by id
I suggest you to use a ROW_NUMBER() like this:
SELECT *
FROM (
SELECT
*, ROW_NUMBER() OVER (ORDER BY id DESC) AS RowNo
FROM
yourTable
) AS t
WHERE
(RowNO < #n)
ORDER BY
id
I have thousands of groups in a table, something like :
1..
1..
2..
2..
2..
2..
3..
3..
.
.
.
10000..
10000..
How can i make a select that give me the Top 3 groups each time.
I Want something like select Top 3 from rows , but it have to return the first three groups not the first three rows.
You can try this :
;with cte as (
select distinct groupId from mytable order by groupid
)
select * from mytable where TheGroupId in (select top 3 groupdid from cte)
You can use DENSE_RANK to assign a number to each group. All members of the same group will have the same number. Then in an outer query, select top 3 groups:
SELECT *
FROM (SELECT *, DENSE_RANK() OVER (ORDER BY id) AS rnk
FROM mytable ) t
WHERE t.rnk <= 3
The above query assumes that id is the column used to group records together.
SQL Fiddle Demo
Use Ranking function Row_Number() :
SELECT *
FROM (SELECT *,
Row_number()
OVER(
partition BY GroupId
ORDER BY GroupId) AS [rn]
FROM YourTable) t
WHERE rn <= 3
Check this MSDN doc for details of all ranking functions.
There is a sql TOP statement that does this
SELECT TOP number|percent column_name(s) FROM table_name;
a description of what it does and how it is used in alternative sql statements for example for mysql and ms access can be found here: http://www.w3schools.com/sql/sql_top.asp
My bad i misread your question, this will return the top rows not groups, could you explain what you are trying to do in more detail?
SELECT *
FROM
(SELECT *
,ROW_NUMBER() OVER (PARTITION BY [Group] ORDER BY [Group] ASC)rn
FROM TableName
)A
WHERE rn <= 3
I used this code to select last 5 rows from table:
SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC
But I want to reorder this result ( the last five rows returned ) By Id ASC
How can I do this?
Use a subquery:
select t.*
from (SELECT TOP(5) * FROM tbl_reg ORDER BY Id DESC) t
order by id ASC;
Here is my query:
SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
This the result:
How can I reverse this table based on date (Column2) by using SQL?
You can use the first query to get the matching ids, and use them as part of an IN clause:
SELECT id, rssi1, date
FROM history
WHERE id IN
(
SELECT TOP 8 id
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC
)
ORDER BY date ASC
You could simply use a sub-query. If you apply a TOP clause the nested ORDER BY is allowed:
SELECT X.* FROM(
SELECT TOP 8 id, Column1, Column2
FROM dbo.History
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) X
ORDER BY Column2
Demo
The SELECT query of a subquery is always enclosed in parentheses. It
cannot include a COMPUTE or FOR BROWSE clause, and may only include an
ORDER BY clause when a TOP clause is also specified.
Subquery Fundamentals
try the below :
select * from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC ) aa order by aa.date DESC
didn't run it, but i think it should go well
WITH cte AS
(
SELECT id, rssi1, date, RANK() OVER (ORDER BY ID DESC) AS Rank
FROM history
WHERE (siteName = 'CCL03412')
)
SELECT id, rssi1, date
FROM cte
WHERE Rank <= 8
ORDER BY Date DESC
I have not run this but i think it will work. Execute and let me know if you face error
select id, rssi1, date from (SELECT TOP 8 id, rssi1, date
FROM history
WHERE (siteName = 'CCL03412')
ORDER BY id DESC) order by date ;
How do I retrieve the second highest value from a table?
select max(val) from table where val < (select max(val) form table)
In MySQL you could for instance use LIMIT 1, 1:
SELECT col FROM tbl ORDER BY col DESC LIMIT 1, 1
See the MySQL reference manual: SELECT Syntax).
The LIMIT clause can be used to constrain the number of rows returned by the SELECT statement. LIMIT takes one or two numeric arguments, which must both be nonnegative integer constants (except when using prepared statements).
With two arguments, the first argument specifies the offset of the first row to return, and the second specifies the maximum number of rows to return. The offset of the initial row is 0 (not 1):
SELECT * FROM tbl LIMIT 5,10; # Retrieve rows 6-15
select top 2 field_name from table_name order by field_name desc limit 1
SELECT E.lastname, E.salary FROM employees E
WHERE 2 = (SELECT COUNT(*) FROM employess E2
WHERE E2.salary > E.salary)
Taken from here
This works in almost all Dbs
Select Top 1 sq.ColumnToSelect
From
(Select Top 2 ColumnToSelect
From MyTable
Order by ColumnToSelect Desc
)sq
Order by sq.ColumnToSelect asc
Cool, this is almost like Code Golf.
Microsoft SQL Server 2005 and higher:
SELECT *
FROM (
SELECT
*,
row_number() OVER (ORDER BY var DESC) AS ranking
FROM table
) AS q
WHERE ranking = 2
Try this
SELECT * FROM
(SELECT empno, deptno, sal,
DENSE_RANK() OVER (PARTITION BY deptno ORDER BY sal DESC NULLS LAST) DENSE_RANK
FROM emp)
WHERE DENSE_RANK = 2;
This works in both Oracle and SQL Server.
Try this
SELECT TOP 1 Column FROM Table WHERE Column < (SELECT MAX(Column) FROM Table)
ORDER BY Column DESC
SELECT TOP 1 Column FROM (SELECT TOP <n> Column FROM Table ORDER BY Column DESC)
ORDER BY ASC
change the n to get the value of any position
Maybe:
SELECT * FROM table ORDER BY value DESC LIMIT 1, 1
one solution would be like this:
SELECT var FROM table ORDER BY var DESC LIMIT 1,1