Question

What is the simplest SQL query to find the second largest integer value in a specific column?

There are maybe duplicate values in the column.

Was it helpful?

Solution

SELECT MAX( col )
  FROM table
 WHERE col < ( SELECT MAX( col )
                 FROM table )

OTHER TIPS

SELECT MAX(col) FROM table WHERE col NOT IN (SELECT MAX(col) FROM table);

In T-Sql there are two ways:

--filter out the max
select max( col )
from [table]
where col < ( 
    select max( col )
    from [table] )

--sort top two then bottom one
select top 1 col 
from (
    select top 2 col 
    from [table]
    order by col) topTwo
order by col desc 

In Microsoft SQL the first way is twice as fast as the second, even if the column in question is clustered.

This is because the sort operation is relatively slow compared to the table or index scan that the max aggregation uses.

Alternatively, in Microsoft SQL 2005 and above you can use the ROW_NUMBER() function:

select col
from (
    select ROW_NUMBER() over (order by col asc) as 'rowNum', col
    from [table] ) withRowNum 
where rowNum = 2

I see both some SQL Server specific and some MySQL specific solutions here, so you might want to clarify which database you need. Though if I had to guess I'd say SQL Server since this is trivial in MySQL.

I also see some solutions that won't work because they fail to take into account the possibility for duplicates, so be careful which ones you accept. Finally, I see a few that will work but that will make two complete scans of the table. You want to make sure the 2nd scan is only looking at 2 values.

SQL Server (pre-2012):

SELECT MIN([column]) AS [column]
FROM (
    SELECT TOP 2 [column] 
    FROM [Table] 
    GROUP BY [column] 
    ORDER BY [column] DESC
) a

MySQL:

SELECT `column` 
FROM `table` 
GROUP BY `column` 
ORDER BY `column` DESC 
LIMIT 1,1

Update:

SQL Server 2012 now supports a much cleaner (and standard) OFFSET/FETCH syntax:

SELECT TOP 2 [column] 
FROM [Table] 
GROUP BY [column] 
ORDER BY [column] DESC
OFFSET 1 ROWS
FETCH NEXT 1 ROWS ONLY;

I suppose you can do something like:

SELECT * FROM Table ORDER BY NumericalColumn DESC LIMIT 1 OFFSET 1

or

SELECT * FROM Table ORDER BY NumericalColumn DESC LIMIT (1, 1)

depending on your database server. Hint: SQL Server doesn't do LIMIT.

you can find the second largest value of column by using the following query

SELECT *
FROM TableName a
WHERE
  2 = (SELECT count(DISTINCT(b.ColumnName))
       FROM TableName b WHERE
       a.ColumnName <= b.ColumnName);

you can find more details on the following link

http://www.abhishekbpatel.com/2012/12/how-to-get-nth-maximum-and-minimun.html

The easiest would be to get the second value from this result set in the application:

SELECT DISTINCT value FROM Table ORDER BY value DESC LIMIT 2

But if you must select the second value using SQL, how about:

SELECT MIN(value) FROM (SELECT DISTINCT value FROM Table ORDER BY value DESC LIMIT 2) AS t

A very simple query to find the second largest value

SELECT `Column` FROM `Table` ORDER BY `Column` DESC LIMIT 1,1;
SELECT MAX(Salary) FROM Employee WHERE Salary NOT IN (SELECT MAX(Salary) FROM Employee )

This query will return the maximum salary, from the result - which not contains maximum salary from overall table.

Old question I know, but this gave me a better exec plan:

 SELECT TOP 1 LEAD(MAX (column)) OVER (ORDER BY column desc)
 FROM TABLE 
 GROUP BY column

select * from (select ROW_NUMBER() over (Order by Col_x desc) as Row, Col_1
    from table_1)as table_new tn inner join table_1 t1
    on tn.col_1 = t1.col_1
where row = 2

Hope this help to get the value for any row.....

Simplest of all

select sal from salary order by sal desc limit 1 offset 1

This is very simple code, you can try this :-

ex : Table name = test

salary 

1000
1500
1450
7500

MSSQL Code to get 2nd largest value

select salary from test order by salary desc offset 1 rows fetch next 1 rows only;

here 'offset 1 rows' means 2nd row of table and 'fetch next 1 rows only' is for show only that 1 row. if you dont use 'fetch next 1 rows only' then it shows all the rows from the second row.

select min(sal) from emp where sal in 
    (select TOP 2 (sal) from emp order by sal desc)

Note

sal is col name
emp is table name

Tom, believe this will fail when there is more than one value returned in select max([COLUMN_NAME]) from [TABLE_NAME] section. i.e. where there are more than 2 values in the data set.

Slight modification to your query will work -

select max([COLUMN_NAME]) from [TABLE_NAME] where [COLUMN_NAME] **IN** 
  ( select max([COLUMN_NAME]) from [TABLE_NAME] )
select max(COL_NAME) from TABLE_NAME where COL_NAME in 
    (select COL_NAME from TABLE_NAME where COL_NAME < (select max(COL_NAME) from TABLE_NAME));

subquery returns all values other than the largest. select the max value from the returned list.

select col_name
from (
    select dense_rank() over (order by col_name desc) as 'rank', col_name
    from table_name ) withrank 
where rank = 2
SELECT 
    * 
FROM 
    table 
WHERE 
    column < (SELECT max(columnq) FROM table) 
ORDER BY 
    column DESC LIMIT 1

It is the most esiest way:

SELECT
      Column name
FROM
      Table name 
ORDER BY 
      Column name DESC
LIMIT 1,1
select age from student group by id having age<(select max(age) from student)order by age limit 1

As you mentioned duplicate values . In such case you may use DISTINCT and GROUP BY to find out second highest value

Here is a table

salary

:

enter image description here

GROUP BY

SELECT  amount FROM  salary 
GROUP by amount
ORDER BY  amount DESC 
LIMIT 1 , 1

DISTINCT

SELECT DISTINCT amount
FROM  salary 
ORDER BY  amount DESC 
LIMIT 1 , 1

First portion of LIMIT = starting index

Second portion of LIMIT = how many value

SELECT MAX(sal) FROM emp
WHERE sal NOT IN (SELECT top 3 sal FROM emp order by sal desc )

this will return the third highest sal of emp table

MSSQL

SELECT  *
  FROM [Users]
    order by UserId desc OFFSET 1 ROW 
FETCH NEXT 1 ROW ONLY;

MySQL

SELECT  *
  FROM Users
    order by UserId desc LIMIT 1 OFFSET 1

No need of sub queries ... just skip one row and select second rows after order by descending

Something like this? I haven't tested it, though:

select top 1 x
from (
  select top 2 distinct x 
  from y 
  order by x desc
) z
order by x

See How to select the nth row in a SQL database table?.

Sybase SQL Anywhere supports:

SELECT TOP 1 START AT 2 value from table ORDER BY value

Using a correlated query:

Select * from x x1 where 1 = (select count(*) from x where x1.a < a)
select * from emp e where 3>=(select count(distinct salary)
    from emp where s.salary<=salary)

This query selects the maximum three salaries. If two emp get the same salary this does not affect the query.

select top 1 MyIntColumn from MyTable
where
 MyIntColumn <> (select top 1 MyIntColumn from MyTable order by MyIntColumn desc)
order by MyIntColumn desc

This works in MS SQL:

select max([COLUMN_NAME]) from [TABLE_NAME] where [COLUMN_NAME] < 
 ( select max([COLUMN_NAME]) from [TABLE_NAME] )

Query to find the 2nd highest number in a row-

select Top 1 (salary) from XYZ
where Salary not in (select distinct TOP 1(salary) from XYZ order by Salary desc)
ORDER BY Salary DESC

By changing the highlighted Top 1 to TOP 2, 3 or 4 u can find the 3rd, 4th and 5th highest respectively.

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