SQL - How to select a row having a column with max value

date                 value

18/5/2010, 1 pm        40
18/5/2010, 2 pm        20
18/5/2010, 3 pm        60
18/5/2010, 4 pm        30
18/5/2010, 5 pm        60
18/5/2010, 6 pm        25 

i need to query for the row having max(value)(i.e. 60). So, here we get two rows. From that, I need the row with the lowest time stamp for that day(i.e 18/5/2010, 3 pm -> 60)

Keywords like TOP, LIMIT, ROWNUM, ...etc are database dependent. Please read this article for more information.

http://en.wikipedia.org/wiki/Select_(SQL)#Result_limits

Oracle: ROWNUM could be used.

select * from (select * from table 
order by value desc, date_column) 
where rownum = 1;

Answering the question more specifically:

select high_val, my_key
from (select high_val, my_key
      from mytable
      where something = 'avalue'
      order by high_val desc)
where rownum <= 1

Analytics! This avoids having to access the table twice:

SELECT DISTINCT
       FIRST_VALUE(date_col)  OVER (ORDER BY value_col DESC, date_col ASC),
       FIRST_VALUE(value_col) OVER (ORDER BY value_col DESC, date_col ASC)
FROM   mytable;

Answer is to add a having clause:

SELECT [columns]
FROM table t1
WHERE value= (select max(value) from table)
AND date = (select MIN(date) from table t2 where t1.value = t2.value)

this should work and gets rid of the neccesity of having an extra sub select in the date clause.

SQL> create table t (mydate,value)
  2  as
  3  select to_date('18/5/2010, 1 pm','dd/mm/yyyy, hh am'), 40 from dual union all
  4  select to_date('18/5/2010, 2 pm','dd/mm/yyyy, hh am'), 20 from dual union all
  5  select to_date('18/5/2010, 3 pm','dd/mm/yyyy, hh am'), 60 from dual union all
  6  select to_date('18/5/2010, 4 pm','dd/mm/yyyy, hh am'), 30 from dual union all
  7  select to_date('18/5/2010, 5 pm','dd/mm/yyyy, hh am'), 60 from dual union all
  8  select to_date('18/5/2010, 6 pm','dd/mm/yyyy, hh am'), 25 from dual
  9  /

Table created.

SQL> select min(mydate) keep (dense_rank last order by value) mydate
  2       , max(value) value
  3    from t
  4  /

MYDATE                   VALUE
------------------- ----------
18-05-2010 15:00:00         60

1 row selected.

Regards, Rob.

Technically, this is the same answer as @Sujee. It also depends on your version of Oracle as to whether it works. (I think this syntax was introduced in Oracle 12??)

SELECT *
FROM   table
ORDER BY value DESC, date_column ASC
FETCH  first 1 rows only;

As I say, if you look under the bonnet, I think this code is unpacked internally by the Oracle Optimizer to read like @Sujee's. However, I'm a sucker for pretty coding, and nesting select statements without a good reason does not qualify as beautiful!! :-P

In Oracle:

This gets the key of the max(high_val) in the table according to the range.

select high_val, my_key
from (select high_val, my_key
      from mytable
      where something = 'avalue'
      order by high_val desc)
where rownum <= 1

In Oracle DB:

create table temp_test1 (id number, value number, description varchar2(20));

insert into temp_test1 values(1, 22, 'qq');
insert into temp_test1 values(2, 22, 'qq');
insert into temp_test1 values(3, 22, 'qq');
insert into temp_test1 values(4, 23, 'qq1');
insert into temp_test1 values(5, 23, 'qq1');
insert into temp_test1 values(6, 23, 'qq1');

SELECT MAX(id), value, description FROM temp_test1 GROUP BY value, description;

Result:
    MAX(ID) VALUE DESCRIPTION
    -------------------------
    6	      23	qq1
    3	      22	qq

The simplest answer would be

--Setup a test table called "t1"

create table t1
(date datetime,
value int)

-- Load the data. -- Note: date format different than in the question

insert into t1
Select '5/18/2010 13:00',40
union all
Select '5/18/2010 14:00',20
union all
Select '5/18/2010 15:00',60 
union all
Select '5/18/2010 16:00',30 
union all
Select '5/18/2010 17:00',60 
union all
Select '5/18/2010 18:00',25 

-- find the row with the max qty and min date.

select *
from t1
where value = 
	(select max(value)  from t1)
and date = 
	(select min(date) 
	from t1
	where value = (select max(value)  from t1))

I know you can do the "TOP 1" answer, but usually your solution gets just complicated enough that you can't use that for some reason.

You can use this function, ORACLE DB

 public string getMaximumSequenceOfUser(string columnName, string tableName, string username)
    {
        string result = "";
        var query = string.Format("Select MAX ({0})from {1} where CREATED_BY = {2}", columnName, tableName, username.ToLower());

        OracleConnection conn = new OracleConnection(_context.Database.Connection.ConnectionString);
        OracleCommand cmd = new OracleCommand(query, conn);
        try
        {
            conn.Open();
            OracleDataReader dr = cmd.ExecuteReader();
            dr.Read();
            result = dr[0].ToString();
            dr.Dispose();
        }
        finally
        {
            conn.Close();
        }
        return result;
    }