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

Question

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
18/5/20         


        

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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 6

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.