SQL - 如何选择具有最大值列的行

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
18/5/2010, 6 pm        25 

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

Answer 1

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 2

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 3

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 4

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.

Answer 5

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 6

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 7

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

Answer 8

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.

Answer 9

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;
    }