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
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.
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;
}
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