Get the latest records per Group By SQL

Question

I have the following table:

CREATE TABLE orders (
    id INT PRIMARY KEY IDENTITY,
    oDate DATE NOT NULL,
    oName VARCHAR(32) NOT NULL,
    oItem INT,
    oQty INT
    -- ...
);


INSERT INTO orders
  VALUES
(1, '2016-01-01', 'A', 1, 2),
(2, '2016-01-01', 'A', 2, 1),
(3, '2016-01-01', 'B', 1, 3),
(4, '2016-01-02', 'B', 1, 2),
(5, '2016-01-02', 'C', 1, 2),
(6, '2016-01-03', 'B', 2, 1),
(7, '2016-01-03', 'B', 1, 4),
(8, '2016-01-04', 'A', 1, 3)
;

I want to get the most recent rows (of which there might be multiple) for each name. For the sample data, the results should be:

id	oDate	oName	oItem	oQty	...
5	2016-01-02	C	1	2
6	2016-01-03	B	2	1
7	2016-01-03	B	1	4
8	2016-01-04	A	1	3

The query might be something like:

SELECT oDate, oName, oItem, oQty, ...
  FROM orders
  WHERE oDate = ???
  GROUP BY oName
  ORDER BY oDate, id

Besides missing the expression (represented by ???) to calculate the desired values for oDate, this statement is invalid as it selects columns that are neither grouped nor aggregates.

Does anyone know how to do get this result?

Bicorn

The result is not "latest record for each name" since there are duplicate names. – Poteen

@Bicorn add primary key buddy it will solve your problem and you can order by that. – Agincourt

@androidGenX, Hi. Just edit the table. I have identity column on my table. Cheers. – Bicorn

@Bicorn check an answer. – Hebron

Possible duplicate of How can I SELECT rows with MAX(Column value), DISTINCT by another column in SQL? – Prostration

Answer 1

The rank window clause allows you to, well, rank rows according to some partitioning, and then you could just select the top ones:

SELECT oDate, oName, oItem, oQty, oRemarks
FROM   (SELECT oDate, oName, oItem, oQty, oRemarks,
               RANK() OVER (PARTITION BY oName ORDER BY oDate DESC) AS rk
        FROM   my_table) t
WHERE  rk = 1

Formaldehyde

RANK could return several rows with the same value (1 in this case), so this subselect could return several rows, what is probably wanted behaviour for the question. But, If you need to get single "latest" record, then ROW_NUMBER instead. Check SO answer: https://mcmap.net/q/66571/-sql-rank-versus-row_number – Hideout

I didn't know about the RANK() function until I saw it here. I'm using it with an IDENTITY column so it works beautifully! – Omland

Answer 2

This is a generic query without using analytical function.

SQLFiddle Demo

SELECT a.*
FROM table1 a
INNER JOIN
    (SELECT max(odate) modate,
          oname,
          oItem
   FROM table1
   GROUP BY oName,
            oItem
    ) 
    b ON a.oname=b.oname
AND a.oitem=b.oitem
AND a.odate=b.modate

Coaster

Answer 3

I think you need a query like this:

SELECT *
FROM (SELECT *,
        ROW_NUMBER() OVER (PARTITION BY oName ORDER BY oDate DESC) seq
    FROM yourTable) t
WHERE (seq <= 2)
ORDER BY oDate; 

Demavend

Answer 4

Add a primary key suppose id field to the table and make it auto increment,. Then order by id you will get it. It is the traditional way. By using your table you can only order by oDate. But is is having same date multiple times, so it also won't solve your problem.

Agincourt

Answer 5

You have to use ROW_NUMBER in following:

select oDate, oName, oItem, oQty, oRemarks
from (
   select *, row_number() over(partition by oName, oItem order by oDate desc) rn  
   from #t
)x
where rn = 1
order by oDate

OUTPUT

oDate   oName   oItem   oQty    oRemarks    
2016-01-01  A   001     2                   
2016-01-01  A   002     1       test        
2016-01-02  C   001     2                   
2016-01-03  B   001     4                   
2016-01-03  B   002     1                   

Coffeecolored