Select newest records that have distinct Name column

Question

I did search around and I found this SQL selecting rows by most recent date with two unique columns Which is so close to what I want but I can't seem to make it work.

I get an error Column 'ID' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause.

I want the newest row by date for each Distinct Name

Select ID,Name,Price,Date
From  table
Group By Name
Order By Date ASC

Here is an example of what I want

Table

ID	Name	Price	Date
0	A	10	2012-05-03
1	B	9	2012-05-02
2	A	8	2012-05-04
3	C	10	2012-05-03
4	B	8	2012-05-01

desired result

ID	Name	Price	Date
2	A	8	2012-05-04
3	C	10	2012-05-03
1	B	9	2012-05-02

I am using Microsoft SQL Server 2008

Answer 1

Select ID,Name, Price,Date
From  temp t1
where date = (select max(date) from temp where t1.name =temp.name)
order by date desc

Here is a SQL Fiddle with a demo of the above

---

Or as Conrad points out you can use an INNER JOIN (another SQL Fiddle with a demo) :

SELECT t1.ID, t1.Name, t1.Price, t1.Date 
FROM   temp t1 
INNER JOIN 
(
    SELECT Max(date) date, name
    FROM   temp 
    GROUP BY name 
) AS t2 
    ON t1.name = t2.name
    AND t1.date = t2.date 
ORDER BY date DESC 

Answer 2

There a couple ways to do this. This one uses ROW_NUMBER. Just partition by Name and then order by what you want to put the values you want in the first position.

WITH cte 
     AS (SELECT Row_number() OVER (partition BY NAME ORDER BY date DESC) RN, 
                id, 
                name, 
                price, 
                date 
         FROM   table1) 
SELECT id, 
       name, 
       price, 
       date 
FROM   cte 
WHERE  rn = 1 

DEMO

Note you should probably add ID (partition BY NAME ORDER BY date DESC, ID DESC) in your actual query as a tie-breaker for date

Answer 3

select * from (
    Select
        ID, Name, Price, Date,
        Rank() over (partition by Name order by Date) RankOrder
    From table
) T
where RankOrder = 1

Answer 4

Use Distinct instead of Group By

Select Distinct ID,Name,Price,Date
From  table
Order By Date ASC

http://technet.microsoft.com/en-us/library/ms187831.aspx