SQL Select newest records that have distinct Name column

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

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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 

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

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