SQL Query for timestamp

Question

I had following Table

CREATE TABLE Customer
    ( `Name` varchar(7), `Address` varchar(55), `City` varchar(15),`Contact` int,`timestamp` int)
;

INSERT INTO Customer
    (`Name`,`Address`, `City`, `Contact`,`timestamp`)
VALUES
    ('Jack','New City','LA',79878458,456125),
    ('Joseph','New Lane23','LA',87458458,794865),
   ('Rosy','Old City','Paris',79878458,215125),
   ('Maria','New City','LA',79878458,699125),
   ('Jack','New City','LA',79878458,456125),
   ('Rosy','Old City','Paris',79878458,845125),
   ('Jack','New Main Street','New York',79878458,555525),
   ('Joseph','Near Bank','SAn Francisco',79878458,984521)

;

I want to get all customer record with highest timestamp without duplication.

Answer 1

I want to get all customer record with highest timestamp without duplication.

Use DISTINCT operator and ORDER BY clause like

select distinct `Name`,`Address`, `City`, `Contact`,`timestamp`
from customer
order by `timestamp` desc;

In that case you can use JOIN query like

select t1.*
from customer t1 join
(select Name, max(`timestamp`) as maxstamp
 from customer
 group by Name) xx 
 on t1.Name = xx.Name
 and t1.`timestamp` = xx.maxstamp;