Select record order by Birthday and assign Row Number in SQL Server

Hi PRA,

Refer below query.

SQL

DECLARE @InfoTable AS TABLE(Id INT IDENTITY(1,1),LastName VARCHAR(25),Name VARCHAR(25),Birthday date,Address VARCHAR(50))
     
INSERT INTO @InfoTable VALUES('Pulodov','Rustam','1987-12-22','city Dushanbe')
INSERT INTO @InfoTable VALUES('Soliev','Firuzjon','1988-06-16','city Hisor')
INSERT INTO @InfoTable VALUES('Sharipov','Sadriddin','1991-12-08','city Dushanbe')
INSERT INTO @InfoTable VALUES('Yakubi','Hasanzod','1991-05-15','city Dushanbe')
INSERT INTO @InfoTable VALUES('Sayfurov','Karomatullo','1980-05-07','city Dushanbe')
INSERT INTO @InfoTable VALUES('Asrori','Yatim','1978-02-06','city Dushanbe')
 
select row_number() over(order by Id) AS Id, LastName, Name, Birthday, CONVERT(VARCHAR,CONVERT(INT,DATEDIFF(DD,Birthday,GETDATE())/365.25))Age, Address 
from @InfoTable 
order by Birthday, Id

Output

SQL

select LastName, Name, Birthday, CONVERT(VARCHAR,CONVERT(INT,DATEDIFF(DD,Birthday,GETDATE())/365.25)) Age, Address 
INTO #Temp
from @InfoTable order by Birthday
SELECT row_number() over(order by (SELECT 0)) AS Id,* FROM #Temp
DROP TABLE #Temp

Output