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I'm trying to get users from a younger country for example I have the following tables.
If there is more than one user of the youngest who have the same age, they should also be shown
You can try this query, get
MIN birthday on subquery then
self join on users table.
select u.idcountry,t.name,u.username, (DATEPART(year, getdate()) - t.years) 'age' from ( SELECT u.idcountry,c.name,DATEPART(year, u.birthday) as 'years',count(*) as 'cnt' FROM users u inner join country c on u.idcountry = c.idcountry group by u.idcountry,c.name,DATEPART(year, u.birthday) ) t inner join users u on t.idcountry = u.idcountry and t.years = DATEPART(year, u.birthday) where t.cnt > 1
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rank() window function:
select ... from ... where rank() over (partition by idcountry order by birthday) = 1
Rows with the same birthday in a country are ranked the same, so this returns all youngest people with if there’s more than one.
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This is a little tricky. I would use window functions -- count the people of a particular age and choose the ones where there are duplicates for the youngest.
You don't specify how to define age, so I'll just use the earliest calendar year:
select u.* from (select u.*, count(*) over (partition by idcountry, year(birthday)) as cnt_cb, rank() over (partition by idcountry order by year(birthday)) as rnk from users u ) u where cnt_cb > 1 and rnk = 1;
I'll let you handle the joins to bring in the country name.
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Your sample data and desired results show the oldest users within each country when more than one of the oldest have the same age. The query below will do that, assuming age is calculated using full birth date.
WITH users AS ( SELECT username , birthday , idcountry , (CAST(CONVERT(char(8),GETDATE(),112) AS int) - CAST(CONVERT(char(8),birthday,112) AS int)) / 10000 AS age , RANK() OVER(PARTITION BY idcountry ORDER BY (CAST(CONVERT(char(8),GETDATE(),112) AS int) - CAST(CONVERT(char(8),birthday,112) AS int)) / 10000 DESC) AS age_rank FROM dbo.Users ) , oldest_users AS ( SELECT username , birthday , idcountry , age , COUNT(*) OVER(PARTITION BY idcountry, age_rank ORDER BY age_rank) AS age_count FROM users WHERE age_rank = 1 ) SELECT c.idcountry , c.name , oldest_users.age , oldest_users.username FROM oldest_users JOIN dbo.Country AS c ON c.idcountry = oldest_users.idcountry WHERE oldest_users.age_count > 1;
SELECT SQL QUERY USING CURRENT DATE TIMETO GET THE , Hi, I have a datable with a Field Birthday which is of Type DateTime and stores the Users Birth date. On my aspx page, I want to display the 99% Match on Sql Server Buy. Start searching with Visymo.com.
SQL Server: Find Users in SQL Server, Is there a query to run in SQL Server that will return all Users created? In SQL Server, there is a system view called sys.sysusers. You can run a query against The CURRENT_USER function returns the name of the current user in the SQL Server database. Technical Details. SQL Server (starting with 2008), Azure SQL Database, Azure SQL Data Warehouse, Parallel Data Warehouse. SQL Server Functions. If you want to report an error, or if you want to make a suggestion, do not hesitate to send us an e-mail:
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- Please don't post your tables as images. Use formatted text for this.
- Show the query you've tried..
- How do you define age?
- @gordon I think we have a language barrier here. Based on the example, I think OP wants the youngest, ie eariest, birthday, so the oldest people.
- @Bohemian . . . I wasn't actually referring to youngest vs. oldest. Ages don't have to be counted in complete years, so the calculation is not precise.
- This query uses birth year rather than age. Users born the same year will have different ages depending on whether the month and day are before or after today.
- Most commonly, age isn't just calculated from the first of the year in which they were born, but also the month and day. If my birthday were today, I would be one year older than I was yesterday.
- @DanGuzman . . . I wouldn't necessarily just use years for counting ages. The OP's example returns as a tie two people born in the same year. The OP should clarify the definition. Note: if the OP uses
datediff()in SQL Server, it is equivalent to the definition I am using.