From this table, I'm trying to determine the nation (s) that have the highest number of teams (a nation X has a team if it has at least one athlete from that country X).
driver(id,name, team, country)
This solution restores all countries in descending order. Would it be possible to ensure that only the one (s) with the most team (s) return and not all of them? I think you should use the 'max' command but I'm not sure.
SELECT (country) ,count(distinct team)
FROM driver
GROUP BY country
order by count(distinct team) DESC;
I would use your query as a CTE and then select from it like this -
WITH t AS
(
SELECT country, count(distinct team) cnt
FROM driver
GROUP BY country
)
SELECT country, cnt FROM t
WHERE cnt = (SELECT max(cnt) FROM t);
You can combine this with a window function:
with counts as (
SELECT country,
count(distinct team) as num_teams,
dense_rank() over (order by count(distinct team) desc) as rnk
FROM driver
GROUP BY country
)
select country, num_teams
from counts
where rnk = 1;
If you are using Postgres 14, you can use fetch first with the option with ties:
SELECT country,
count(distinct team) as num_teams
FROM driver
GROUP BY country
order by count(distinct team) desc
fetch first 1 rows with ties
If two countries have the same highest number of drivers, this would return both. Without the with ties option (which was introduced in Postgres 14) only one of them would be returned.
Related
Im trying to find the distributor with the highest and lowest quantity for each country
in two columns distributor with minimum quantity and maximum quantity
I have been able to get the information from other posts but it is in a column however I want it on a row per country
See http://sqlfiddle.com/#!17/448f6/2
Desired result
"country" "min_qty_name" "max_qty_name"
1. "Madagascar" "Leonard Cardenas" "Gwendolyn Mccarty"
2. "Malaysia" "Arsenio Knowles" "Yael Carter"
3. "Palau" "Brittany Burris" "Clark Weaver"
4. "Tanzania" "Levi Douglas" "Levi Douglas"
You can use subqueries:
select distinct country,
(select distributor_name
from product
where country = p.country
order by quantity limit 1) as min_qty_name,
(select distributor_name
from product
where country = p.country
order by quantity desc limit 1) as max_qty_name
from product p;
Fiddle
You can do it with cte too (result here)
WITH max_table AS
(
SELECT ROW_NUMBER() OVER (partition by country order by country,quantity DESC) AS rank,
country, quantity,distributor_name
FROM
product
),
min_table AS
(
SELECT ROW_NUMBER() OVER (partition by country order by country,quantity) AS rank,
country, quantity,distributor_name
FROM
product
)
SELECT m1.country,m2.distributor_name,m1.distributor_name
from max_table m1, min_table m2
where m1.country = m2.country
and m1.rank = 1 and m2.rank = 1
You can do this with a single sort and pass through the data as follows:
with min_max as (
select distinct country,
first_value(distributor_name) over w as min_qty_name,
last_value(distributor_name) over w as max_qty_name
from product
window w as (partition by country
order by quantity
rows between unbounded preceding
and unbounded following)
)
select *
from min_max
order by min_max;
Updated Fiddle
I have two tables. One is Transactions and the other is Tickets. In Tickets I have the Ticket_Number,the name of the Category(Theater,Cinema,Concert), the Price of the Ticket. In Transactions I also have the Ticket_Number. What i want to do is to Get a SUM of money for each Category, and then with that data I want to Select the Category with the most money.
I already managed to get the SUM for each category but I am stuck here
SELECT category, SUM (Tickets.Price) AS Price
FROM Tickets,Transactions
WHERE Tickets.ticket_num=Transactions.ticket_num
GROUP BY Category
ORDER BY Price DESC;
I know i can add LIMIT 1 but I know it's not correct because 2 or more values can be the same
Using ROW_NUMBER to generate a sequence based on the sum of the price. Then, restrict to only the matching aggregated row with the highest total price.
WITH cte AS (
SELECT category, SUM(t1.Price) AS Price,
ROW_NUMBER() OVER (ORDER BY SUM(t1.Price) DESC) rn
FROM Tickets t1
INNER JOIN Transactions t2
ON t1.ticket_num = t2.ticket_num
GROUP BY Category
)
SELECT category, Price
FROM cte
WHERE rn = 1
ORDER BY Price DESC;
Note that if you want to capture all categories tied for the highest price, should a tie occur, then replace ROW_NUMBER in the above CTE with RANK, keeping everything else the same.
What you are looking for is a window function DENSE_RANK() which will handle ties properly.
RANK() will also work for your case, but if you would like to extend it to get TOP N places with ties (where N > 1), dense rank is the way to go.
SELECT Category, Price
FROM (
SELECT
Category,
SUM(ti.Price) AS Price,
DENSE_RANK() OVER (ORDER BY SUM(ti.Price) DESC) AS rnk
FROM Tickets ti
INNER JOIN Transactions tr ON
ti.ticket_num = tr.ticket_num
GROUP BY Category
) t
WHERE rnk = 1
I've also replaced the old style and not recommended joining of tables as comma separated list in FROM clause to a proper INNER JOIN clause and assigned aliases to tables.
You can use rank() to rank the sums of the prices, more expensive first.
SELECT category,
price
FROM (SELECT category,
sum(tickets.price) price,
rank() OVER (ORDER BY sum(tickets.price) DESC) r
FROM tickets
INNER JOIN transactions
ON transactions.ticket_num = tickets.ticket_num
GROUP BY category) x
WHERE r = 1;
I also took the liberty to rewrite your join from the ancient comma style to a modern, clearer version.
I'v provided following query to find common records in 2 data sets but it's difficult for me to make sure about correctness of my query because of that I have a lot of data records in my DB.
Is it OK to implement Intersect between "Customers" & "Employees" tables using UNION ALL and apply GROUP BY on the result like below?
SELECT D.Country, D.Region, D.City
FROM (SELECT DISTINCT Country, Region, City
FROM Customers
UNION ALL
SELECT DISTINCT Country, Region, City
FROM Employees) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
So can we say that any record which exists in the result of this query also exists in the Intersect set between "Customers & Employees" tables AND any record that exists in Intersect set between "Customers & Employees" tables will be in the result of this query too?
So is it right to say any record in result of this query is in
"Intersect" set between "Customers & Employees" "AND" any record that
exist in "Intersect" set between "Customers & Employees" is in result
of this query too?
YES.
... Yes, but it won't be as efficient because you are filtering out duplicates three times instead of once. In your query you're
Using DISTINCT to pull unique records from employees
Using DISTINCT to pull unique records from customers
Combining both queries using UNION ALL
Using GROUP BY in your outer query to to filter the records you retrieved in steps 1,2 and 3.
Using INTERSECT will return identical results but more efficiently. To see for yourself you can create the sample data below and run both queries:
use tempdb
go
if object_id('dbo.customers') is not null drop table dbo.customers;
if object_id('dbo.employees') is not null drop table dbo.employees;
create table dbo.customers
(
customerId int identity,
country varchar(50),
region varchar(50),
city varchar(100)
);
create table dbo.employees
(
employeeId int identity,
country varchar(50),
region varchar(50),
city varchar(100)
);
insert dbo.customers(country, region, city)
values ('us', 'N/E', 'New York'), ('us', 'N/W', 'Seattle'),('us', 'Midwest', 'Chicago');
insert dbo.employees
values ('us', 'S/E', 'Miami'), ('us', 'N/W', 'Portland'),('us', 'Midwest', 'Chicago');
Run these queries:
SELECT D.Country, D.Region, D.City
FROM
(
SELECT DISTINCT Country, Region, City
FROM Customers
UNION ALL
SELECT DISTINCT Country, Region, City
FROM Employees
) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
SELECT Country, Region, City
FROM dbo.customers
INTERSECT
SELECT Country, Region, City
FROM dbo.employees;
Results:
Country Region City
----------- ---------- ----------
us Midwest Chicago
Country Region City
----------- ---------- ----------
us Midwest Chicago
If using INTERSECT is not an option OR you want a faster query you could improve the query you posted a couple different ways, such as:
Option 1: let GROUP BY handle ALL the de-duplication like this:
This is the same as what you posted but without the DISTINCTS
SELECT D.Country, D.Region, D.City
FROM
(
SELECT Country, Region, City
FROM Customers
UNION ALL
SELECT Country, Region, City
FROM Employees
) AS D
GROUP BY D.Country, D.Region, D.City
HAVING COUNT(*) = 2;
Option 2: Use ROW_NUMBER
This would be my preference and will likely be most efficient
SELECT Country, Region, City
FROM
(
SELECT
rn = row_number() over (partition by D.Country, D.Region, D.City order by (SELECT null)),
D.Country, D.Region, D.City
FROM
(
SELECT Country, Region, City
FROM Customers
UNION ALL
SELECT Country, Region, City
FROM Employees
) AS D
) uniquify
WHERE rn = 2;
Can you let me know how to select only two employees from every department? The table has deptname, ssn, name . I am doing a sampling and I need only two ssns for every department name. Can someone help?
You can accomplish this with an "OLAP expression" row_number()
with e as
( select deptname, ssn, empname,
row_number() over (partition by dptname order by empname) as pick
from employees
)
select deptname, ssn, empname
from e
where pick < 3
order by deptname, ssn
This example will give you the two employees with the lowest order names, because that is what is specified in the row_number() (order by) expression.
Try this:
select *
from t t1
where (
select count(*)
from t t2
where
t2.deptname = t1.deptname
and
t2.ssn <= t1.ssn) <= 2
order by deptname, ssn,name;
The above will give "smallest" two ssn.
If you want top 2, change to t2.ssn >= t1.ssn
sqlfiddle
The data:
The result from query:
select * from
( select rank() over (partition by dptname order by empname) as count , *
from employees
)
where count<=2
order by deptname, ssn,name;
I am selecting column used in group by and count, and query looks something like
SELECT s.country, count(*) AS posts_ct
FROM store s
JOIN store_post_map sp ON sp.store_id = s.id
GROUP BY 1;
However, I want to select some more fields, like store name or store address from store table where count is max, but I don't to include that in group by clause.
For instance, to get the stores with the highest post-count per country:
SELECT DISTINCT ON (s.country)
s.country, s.store_id, s.name, sp.post_ct
FROM store s
JOIN (
SELECT store_id, count(*) AS post_ct
FROM store_post_map
GROUP BY store_id
) sp ON sp.store_id = s.id
ORDER BY s.country, sp.post_ct DESC
Add any number of columns from store to the SELECT list.
Details about this query style in this related answer:
Select first row in each GROUP BY group?
Reply to comment
This produces the count per country and picks (one of) the store(s) with the highest post-count:
SELECT DISTINCT ON (s.country)
s.country, s.store_id, s.name
,sum(post_ct) OVER (PARTITION BY s.country) AS post_ct_for_country
FROM store s
JOIN (
SELECT store_id, count(*) AS post_ct
FROM store_post_map
GROUP BY store_id
) sp ON sp.store_id = s.id
ORDER BY s.country, sp.post_ct DESC;
This works because the window function sum() is applied before DISTINCT ON per definition.