Select persons who do only the project with the same city - tsql

Question

So I have 3 tables: Management (…, personID, projectID //personID and projectID can dublicate) Project(id, city, …) and Person (id, name, city, …)
I need to output persons who only do projects with the same city.
Will be greatfull for any help or idea.

Answer 1

Here's a simple query with joins, group by, having, and count(distinct) that gets the desired results.
First, create and populate sample tables (Please save us this step in your future questions):
DECLARE #Project as table
(
id int,
city varchar(20)
);
DECLARE #Person as table
(
id int,
firstName varchar(10),
lastName varchar(10)
);
DECLARE #Managment as table
(
projectId int,
personId int
);
INSERT INTO #Project (id, city) VALUES
(1, 'Abilene'),
(2, 'Akron'),
(3, 'Albuquerque'),
(4, 'Alexandria'),
(5, 'Allentown');
INSERT INTO #Person (Id, firstName, lastName) VALUES
(1, 'Aaron', 'Carter'),
(2, 'Aaron', 'Eckhart'),
(3, 'Abbie', 'Cornish'),
(4, 'Ace', 'Young'),
(5, 'Adam', 'Brody');
INSERT INTO #Managment (projectId, personId) VALUES
(1, 1),
(1, 2),
(3, 3),
(3, 2),
(3, 5),
(4, 4),
(4, 1),
(5, 5);
The query:
SELECT personId, firstName, lastName, max(city) as city
FROM #Person As pe
JOIN #Managment As m
ON pe.Id = m.personId
JOIN #Project As pr
ON pr.Id = m.projectId
GROUP BY personId, firstName, lastName
HAVING COUNT(DISTINCT city) = 1
(I've added the city's name as a bonus if you wanted two cities you could go with max and min, more than that you can't) - note the usage of count(distinct city) in the having clause.
Results:
personId firstName lastName city
3 Abbie Cornish Albuquerque
4 Ace Young Alexandria