I am doind a free Stanford online course (which is pretty cool, you should check that out) and I've been racking my brains for the lest 2 days and can't find an answer to the following problem. Please help.
Question 4 Find names and grades of students who only have friends in the same grade. Return the result sorted by grade, then by name within each grade.
When I finally thought that I had the answer my query returned all the values from the table Friend.
This is the best I could come up with.
select h1.id, h1.name, h1.grade, h2.id, h2.name, h2.grade
from friend f1
join highschooler h1 on f1.id1 = h1.id
join highschooler h2 on f1.id2 = h2.id
where h1.grade = any (select h3.grade from friend f2
join highschooler h3 on f2.id1 = h3.id
where h3.id = f1.id1)
I'm required to run the query in SQL Lite. I'm using http://sqlfiddle.com to test my queries in SQL Lite and here is the sample data I'm using.
/* Create the schema for our tables */
create table Highschooler(ID int, name text, grade int);
create table Friend(ID1 int, ID2 int);
create table Likes(ID1 int, ID2 int);
/* Populate the tables with our data */
insert into Highschooler values (1510, 'Jordan', 9);
insert into Highschooler values (1689, 'Gabriel', 9);
insert into Highschooler values (1381, 'Tiffany', 9);
insert into Highschooler values (1709, 'Cassandra', 9);
insert into Highschooler values (1101, 'Haley', 10);
insert into Highschooler values (1782, 'Andrew', 10);
insert into Highschooler values (1468, 'Kris', 10);
insert into Highschooler values (1641, 'Brittany', 10);
insert into Highschooler values (1247, 'Alexis', 11);
insert into Highschooler values (1316, 'Austin', 11);
insert into Highschooler values (1911, 'Gabriel', 11);
insert into Highschooler values (1501, 'Jessica', 11);
insert into Highschooler values (1304, 'Jordan', 12);
insert into Highschooler values (1025, 'John', 12);
insert into Highschooler values (1934, 'Kyle', 12);
insert into Highschooler values (1661, 'Logan', 12);
insert into Friend values (1510, 1381);
insert into Friend values (1510, 1689);
insert into Friend values (1689, 1709);
insert into Friend values (1381, 1247);
insert into Friend values (1709, 1247);
insert into Friend values (1689, 1782);
insert into Friend values (1782, 1468);
insert into Friend values (1782, 1316);
insert into Friend values (1782, 1304);
insert into Friend values (1468, 1101);
insert into Friend values (1468, 1641);
insert into Friend values (1101, 1641);
insert into Friend values (1247, 1911);
insert into Friend values (1247, 1501);
insert into Friend values (1911, 1501);
insert into Friend values (1501, 1934);
insert into Friend values (1316, 1934);
insert into Friend values (1934, 1304);
insert into Friend values (1304, 1661);
insert into Friend values (1661, 1025);
insert into Friend select ID2, ID1 from Friend;
insert into Likes values(1689, 1709);
insert into Likes values(1709, 1689);
insert into Likes values(1782, 1709);
insert into Likes values(1911, 1247);
insert into Likes values(1247, 1468);
insert into Likes values(1641, 1468);
insert into Likes values(1316, 1304);
insert into Likes values(1501, 1934);
insert into Likes values(1934, 1501);
insert into Likes values(1025, 1101);
Thank you in advance.
Regards.
Cesar
So we want to find students for whom there are no students in other grades they have a friendship relationship, right? This is one way to express that:
select * from highschooler h
where not exists
(select 1 from highschooler h2 where h2.grade != h.grade and exists
(select 1 from friends f where (f.id1 = h.id or f.id2 = h.id) and (f.id1 = h2.id or f.id2 = h2.id)))
order by grade, name
EDIT: If you also require them to have at least one friend, you'll need to check for that too