SQL - 查找僅喜歡同一年級學生的學生的成績

Question

我在做一個免費的斯坦福大學在線課程（這很酷，你應該檢查一下），我一直在絞盡腦汁，天，無法找到以下問題的答案。請幫忙。

問題4 查找只有朋友在同一年級的學生的名字和成績。返回按年級排序的結果，然後按照每個年級的名稱進行排序。

當我終於認爲我有答案時，我的查詢返回表Friend中的所有值。

這是我能想到的最好的。

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) 

我需要在SQL Lite中運行查詢。 我使用http://sqlfiddle.com在SQL Lite中測試我的查詢，這裏是我正在使用的示例數據。

/* 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); 

在此先感謝您。

問候。

塞薩爾

Answer 1

因此，我們要找到的人有在他們有一個友誼關係的其他牌號無生的學生，對吧？這是一個表達方式：

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 

編輯：如果你還要求他們至少有一個朋友，你需要檢查太

Answer 2

我的解決辦法：

 
SELECT name, grade 
FROM Highschooler 
WHERE ID NOT IN 
(SELECT ID1 
FROM Friend F1 JOIN Highschooler H1 
ON H1.ID = F1.ID1 
JOIN Highschooler H2 
ON H2.ID = F1.ID2 
WHERE H1.grade <> H2.grade) 
ORDER BY grade, name 

實質上，內部子查詢返回學生與具有不同成績的朋友的關係（其中H1.grade < >爲H2.grade）。然後，外部查詢只列出了這個內部關係中沒有特徵的所有學生。