Question

我正在做斯坦福免费在线课程（我推荐！），我遇到了以下问题。请帮忙。

找到与名为Gabriel的人成为朋友的所有学生的姓名。

这是我能想到的最好的。

select name 
from highschooler h 
where h.id exists in (
    select friend.id1 
    from friend 
    where friend.id2 exists in (
        select h.id 
        from highschooler h2 
        where h2.name="gabriel"
    )
);

我需要在SQL Lite中运行查询，尽管我使用MySQL工作台作为测试位置（我已经理解它们对于像这样的基本查询它们是相对类似的）。

/* 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);

非常感谢任何帮助！

Answer 1

你需要两次引用Highschooler表，基本上建立关系Highschooler - ＆gt;朋友 - ＆gt; Highschooler

以下是如何操作：

select h1.name
from Highschooler h1
inner join Friend f on f.ID1 = h1.ID
inner join Highschooler h2 on f.ID2 = h2.ID
where h2.name = 'Gabriel'

这是一个SQL Fiddle，上面有一个查询。

Answer 2

我建议你做的是把它分解成几块并放回原处。

首先，让我们获取所有名为gabriel的人的身份证明：

SELECT s.id
FROM highschooler s
WHERE s.name = 'Gabriel';

现在，我们可以找到此id存在的所有友谊列表，如id1：

SELECT f.id2
FROM friend f
WHERE f.id1 IN(SELECT s.id
                FROM highschooler s
                WHERE s.name = 'Gabriel');

由于您已经设置了表格以包含列出同一个朋友的两种方式，因此您不必担心相反的情况（f.id2是Gabriel）。

现在，您可以使用highschooler表加入上述查询以获取其名称。

SELECT s.name
FROM highschooler s
JOIN(SELECT f.id2
    FROM friend f
    WHERE f.id1 IN(SELECT s.id
                    FROM highschooler s
                    WHERE s.name = 'Gabriel')
     ) t
ON t.id2 = s.id;

这是。{/ p>的SQL Fiddle示例

具有双嵌套级别的SQL查询

2 个答案: