5

我有一个表模式

create table Location(
id int primary key,
city varchar(255),
state varchar(100),
country varchar(255)
);
create table Person( 
id int primary key,
name varchar(100)
);
create table Photographer(
id int primary key references Person(id) on update cascade on delete cascade,
livesIn int not null references Location(id) on update cascade on delete no action
);
create table Specialty(
photographer int references Photographer(id) on update cascade on delete cascade,
type enum('portrait','landscape','sport'),
primary key(photographer, type)
);
create table Photo(
id int primary key,
takenAt timestamp not null,
takenBy int references Photographer(id) on update cascade on delete no action,
photographedAt int references Location(id) on update cascade on delete no action
);
create table Appearance(
  shows int references Person(id) on update cascade on delete cascade,
isShownIn int references Photo(id) on update cascade on delete cascade,
primary key(shows, isShownIn)
);

我被两个问题困住了：

1) 照片仅显示居住在同一地点的摄影师。列出每张照片一次。也就是说，照片必须有摄影师的人，他们都需要住在同一个地方。

2) 哪些地点具有该地点的每张照片均由一位未出现在马萨诸塞州任何照片中的摄影师拍摄的属性？每个位置只显示城市，每个位置只显示一次。

我的尝试：1）

SELECT ph.id, ph.takenAt, ph.takenBy, ph.photographedAt FROM 
(SELECT * FROM Photo p, Appearance ap WHERE p.id = ap.isShownIn 
HAVING ap.shows IN (SELECT person.id FROM Person,Photographer WHERE person.id
photographer.id)) ph
WHERE ph.photographedAt = (SELECT location.id FROM location WHERE location.id = 
(SELECT livesIn FROM Photographer WHERE id = ph.takenBy))

2)

select distinct city from location where location.id in (
select photographedAt from photo, (select * from appearance where appearance.shows in
(select photographer.id from photographer)) ph
where photo.id = ph.isShownIn )
and location.state <> 'Massachusetts'

任何人都可以帮助创建这些查询吗？

4

2 回答 2

1

您的查询都是“列出具有 X 和 Y 属性的单个项目，其中 X 和 Y 在不同的表中”的种类。

这些类型的问题通常使用带有EXISTS和的相关子查询来解决NOT EXISTS。

使用EXISTS负责“仅显示每个项目一次”部分。否则，您将需要将分组与复杂的联接结合使用，这很快就会变得混乱。

问题 1 要求：

[...] 照片必须有摄影师，而且他们都需要住在同一个地方。

请注意，此定义并没有说“如果照片也包含其他人，则不要显示照片”。如果这就是您的真正意思，那么您可以从下面的 SQL 中得出结论，并在下次编写更好的定义。;)

SELECT
  *
FROM
  Photo p
WHERE
  EXISTS (
    -- ...that has at least one appearance of a photographer
    SELECT 
      1 
    FROM
      Appearance              a 
      INNER JOIN Photographer r ON r.id = a.shows
      INNER JOIN Location     l ON l.id = r.livesIn
    WHERE
      a.isShownIn = p.id
      -- AND l.id = <optional location filter would go here>
      AND NOT EXISTS (
        -- ...that does not have an appearance of a photographer from 
        --    some place else
        SELECT 
          1 
        FROM
          Appearance              a1 
          INNER JOIN Photographer r1 ON r1.id = a1.shows
          INNER JOIN Location     l1 ON l1.id = r1.livesIn
        WHERE
          a1.isShownIn = p.Id
          AND l1.id <> l.id
      )
  )

第二个问题是

[...] 地点是否拥有该地点的每张照片均由一位未出现在马萨诸塞州任何照片中的摄影师拍摄的财产？每个位置只显示城市，每个位置只显示一次。

相应的 SQL 如下所示：

SELECT
  city
FROM
  Location l
WHERE
  NOT EXISTS (
    -- ...a photo at this location taken by a photographer who makes 
    --    an apperance on another photo which which was taken in Massachusetts
    SELECT
      1
    FROM
      Photo                    p
      INNER JOIN Photographer  r ON r.id = p.takenBy
      INNER JOIN Appearance    a ON a.shows = r.id
      INNER JOIN Photo        p1 ON p1.id = a.isShownIn
    WHERE
      p.photographedAt = l.Id
      AND p1.photographedAt = <the location id of Massachusetts>
  )
于 2012-10-08T07:27:02.377 回答
0

我对 Query1 的尝试。显示居住在同一城市的摄影师的照片。

select ph.id, ph.takenAt, ph.takenBy, ph.photographedAt from Photo as ph 
   join Appearance as a on ph.id = a.isShownIn 
   join Photographer as p on a.shows = p.id where p.livesIn in 
      (select p1.id from Photographer as p1, Photographer as p2 
         where p1.id != p2.id and p1.livesIn = p2.livesIn);

我对 Query2 的尝试。参考在马萨诸塞州拍摄的照片中的人物，然后列出所有不是该人拍摄的照片。

select * from Photo where takenBy not in 
   (select a.shows from Photo as ph 
      join Location as l on ph.photographedAt = l.id 
      join Appearance as a on a.isShownIn = ph.id 
      where city = 'Massachusets');

希望有帮助。

于 2012-10-07T22:26:12.103 回答