回答 sql ex ru 编号 57

Question

您好，我正在尝试解决这个特定问题：

对于具有不可挽回的战斗损失且数据库中至少有三艘船的班级，显示班级的名称和沉没的船只数量。

基于此数据库： 参与二战的海军舰艇数据库正在考虑中。数据库由以下关系组成： 类（类，类型，国家，numGuns，孔，位移） 船舶（名称、等级、发射） 战斗（名称，日期） 结果（船，战斗，结果） 各类船舶都具有相同的总体设计。一个类通常被分配一个根据相应设计建造的第一艘船的名称，或者一个不同于数据库中任何船名的名称。其名称被分配到一个级别的船称为领导船。 Classes 关系包括类的名称、类型（可以是 bb 表示战舰，或 bc 表示战列巡洋舰）、舰艇建​​造的国家、主炮数量、火炮口径（口径以英寸为单位） , 和排量（重量，以吨为单位）。 Ships 关系包含有关船舶名称、相应类别的名称以及船舶下水年份的信息。 Battles 关系包含船只参与的战斗的名称和日期，而 Outcomes 关系 - 给定船只的战斗结果（可能是沉没、损坏或 OK，最后一个值表示船只在战斗中幸存下来并没有受到伤害）。 注意： 1) 结果关系可能包含船舶关系中不存在的船舶。 2）沉船无法参加后期战斗。 3) 由于历史原因，在许多练习中将领头船称为头船。4) 在 Outcomes 表中找到但不在 Ships 表中的船仍被考虑在数据库中。即使沉没也是如此。

这是我提交的答案，它确实生成了正确的结果，但它说“您的查询在主数据库上产生了正确的结果集，但第二次测试失败，正在检查数据库 * 错误的记录数（减 1）":

SELECT DISTINCT C.class, count(result)
  FROM classes c JOIN ships s ON
  c.class = s.class left join outcomes o on
   o.ship = s.name
WHERE result='sunk' 
   and c.class IN (
            SELECT DISTINCT c.class
            FROM classes c JOIN ships s ON
            c.class = s.class left join outcomes o on o.ship = s.name
           GROUP BY c.class
           HAVING count(name)>=3)
           group by c.class
           HAVING count(result) is not NULL

我不确定我哪里出错了://

Answer 1

我想我明白了。

由于您的查询会带来 T_Outcomes.C_Ships 中可能不存在于 T_Ships.C_Ships 中的信息，因此使用内连接可能会带来 T_Ships 和 T_Class 中存在但不存在于 T_Outcomes 中的信息。您的主表应该是 T_Outcomes，因为它应该包含所有 C_Ships。但是，您将拥有未分类的船只。也许：

SELECT COALESCE(TC.C_Class, 'No_Class') AS Class, COUNT(*) AS ShipsSunk FROM T_Outcomes T0 LEFT JOIN T_Ships TS ON T0.C_Ships = TS.C_Ships LEFT JOIN T_Classes TC on TS.C_Class = TC.C_Class WHERE C_Result = 'Sunk' GROUP BY COALESCE(TC.C_Class, 'No_Class') HAVING COUNT(*) >= 3;

Answer 2

根据问题描述

For classes having irreparable combat losses and at least three ships in the database, display the name of the class and the number of ships sunk.

• 适用于具有无法弥补的战斗损失的职业

这意味着该类应该至少有一艘沉船，您需要以下查询

  SELECT C.class, S.name, O.result FROM classes as C
  LEFT JOIN ships as S ON (C.class = S.class)
  LEFT JOIN outcomes as O ON (O.ship = S.name)
  WHERE O.result = 'sunk'
  UNION
  SELECT C.class, O.ship, O.result FROM classes as C
  LEFT JOIN outcomes as O ON (O.ship = C.class)
  WHERE O.result = 'sunk'

第一个查询将类与船只连接起来，并过滤船只以仅获取沉没的船只。

第二个查询获取未包含在船表中的主要船只，并过滤船只以仅获取沉没的船只。

稍后我们需要按类别对船只进行分组并计数

WITH sunk_ships AS (
  SELECT C.class, S.name, O.result FROM classes as C
  LEFT JOIN ships as S ON (C.class = S.class)
  LEFT JOIN outcomes as O ON (O.ship = S.name)
  WHERE O.result = 'sunk'
  UNION
  SELECT C.class, O.ship, O.result FROM classes as C
  LEFT JOIN outcomes as O ON (O.ship = C.class)
  WHERE O.result = 'sunk'
),
clasess_with_sunk_ships AS (
 SELECT sships.class, COUNT(sships.result) sunk_count FROM sunk_ships sships
 GROUP BY sships.class
)
SELECT * FROM clasess_with_sunk_ships

• 数据库中至少包含三艘船的类别

首先我们需要一份所有船只的清单

 SELECT C.class, S.name FROM classes as C
 JOIN ships as S ON (C.class = S.class)
 UNION
 SELECT C.class, O.ship FROM classes as C
 JOIN outcomes O ON (O.ship = C.class)

第一个查询将类与 ship 表连接起来。

第二个查询获取未包含在船舶表中的主要船舶

有了这个列表，我们需要按类别对船进行分组，并过滤​​至少有 3 艘船的类别

all_ships AS (
 SELECT C.class, S.name FROM classes as C
 JOIN ships as S ON (C.class = S.class)
 UNION
 SELECT C.class, O.ship FROM classes as C
 JOIN outcomes O ON (O.ship = C.class)
),
classes_with_valid_count AS (
 SELECT aships.class, COUNT(aships.name) count FROM all_ships as aships
 GROUP BY aships.class
 HAVING COUNT(aships.name) >= 3
)
SELECT * FROM classes_with_valid_count

clasess_with_sunk_ships和classes_with_valid_count的加入会给我们正确的答案

WITH sunk_ships AS (
  SELECT C.class, S.name, O.result FROM classes as C
  LEFT JOIN ships as S ON (C.class = S.class)
  LEFT JOIN outcomes as O ON (O.ship = S.name)
  WHERE O.result = 'sunk'
  UNION
  SELECT C.class, O.ship, O.result FROM classes as C
  LEFT JOIN outcomes as O ON (O.ship = C.class)
  WHERE O.result = 'sunk'
),
clasess_with_sunk_ships AS (
 SELECT sships.class, COUNT(sships.result) sunk_count FROM sunk_ships sships
 GROUP BY sships.class
),
all_ships AS (
 SELECT C.class, S.name FROM classes as C
 JOIN ships as S ON (C.class = S.class)
 UNION
 SELECT C.class, O.ship FROM classes as C
 JOIN outcomes O ON (O.ship = C.class)
),
classes_with_valid_count AS (
 SELECT aships.class, COUNT(aships.name) count FROM all_ships as aships
 GROUP BY aships.class
 HAVING COUNT(aships.name) >= 3
)
SELECT cwsships.class, cwsships.sunk_count FROM clasess_with_sunk_ships cwsships
JOIN classes_with_valid_count cwvcount ON (cwvcount.class = cwsships.class)