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中等
数据库

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题目描述

表: Teams

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| team_id        | int     |
| team_name      | varchar |
+----------------+---------+
team_id 是该表主键.
每一行都包含了一个参加联赛的队伍信息.

 

表: Matches

+-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| home_team_id    | int     |
| away_team_id    | int     |
| home_team_goals | int     |
| away_team_goals | int     |
+-----------------+---------+
(home_team_id, away_team_id) 是该表主键.
每一行包含了一次比赛信息.
home_team_goals 代表主场队得球数.
away_team_goals 代表客场队得球数.
获得球数较多的队伍为胜者队伍.

 

写一段SQL,用来报告联赛信息. 统计数据应使用已进行的比赛来构建,其中 获胜 球队获得 三分 ,而失败球队获得 零分 。如果 打平 ,两支球队都得 一分 

result 表的每行应包含以下信息:

  • team_name - Teams 表中的队伍名字
  • matches_played - 主场与客场球队进行的比赛次数.
  • points - 球队获得的总分数.
  • goal_for - 球队在所有比赛中获取的总进球数
  • goal_against - 球队在所有比赛中,他的对手球队的所有进球数
  • goal_diff - goal_for - goal_against.

points 降序 返回结果表。 如果两队或多队得分相同,则按 goal_diff 降序 排列。 如果仍然存在平局,则以 team_name 按字典顺序 排列它们。

查询的结果格式如下例所示。

 

示例 1:

输入:
Teams 表:
+---------+-----------+
| team_id | team_name |
+---------+-----------+
| 1       | Ajax      |
| 4       | Dortmund  |
| 6       | Arsenal   |
+---------+-----------+
Matches 表:
+--------------+--------------+-----------------+-----------------+
| home_team_id | away_team_id | home_team_goals | away_team_goals |
+--------------+--------------+-----------------+-----------------+
| 1            | 4            | 0               | 1               |
| 1            | 6            | 3               | 3               |
| 4            | 1            | 5               | 2               |
| 6            | 1            | 0               | 0               |
+--------------+--------------+-----------------+-----------------+
输出:
+-----------+----------------+--------+----------+--------------+-----------+
| team_name | matches_played | points | goal_for | goal_against | goal_diff |
+-----------+----------------+--------+----------+--------------+-----------+
| Dortmund  | 2              | 6      | 6        | 2            | 4         |
| Arsenal   | 2              | 2      | 3        | 3            | 0         |
| Ajax      | 4              | 2      | 5        | 9            | -4        |
+-----------+----------------+--------+----------+--------------+-----------+
解释:
Ajax (team_id=1) 有4场比赛: 2败2平. 总分数 = 0 + 0 + 1 + 1 = 2.
Dortmund (team_id=4) 有2场比赛: 2胜. 总分数 = 3 + 3 = 6.
Arsenal (team_id=6) 有2场比赛: 2平. 总分数 = 1 + 1 = 2.
Dortmund 是积分榜上的第一支球队. Ajax和Arsenal 有同样的分数, 但Arsenal的goal_diff高于Ajax, 所以Arsenal在表中的顺序在Ajaxzhi'qian.

解法

方法一

MySQL

# Write your MySQL query statement below
WITH
    Scores AS (
        SELECT
            home_team_id AS team_id,
            CASE
                WHEN home_team_goals > away_team_goals THEN 3
                WHEN home_team_goals < away_team_goals THEN 0
                ELSE 1
            END AS score,
            home_team_goals AS goals,
            away_team_goals AS away_goals
        FROM Matches
        UNION ALL
        SELECT
            away_team_id AS team_id,
            CASE
                WHEN home_team_goals > away_team_goals THEN 0
                WHEN home_team_goals < away_team_goals THEN 3
                ELSE 1
            END AS score,
            away_team_goals AS goals,
            home_team_goals AS away_goals
        FROM Matches
    )
SELECT
    team_name,
    COUNT(1) AS matches_played,
    SUM(score) AS points,
    SUM(goals) AS goal_for,
    SUM(away_goals) AS goal_against,
    (SUM(goals) - SUM(away_goals)) AS goal_diff
FROM
    Scores AS s
    JOIN Teams AS t ON s.team_id = t.team_id
GROUP BY s.team_id
ORDER BY points DESC, goal_diff DESC, team_name;