文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

Table: Teams

+---------------+----------+ | Column Name | Type | +---------------+----------+ | team_id | int | | team_name | varchar | +---------------+----------+ 此表的主鍵是 team_id，表中的每一行都代表一支獨(dú)立足球隊(duì)。

Table: Matches

+---------------+---------+ | Column Name | Type | +---------------+---------+ | match_id | int | | host_team | int | | guest_team | int | | host_goals | int | | guest_goals | int | +---------------+---------+ 此表的主鍵是 match_id，表中的每一行都代表一場(chǎng)已結(jié)束的比賽， 比賽的主客隊(duì)分別由它們自己的 id 表示， 他們的進(jìn)球由 host_goals 和 guest_goals 分別表示。積分規(guī)則如下：贏一場(chǎng)得三分； 平一場(chǎng)得一分； 輸一場(chǎng)不得分。

寫(xiě)出一條SQL語(yǔ)句以查詢每個(gè)隊(duì)的 team_id，team_name 和 num_points。
結(jié)果根據(jù) num_points 降序排序，如果有兩隊(duì)積分相同，那么這兩隊(duì)按 team_id 升序排序。

查詢結(jié)果格式如下：

Teams table: +-----------+--------------+ | team_id | team_name | +-----------+--------------+ | 10 | Leetcode FC | | 20 | NewYork FC | | 30 | Atlanta FC | | 40 | Chicago FC | | 50 | Toronto FC | +-----------+--------------+Matches table: +------------+--------------+---------------+-------------+--------------+ | match_id | host_team | guest_team | host_goals | guest_goals | +------------+--------------+---------------+-------------+--------------+ | 1 | 10 | 20 | 3 | 0 | | 2 | 30 | 10 | 2 | 2 | | 3 | 10 | 50 | 5 | 1 | | 4 | 20 | 30 | 1 | 0 | | 5 | 50 | 30 | 1 | 0 | +------------+--------------+---------------+-------------+--------------+Result table: +------------+--------------+---------------+ | team_id | team_name | num_points | +------------+--------------+---------------+ | 10 | Leetcode FC | 7 | | 20 | NewYork FC | 3 | | 50 | Toronto FC | 3 | | 30 | Atlanta FC | 1 | | 40 | Chicago FC | 0 | +------------+--------------+---------------+

來(lái)源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/team-scores-in-football-tournament
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2. 解題

• 按照主隊(duì)，客隊(duì)分別計(jì)算完 合并，再group by，left join 隊(duì)名表

# Write your MySQL query statement below select Teams.team_id, Teams.team_name, ifnull(Scores.score, 0) num_points from Teams left join (select team, sum(score) scorefrom((select host_team team, (case when host_goals > guest_goals then 3when host_goals = guest_goals then 1else 0 end) scorefrom Matches)union all(select guest_team team, (case when host_goals > guest_goals then 0when host_goals = guest_goals then 1else 3 end) scorefrom Matches)) tempgroup by team ) Scores on Teams.team_id = Scores.team order by num_points desc, team_id

690 ms

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