使用 MySQL 为每个返回的行执行相关子查询
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【中文标题】使用 MySQL 为每个返回的行执行相关子查询【英文标题】:Execute a correlated subquery for each returned row with MySQL 【发布时间】:2017-05-31 08:43:42 【问题描述】:我正在尝试使用 mysql 计算一些体育运动员的统计数据。该数据库有 3 个表。
相关Rextester:http://rextester.com/SNAL27886
玩家
球员名单。
+----+---------+-----------+
| id | team_id | lastname |
+----+---------+-----------+
| 1 | 1 | Moubandje |
| 2 | 2 | Rüfli |
| 3 | 3 | Selnaes |
| 4 | 1 | Somália |
| 5 | 4 | Kerbrat |
+----+---------+-----------+
匹配
团队列表。
+----+--------------+--------------+-----+
| id | home_team_id | away_team_id | day |
+----+--------------+--------------+-----+
| 1 | 1 | 2 | 1 |
| 2 | 2 | 1 | 2 |
| 3 | 2 | 3 | 3 |
| 4 | 3 | 4 | 4 |
| 5 | 3 | 5 | 5 |
+----+--------------+--------------+-----+
player_match
每场比赛球员的统计数据。
+-----------+----------+-----------+------------+-------+
| player_id | match_id | rating | substitute | goals |
+-----------+----------+-----------+------------+-------+
| 1 | 1 | 6 | 0 | 2 |
| 2 | 2 | 5 | 1 | 0 |
| 1 | 10 | 3 | 0 | 0 |
+-----------+----------+---------+-----------+----------+
这是我的查询,用于计算有关球员的各种统计数据(例如他的进球数或他的全球平均评分):
SELECT
p.id AS p_id,
p.lastname AS lastname,
p.team_id as team_id,
AVG(pm.rating) AS avg_rating,
COUNT(pm.player_id) AS nb_matches,
SUM(pm.substitute) AS nb_matches_substitute,
SUM(pm.goals) AS goals,
(SUM(pm.goals) / COUNT(pm.player_id)) AS goals_per_matches
FROM
player p
INNER JOIN player_match pm ON pm.player_id = p.id
INNER JOIN `match` m ON pm.match_id = m.id AND (m.home_team_id = p.team_id OR m.away_team_id = p.team_id)
GROUP BY
p.id,
p.lastname,
p.team_id
ORDER BY
avg_rating DESC, lastname ASC
;
我还想计算该球员在他的球队最近 5 场比赛中的平均评分(如果该球员没有参加过比赛,他的评分必须为 0)。然后,我想按这个特定的平均评分对列表的结果进行排序。
这是我的查询,用于检索给定球员在他的球队最近 5 场比赛中的平均评分以及每场比赛的评分作为字符串:
SELECT
SUM(pm1.rating) / COUNT(m1.id) last_5_matches_rating,
GROUP_CONCAT(CONCAT(m1.day, '=', COALESCE(pm1.rating, '~')) ORDER BY m1.day)
FROM
`match` m1
INNER JOIN (
SELECT m2.id, m2.home_team_id, m2.away_team_id
FROM `match` m2
AND (m2.home_team_id=1 OR m2.away_team_id=1)
ORDER BY m2.day DESC
LIMIT 5
) last_5_games ON m1.id = last_5_games.id
LEFT JOIN player_match pm1 ON m1.id = pm1.match_id AND pm1.player_id=4
这是第 1 队的第 4 名球员的结果。
有没有办法将最后一个查询作为前一个查询的子查询执行,并按last_5_matches_rating
排序结果?
我期望的是以下列的结果:
| id | lastname | team_id | avg_rating | last_5_matches_rating | nb_matches | ...
【问题讨论】:
见:Why should I provide an MCVE for what seems to me to be a very simple SQL query? @Strawberry MVCE 添加。 感谢进度报告 【参考方案1】:我仍然无法获得返回预期结果的查询,但我已改用另一种明显的方式来完成这项工作。
我在播放器表中添加了两个计算列:last_five_matches_avg_rating
和 last_five_matches_ratings
。
每次数据库更新后,都会执行一个脚本:它遍历所有玩家,执行第二个查询以检索统计数据并将这些计算的统计数据存储在相应的列中。
这是这个 php 脚本(它使用 Doctrine ORM):
<?php
namespace App;
use App\Entity\Player;
use Doctrine\Common\Persistence\ManagerRegistry;
class StatsComputer
private $doctrine;
public function __construct(ManagerRegistry $doctrine)
$this->doctrine = $doctrine;
public function update()
$manager = $this->doctrine->getManager();
$stmt = $manager->getConnection()->prepare(<<<SQL
SELECT
SUM(pm1.rating) / COUNT(m1.id) last_5_matches_avg_rating,
GROUP_CONCAT(COALESCE(pm1.rating, '0')) ORDER BY m1.day) last_5_matches_ratings
FROM
`match` m1
INNER JOIN (
SELECT m2.id, m2.home_team_id, m2.away_team_id
FROM `match` m2
AND (m2.home_team_id=:team_id OR m2.away_team_id=:team_id)
ORDER BY m2.day DESC
LIMIT 5
) last_5_games ON m1.id = last_5_games.id
LEFT JOIN player_match pm1 ON m1.id = pm1.match_id AND pm1.player_id=:player_id
SQL
);
foreach ($this->doctrine->getRepository(Player::class)->findAll() as $player)
/**
* @var $player Player
*/
$stmt->execute(['team_id' => $player->team->id, 'player_id' => $player->id]);
$results = $stmt->fetch();
$player->last5MatchesAvgRating = $results['last_5_matches_avg_rating'];
$player->last5MatchesRatings = array_map('intval', explode(',', $results['last_5_matches_ratings']));
$manager->flush();
以及第一个查询的更新版本:
SELECT
p.id AS p_id,
p.lastname AS lastname,
p.team_id as team_id,
p.last_5_matches_avg_rating as last_5_matches_avg_rating,
p.last_5_matches_ratings as last_5_matches_ratings,
AVG(pm.rating) AS avg_rating,
COUNT(pm.player_id) AS nb_matches,
SUM(pm.substitute) AS nb_matches_substitute,
SUM(pm.goals) AS goals,
(SUM(pm.goals) / COUNT(pm.player_id)) AS goals_per_matches
FROM
player p
INNER JOIN player_match pm ON pm.player_id = p.id
INNER JOIN `match` m ON pm.match_id = m.id AND (m.home_team_id = p.team_id OR m.away_team_id = p.team_id)
GROUP BY
p.id,
p.lastname,
p.team_id,
p.last_5_matches_avg_rating,
p.last_5_matches_ratings
ORDER BY
last_5_matches_avg_rating DESC, avg_rating DESC, lastname ASC
;
这并不是我最初想要的,但它确实有效(这个数据库的更新次数很少)而且速度很快。
我希望它可以帮助某人。
【讨论】:
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