我要去拿那个 trophies
由一个团队赢得,每一个奖杯都是在一个特定赛季的比赛中获得的,但是一个团队可以在不同的赛季为同一个比赛赢得一个奖杯。
我写了一个从数据库中提取奖杯的查询:
$sql = $this->db->prepare("SELECT t.*, s.*, c.*,
t.team_id as team_id,
s.id as season_id,
s.name as season_name,
c.id as competition_id,
c.name as competition_name
FROM team_trophies t
INNER JOIN competition_seasons s ON s.id = t.season_id
INNER JOIN competition c ON c.id = s.competition_id
WHERE team_id = :team_id
");
基本上我选择了所有 trophies
一个特定的团队 team_trophies
加入了 competition_seasons
对于检索奖杯的赛季细节,对于 competition
table。
这很有效,我得到:
[
{
"team_id": "23291",
"season_id": "2",
"position": "Winner",
"wins": "4",
"id": "1093",
"competition_id": "1093",
"name": "Premier League",
"update_at": "2018-06-04 12:12:30",
"country_id": "1",
"category": "1",
"season_name": "2017",
"competition_name": "Premier League"
},
{
"team_id": "23291",
"season_id": "3",
"position": "Runner-up",
"wins": "1",
"id": "1093",
"competition_id": "1093",
"name": "Premier League",
"update_at": "2018-06-04 12:14:39",
"country_id": "1",
"category": "1",
"season_name": "2015",
"competition_name": "Premier League"
}
]
但我会返回这样一个结果:
[
{
"team_id": "23291",
"position": "Winner",
"wins": "4",
"id": "1093",
"competition_id": "1093",
"name": "Premier League",
"update_at": "2018-06-04 12:12:30",
"country_id": "1",
"category": "1",
"seasons": [
["season_name":"2017", "season_id":"2"],
["season_name":"2015", "season_id":"3"],
]
"competition_name": "Premier League"
}
]
如您所见,结果中有一行包含 seasons
把奖杯当作 array
,这样可读性更强,避免了冗余。
是否可以使用 sql
? 或者我需要解决问题 php
?
谢谢。
更新-表结构
CREATE TABLE IF NOT EXISTS `swp`.`competition` (
`id` INT NOT NULL,
`name` VARCHAR(255) NULL,
PRIMARY KEY (`id`),
ENGINE = InnoDB;
CREATE TABLE IF NOT EXISTS `swp`.`competition_seasons` (
`id` INT NOT NULL AUTO_INCREMENT,
`competition_id` INT NOT NULL,
`season_id` INT NULL,
`name` VARCHAR(45) NOT NULL,
`update_at` DATETIME NULL,
PRIMARY KEY (`id`),
INDEX `FK_competition_competition_seasons_competition_id_idx` (`competition_id` ASC),
CONSTRAINT `FK_competition_competition_seasons_competition_id`
FOREIGN KEY (`competition_id`)
REFERENCES `swp`.`competition` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
CREATE TABLE IF NOT EXISTS `swp`.`team_trophies` (
`team_id` INT NOT NULL,
`season_id` INT NOT NULL,
`position` VARCHAR(255) NOT NULL,
`wins` INT NOT NULL,
INDEX `FK_team_team_trophies_team_id_idx` (`team_id` ASC),
INDEX `FK_season_team_trophies_season_id_idx` (`season_id` ASC),
CONSTRAINT `FK_team_team_trophies_team_id`
FOREIGN KEY (`team_id`)
REFERENCES `swp`.`team` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION,
CONSTRAINT `FK_season_team_trophies_season_id`
FOREIGN KEY (`season_id`)
REFERENCES `swp`.`competition_seasons` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
-- Some INSERTs
INSERT INTO `team_trophies` (`team_id`, `season_id`, `position`, `wins` VALUES (23291, 2, 'Winner', 4), (23291, 3, 'Runner-up', 1);
INSERT INTO `competition` (`id`, `country_id`, `name`, `category`) VALUES (1093, 1, 'Premier League', 1);
INSERT INTO `competition_seasons` (`id`, `competition_id`, `season_id`,
`name`, `update_at`) VALUES
(1, 1093, 14963, '2018', '2018-06-04 12:10:28'),
(2, 1093, 13198, '2017', '2018-06-04 12:12:30');
1条答案
按热度按时间uyto3xhc1#
承蒙一位叫做ghost的大师的帮助,这里有一个简单的例子,可以将一维数组转换为多维数组。。。
输出: