Php 如何仅选择最新添加的记录以避免重复

我需要从表

competition\u rounds

中选择链接到

competition\u seasures

表的

seasures\u id

列的最新添加记录，例如：

比赛\u轮

id |    name     | season_id
 1    Round A        20
 2    Round B        20
 3    Round A        21
 4    Round B        21

竞争季节

id | name      | competition_id
 20  2017/2018      40
 21  2018/2019      40

我只想退货：

  round_id  |   round_name   |  season_id | season_name
   3           Round A           21            2018/2019
   4           Round B           21            2018/2019

问题在于，我的查询返回所有可用回合：

$sql = $this->db->prepare("SELECT max(r.id) AS round_id,
    r.name as round_name, r.season_id AS season_id, s.name AS season_name
    FROM competition_rounds r
    JOIN competition_seasons s ON r.season_id = s.id
    JOIN competition c ON s.competition_id = c.id
    WHERE c.id = :competition_id 
    GROUP BY r.id
    ORDER BY max(r.season_id) DESC");

$sql->bindParam("competition_id", 40);
$sql->execute();
$rounds = $sql->fetchAll();
return $response->withJson($rounds);

---

注意：表

competition

仅包含一个竞赛参考列表。

您在

SELECT

中没有任何

competition

列，因此您可以从查询中省略它。 就给定的

：竞赛\u id

SELECT r.id AS round_id,
    r.name as round_name, r.season_id AS season_id, s.name AS season_name
    FROM competition_rounds r
    JOIN competition_seasons s ON r.season_id = s.id    
    WHERE s.id = (SELECT max(cs.id) 
                   FROM competition_seasons cs 
                   WHERE cs.competition_id = :competition_id)

---

如果我理解正确，您可以使用子查询从seasons表中仅返回最新的季节：

SELECT r.id AS round_id,
       r.name as round_name, r.season_id AS season_id, s.name AS season_name
FROM competition_rounds r JOIN
     (SELECT s.*
      FROM competition_seasons s
      WHERE s.competition_id = :competition_id
      ORDER BY s.id DESC
      LIMIT 1
     ) s
     ON r.season_id = s.id ;

您的问题只提到两个表，尽管查询有三个表。这只是基于这个问题