使用mysql/php中的id/parent_id模型获取记录的所有父项的最简单方法是什么？

Question

我正在寻找使用邻接列表/单表继承模型(id, parent_id)以递归方式从数据库中获取所有父元素的最简单方法.

我的选择目前看起来像这样:

$sql = "SELECT
             e.id,
             TIME_FORMAT(e.start_time, '%H:%i') AS start_time,
             $title AS title,
             $description AS description,
             $type AS type,
             $place_name AS place_name,
             p.parent_id AS place_parent_id,
             p.city AS place_city,
             p.country AS place_country
         FROM event AS e
         LEFT JOIN place AS p ON p.id = e.place_id                          
         LEFT JOIN event_type AS et ON et.id = e.event_type_id
         WHERE e.day_id = '$day_id'
         AND e.private_flag = 0
         ORDER BY start_time";

每个event都与a相关联place,每个都place可以是另一个孩子place(最多约5级)

使用mysql可以在单个选择中实现吗？

目前我认为它可能是一个单独的函数,它循环返回的$events数组,添加place_parent_X元素,但不知道如何实现它.

Answer 1

可以这样做MySQL,但您需要创建一个函数并在查询中使用.

有关详细说明,请参阅我的博客中的此条目:

• 中的分层查询 MySQL

这是函数和查询:

CREATE FUNCTION hierarchy_connect_by_parent_eq_prior_id(value INT) RETURNS INT
NOT DETERMINISTIC
READS SQL DATA
BEGIN
        DECLARE _id INT;
        DECLARE _parent INT;
        DECLARE _next INT;
        DECLARE CONTINUE HANDLER FOR NOT FOUND SET @id = NULL;

        SET _parent = @id;
        SET _id = -1;

        IF @id IS NULL THEN
                RETURN NULL;
        END IF;

        LOOP
                SELECT  MIN(id)
                INTO    @id
                FROM    place
                WHERE   parent = _parent
                        AND id > _id;
                IF @id IS NOT NULL OR _parent = @start_with THEN
                        SET @level = @level + 1;
                        RETURN @id;
                END IF;
                SET @level := @level - 1;
                SELECT  id, parent
                INTO    _id, _parent
                FROM    place
                WHERE   id = _parent;
        END LOOP;
END

SELECT  id, parent
FROM    (
        SELECT  hierarchy_connect_by_parent_eq_prior_id(id) AS id, @level AS level
        FROM    (
                SELECT  @start_with := 0,
                        @id := @start_with,
                        @level := 0
                ) vars, t_hierarchy
        WHERE   @id IS NOT NULL
        ) ho
JOIN    place hi
ON      hi.id = ho.id

后一个查询将选择给定节点的所有后代(您应该在@start_with变量中设置)

要查找给定节点的所有祖先,您可以使用不带函数的简单查询:

SELECT  @r AS _id,
        @r := (
        SELECT  parent
        FROM    place
        WHERE   id = _id
        ) AS parent
FROM    (
        SELECT  @r := @node_id
        ) vars,
        place

我的博客中的这篇文章更详细地描述了这个查询:

• 排序列表

对于这两种解决方案,在合理的时间工作,你需要有两个指标id和parent.

确保您id的定义为a PRIMARY KEY并且您有一个seconday索引parent.