2

试图收集一些层次数据发送给第三方，并被定向到这个帖子。

在尝试将其调整到我在SQL Fiddle上的用例之后，存储过程一直超时。
所以我在本地尝试了两次（通过 PhpMyAdmin）。
当我在调用存储过程后尝试在浏览器中重新加载 PMA 时，我只会得到一个永恒的“等待响应”微调器（超过 10 或 20 分钟）。

我假设我的 SP 代码有问题？？？

CREATE TABLE foo
    (`id` int, `name` varchar(100), `parentId` int, `path` varchar(100))
//

INSERT INTO foo (`id`, `name`, `parentId`, `path`)
VALUES (1, 'discrete', 0, NULL),
       (2, 'res', 1, NULL),
       (3, 'smt', 2, NULL),
       (4, 'cap', 1, NULL),
       (5, 'ind', 1, NULL),
       (6, 'smt', 4, NULL),
       (7, 'tant', 6, NULL),
       (8, 'cer', 6, NULL)
//

CREATE PROCEDURE updatePath()
BEGIN
DECLARE cnt, n int;
    SELECT COUNT(*) INTO n FROM foo WHERE parentId = 0;
    UPDATE foo a, foo b SET a.path = b.name WHERE b.parentId IS NULL AND a.parentId = b.id;
    SELECT count(*) INTO cnt FROM foo WHERE path IS NULL;
    while cnt > n do
        UPDATE foo a, foo b SET a.path = concat(b.path, '|', b.id) WHERE b.path IS NOT NULL AND a.parentId = b.id;
        SELECT count(*) INTO  cnt FROM foo WHERE path IS NULL;
    end while;
END//

编辑
预期结果：

VALUES (1, 'discrete', 0, '1'),
       (2, 'res', 1, '1|2'),
       (3, 'smt', 2, '1|2|3'),
       (4, 'cap', 1, '1|4'),
       (5, 'ind', 1, '1|5'),
       (6, 'smt', 4, '1|4|6'),
       (7, 'tant', 6, '1|4|6|7'),
       (8, 'cer', 6, '1|4|6|8');
4

2 回答 2

1

睡了一夜好觉后，我接受了@Drew 的带领，一次只用了一台电脑。
得到它的工作。这是我要离开的地方：

CREATE TABLE foo
    (`id` int, `name` varchar(100), `parentId` int, `path` varchar(100))
//

INSERT INTO foo
    (`id`, `name`, `parentId`, `path`)
VALUES
    (1, 'dscr', 0, NULL),
    (2, 'res', 1, NULL),
    (3, 'smt', 2, NULL),
    (4, 'cap', 1, NULL),
    (5, 'ind', 1, NULL),
    (6, 'chp', 4, NULL),
    (7, 'tant', 6, NULL),
    (8, 'cer', 6, NULL)
//

CREATE PROCEDURE updatePath()
BEGIN
DECLARE cnt, n int;

    SELECT COUNT(*) INTO n FROM foo WHERE parentId = 0; -- n is now 1
    SELECT COUNT(*) INTO cnt FROM foo WHERE path IS NULL; -- cnt is now 8

    UPDATE foo child, foo parent        -- each child now has its parent and own ID's in the path
      SET child.path = CONCAT(parent.id, '|', child.id)
      WHERE parent.parentId = 0
      AND child.parentId = parent.id;

    WHILE cnt > n DO
        UPDATE foo child, foo parent    -- concat parent's path and own ID into each empty child's path
          SET child.path = concat( parent.path,'|',child.id ) 
          WHERE parent.path IS NOT NULL 
          AND child.parentId = parent.id;

        SELECT COUNT(*) INTO  cnt  -- decrement cnt
          FROM foo 
          WHERE path IS NULL;

    END WHILE;

    UPDATE foo  -- set path for any top-level categories
      SET path = id
      WHERE path IS NULL;

END//

call updatePath()//

随意批评。
希望这对其他人有所帮助。

于 2016-06-03T20:46:46.290 回答
0

您是否尝试进行自引用连接来创建层次结构？

select a.name, parentName = b.name from foo a, 外连接 foo b on (a.id = b.parentId)

于 2016-06-02T23:46:27.763 回答