如何使用SQL修复数据库中的已损坏的MPTT树(嵌套集)?

I have an MPTT tree of over 100,000 records stored in MySQL using lft, rght and parent_id columns. Now the left/right values became corrupted, while the parent ids are still intact. It would require tons of queries to repair it in the application layer. Is there a good way to put the burden on the database and have it recalculate the left/right values using only SQL?

我使用lft、rght和parent_id列在MySQL中存储了超过100,000条记录的MPTT树。现在，左/右值已被破坏，而父id仍然完好无损。在应用层中修复它需要大量的查询。是否有一种很好的方法来将这个负担放在数据库上，并让它仅使用SQL重新计算左/右值?

---

Just to clarify, I need to recalculate the numeric lft/rght values of a nested set, not the ids of neighboring records.

为了澄清，我需要重新计算嵌套集合的数值lft/rght值，而不是相邻记录的id。

The Nested Set http://dev.mysql.com/tech-resources/articles/hierarchical-data-4.png

嵌套集http://dev.mysql.com/tech resources/articles/hierarchical -数据- 4. png

4 个解决方案

#1

Using SQL Server, following script seems to be working for me.

使用SQL Server，下面的脚本似乎对我有用。

Output testscript

category_id name                 parent      lft         rgt         lftcalc     rgtcalc
----------- -------------------- ----------- ----------- ----------- ----------- -----------
1           ELECTRONICS          NULL        1           20          1           20
2           TELEVISIONS          1           2           9           2           9
3           TUBE                 2           3           4           3           4
4           LCD                  2           5           6           5           6
5           PLASMA               2           7           8           7           8
6           PORTABLE ELECTRONICS 1           10          19          10          19
7           MP3 PLAYERS          6           11          14          11          14
8           FLASH                7           12          13          12          13
9           CD PLAYERS           6           15          16          15          16
10          2 WAY RADIOS         6           17          18          17          18

Script

SET NOCOUNT ON
GO

DECLARE @nested_category TABLE (
 category_id INT PRIMARY KEY,
 name VARCHAR(20) NOT NULL,
 parent INT,
 lft INT,
 rgt INT
);

DECLARE @current_Category_ID INTEGER
DECLARE @current_parent INTEGER
DECLARE @SafeGuard INTEGER
DECLARE @myLeft INTEGER
SET @SafeGuard = 100

INSERT INTO @nested_category 
SELECT           1,'ELECTRONICS',NULL,NULL,NULL
UNION ALL SELECT 2,'TELEVISIONS',1,NULL,NULL
UNION ALL SELECT 3,'TUBE',2,NULL,NULL
UNION ALL SELECT 4,'LCD',2,NULL,NULL
UNION ALL SELECT 5,'PLASMA',2,NULL,NULL
UNION ALL SELECT 6,'PORTABLE ELECTRONICS',1,NULL,NULL
UNION ALL SELECT 7,'MP3 PLAYERS',6,NULL,NULL
UNION ALL SELECT 8,'FLASH',7,NULL,NULL
UNION ALL SELECT 9,'CD PLAYERS',6,NULL,NULL
UNION ALL SELECT 10,'2 WAY RADIOS',6,NULL,NULL

/* Initialize */
UPDATE  @nested_category 
SET     lft = 1
        , rgt = 2
WHERE   parent IS NULL

UPDATE  @nested_category 
SET     lft = NULL
        , rgt = NULL
WHERE   parent IS NOT NULL

WHILE EXISTS (SELECT * FROM @nested_category WHERE lft IS NULL) AND @SafeGuard > 0
BEGIN
  SELECT  @current_Category_ID = MAX(nc.category_id)
  FROM    @nested_category nc
          INNER JOIN @nested_category nc2 ON nc2.category_id = nc.parent
  WHERE   nc.lft IS NULL
          AND nc2.lft IS NOT NULL

  SELECT  @current_parent = parent
  FROM    @nested_category
  WHERE   category_id = @current_category_id

  SELECT  @myLeft = lft
  FROM    @nested_category
  WHERE   category_id = @current_parent

  UPDATE @nested_category SET rgt = rgt + 2 WHERE rgt > @myLeft;
  UPDATE @nested_category SET lft = lft + 2 WHERE lft > @myLeft;
  UPDATE @nested_category SET lft = @myLeft + 1, rgt = @myLeft + 2 WHERE category_id = @current_category_id

  SET @SafeGuard = @SafeGuard - 1
END

SELECT * FROM @nested_category ORDER BY category_id

SELECT  COUNT(node.name), node.name, MIN(node.lft)
FROM    @nested_category AS node,
        @nested_category AS parent
WHERE   node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY 
        node.name
ORDER BY
        3, 1

Testscript ##

SET NOCOUNT ON
GO

DECLARE @nested_category TABLE (
 category_id INT PRIMARY KEY,
 name VARCHAR(20) NOT NULL,
 parent INT,
 lft INT,
 rgt INT, 
 lftcalc INT,
 rgtcalc INT
);

INSERT INTO @nested_category 
SELECT           1,'ELECTRONICS',NULL,1,20,NULL,NULL
UNION ALL SELECT 2,'TELEVISIONS',1,2,9,NULL,NULL
UNION ALL SELECT 3,'TUBE',2,3,4,NULL,NULL
UNION ALL SELECT 4,'LCD',2,5,6,NULL,NULL
UNION ALL SELECT 5,'PLASMA',2,7,8,NULL,NULL
UNION ALL SELECT 6,'PORTABLE ELECTRONICS',1,10,19,NULL,NULL
UNION ALL SELECT 7,'MP3 PLAYERS',6,11,14,NULL,NULL
UNION ALL SELECT 8,'FLASH',7,12,13,NULL,NULL
UNION ALL SELECT 9,'CD PLAYERS',6,15,16,NULL,NULL
UNION ALL SELECT 10,'2 WAY RADIOS',6,17,18,NULL,NULL

/* Initialize */
UPDATE  @nested_category 
SET     lftcalc = 1
        , rgtcalc = 2
WHERE   parent IS NULL

DECLARE @current_Category_ID INTEGER
DECLARE @current_parent INTEGER
DECLARE @SafeGuard INTEGER
DECLARE @myRight INTEGER
DECLARE @myLeft INTEGER
SET @SafeGuard = 100
WHILE EXISTS (SELECT * FROM @nested_category WHERE lftcalc IS NULL) AND @SafeGuard > 0
BEGIN
  SELECT  @current_Category_ID = MAX(nc.category_id)
  FROM    @nested_category nc
          INNER JOIN @nested_category nc2 ON nc2.category_id = nc.parent
  WHERE   nc.lftcalc IS NULL
          AND nc2.lftcalc IS NOT NULL

  SELECT  @current_parent = parent
  FROM    @nested_category
  WHERE   category_id = @current_category_id

  SELECT  @myLeft = lftcalc
  FROM    @nested_category
  WHERE   category_id = @current_parent

  UPDATE @nested_category SET rgtcalc = rgtcalc + 2 WHERE rgtcalc > @myLeft;
  UPDATE @nested_category SET lftcalc = lftcalc + 2 WHERE lftcalc > @myLeft;
  UPDATE @nested_category SET lftcalc = @myLeft + 1, rgtcalc = @myLeft + 2 WHERE category_id = @current_category_id

  SELECT * FROM @nested_category WHERE category_id = @current_parent
  SELECT * FROM @nested_category ORDER BY category_id
  SET @SafeGuard = @SafeGuard - 1
END

SELECT * FROM @nested_category ORDER BY category_id

SELECT  COUNT(node.name), node.name, MIN(node.lft)
FROM    @nested_category AS node,
        @nested_category AS parent
WHERE   node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY 
        node.name
ORDER BY
        3, 1

#2

Here's what I have adapted from @Lieven's answer, incorporating feedback from here for better performance:

以下是我根据@Lieven的回答改编而成的，结合了来自这里的反馈以获得更好的性能:

DROP PROCEDURE IF EXISTS tree_recover;

DELIMITER //

CREATE PROCEDURE tree_recover ()
MODIFIES SQL DATA
BEGIN

    DECLARE currentId, currentParentId  CHAR(36);
    DECLARE currentLeft                 INT;
    DECLARE startId                     INT DEFAULT 1;

    # Determines the max size for MEMORY tables.
    SET max_heap_table_size = 1024 * 1024 * 512;

    START TRANSACTION;

    # Temporary MEMORY table to do all the heavy lifting in,
    # otherwise performance is simply abysmal.
    CREATE TABLE `tmp_tree` (
        `id`        char(36) NOT NULL DEFAULT '',
        `parent_id` char(36)          DEFAULT NULL,
        `lft`       int(11)  unsigned DEFAULT NULL,
        `rght`      int(11)  unsigned DEFAULT NULL,
        PRIMARY KEY      (`id`),
        INDEX USING HASH (`parent_id`),
        INDEX USING HASH (`lft`),
        INDEX USING HASH (`rght`)
    ) ENGINE = MEMORY
    SELECT `id`,
           `parent_id`,
           `lft`,
           `rght`
    FROM   `tree`;

    # Leveling the playing field.
    UPDATE  `tmp_tree`
    SET     `lft`  = NULL,
            `rght` = NULL;

    # Establishing starting numbers for all root elements.
    WHILE EXISTS (SELECT * FROM `tmp_tree` WHERE `parent_id` IS NULL AND `lft` IS NULL AND `rght` IS NULL LIMIT 1) DO

        UPDATE `tmp_tree`
        SET    `lft`  = startId,
               `rght` = startId + 1
        WHERE  `parent_id` IS NULL
          AND  `lft`       IS NULL
          AND  `rght`      IS NULL
        LIMIT  1;

        SET startId = startId + 2;

    END WHILE;

    # Switching the indexes for the lft/rght columns to B-Trees to speed up the next section, which uses range queries.
    DROP INDEX `lft`  ON `tmp_tree`;
    DROP INDEX `rght` ON `tmp_tree`;
    CREATE INDEX `lft`  USING BTREE ON `tmp_tree` (`lft`);
    CREATE INDEX `rght` USING BTREE ON `tmp_tree` (`rght`);

    # Numbering all child elements
    WHILE EXISTS (SELECT * FROM `tmp_tree` WHERE `lft` IS NULL LIMIT 1) DO

        # Picking an unprocessed element which has a processed parent.
        SELECT     `tmp_tree`.`id`
          INTO     currentId
        FROM       `tmp_tree`
        INNER JOIN `tmp_tree` AS `parents`
                ON `tmp_tree`.`parent_id` = `parents`.`id`
        WHERE      `tmp_tree`.`lft` IS NULL
          AND      `parents`.`lft`  IS NOT NULL
        LIMIT      1;

        # Finding the element's parent.
        SELECT  `parent_id`
          INTO  currentParentId
        FROM    `tmp_tree`
        WHERE   `id` = currentId;

        # Finding the parent's lft value.
        SELECT  `lft`
          INTO  currentLeft
        FROM    `tmp_tree`
        WHERE   `id` = currentParentId;

        # Shifting all elements to the right of the current element 2 to the right.
        UPDATE `tmp_tree`
        SET    `rght` = `rght` + 2
        WHERE  `rght` > currentLeft;

        UPDATE `tmp_tree`
        SET    `lft` = `lft` + 2
        WHERE  `lft` > currentLeft;

        # Setting lft and rght values for current element.
        UPDATE `tmp_tree`
        SET    `lft`  = currentLeft + 1,
               `rght` = currentLeft + 2
        WHERE  `id`   = currentId;

    END WHILE;

    # Writing calculated values back to physical table.
    UPDATE `tree`, `tmp_tree`
    SET    `tree`.`lft`  = `tmp_tree`.`lft`,
           `tree`.`rght` = `tmp_tree`.`rght`
    WHERE  `tree`.`id`   = `tmp_tree`.`id`;

    COMMIT;

    DROP TABLE `tmp_tree`;

END//

DELIMITER ;

Worked well with some test data, but it's still running on my 100,000 records tree, so I can't give any final judgement yet. The naïve script working directly on the physical table has abysmal performance, running for at least hours, more likely days. Switching to a temporary MEMORY table brought this time down to about an hour, choosing the right indexes cut it down to 10 mins.

一些测试数据处理得很好，但是它仍然在我的100,000个记录树上运行，所以我还不能给出任何最终的判断。直接在物理表上工作的幼稚脚本的性能非常糟糕，至少运行数小时，更可能运行数天。切换到临时的内存表可以将时间缩短到大约一个小时，选择合适的索引可以将时间缩短到10分钟。

#3

You are rescue me!!! I use mixed tree model, so when the day is coming, my tree (30000+) was corrupted. I learn from both of your tech, but is not recovery, just fully rebuild with lost all of sorting and reverse tree... I think, need to keep in mind older cat_left.... Just for possible integrity. So, it maybe looks like...

你救我! ! !我使用混合树模型，所以当那天到来时，我的树(30000+)被破坏了。我从你的两个技术中学习，但不是恢复，只是完全重建失去的排序和反向树……我认为,需要记住老cat_left ....只是为了可能的完整性。所以，也许看起来……

DROP PROCEDURE IF EXISTS tree_recover;

DELIMITER |

CREATE PROCEDURE tree_recover ()
MODIFIES SQL DATA
BEGIN

    DECLARE currentId, currentParentId  CHAR(36);
    DECLARE currentLeft                 INT;
    DECLARE startId                     INT DEFAULT 1;

    # Determines the max size for MEMORY tables.
    SET max_heap_table_size = 1024 * 1024 * 512;

    START TRANSACTION;

    # Temporary MEMORY table to do all the heavy lifting in,
    # otherwise performance is simply abysmal.
    DROP TABLE IF EXISTS `tmp_cat`;
    CREATE TABLE `tmp_cat` (
        `cat_id`        char(36) NOT NULL DEFAULT '',
        `cat_parent` char(36)          DEFAULT NULL,
        `cat_left`       int(11)  unsigned DEFAULT NULL,
        `cat_right`      int(11)  unsigned DEFAULT NULL,
        `cat_left_old`   int(11)  unsigned DEFAULT NULL,
        PRIMARY KEY      (`cat_id`),
        INDEX USING HASH (`cat_parent`),
        INDEX USING HASH (`cat_left`),
        INDEX USING HASH (`cat_right`),
    INDEX USING HASH (`cat_left_old`)
    ) ENGINE = MEMORY
    SELECT `cat_id`,
           `cat_parent`,
           `cat_left`,
           `cat_right`,
       `cat_left` as cat_left_old
    FROM   `catalog`;

    # Leveling the playing field.
    UPDATE  `tmp_cat`
    SET     `cat_left`  = NULL,
            `cat_right` = NULL;

    # Establishing starting numbers for all root elements.
    WHILE EXISTS (SELECT * FROM `tmp_cat` WHERE `cat_parent` IS NULL AND `cat_left` IS NULL AND `cat_right` IS NULL ORDER BY cat_left_old LIMIT 1) DO

        UPDATE `tmp_cat`
        SET    `cat_left`  = startId,
               `cat_right` = startId + 1
        WHERE  `cat_parent` IS NULL
          AND  `cat_left`       IS NULL
          AND  `cat_right`      IS NULL
        LIMIT  1;

        SET startId = startId + 2;

    END WHILE;

    # Switching the indexes for the cat_left/rght columns to B-Trees to speed up the next section, which uses range queries.
    DROP INDEX `cat_left`  ON `tmp_cat`;
    DROP INDEX `cat_right` ON `tmp_cat`;
    DROP INDEX `cat_left_old` ON `tmp_cat`;

    CREATE INDEX `cat_left`  USING BTREE ON `tmp_cat` (`cat_left`);
    CREATE INDEX `cat_right` USING BTREE ON `tmp_cat` (`cat_right`);
    CREATE INDEX `cat_left_old` USING BTREE ON `tmp_cat` (`cat_left_old`);

    # Numbering all child elements
    WHILE EXISTS (SELECT * FROM `tmp_cat` WHERE `cat_left` IS NULL ORDER BY cat_left_old LIMIT 1) DO

        # Picking an unprocessed element which has a processed parent.
        SELECT     `tmp_cat`.`cat_id`
          INTO     currentId
        FROM       `tmp_cat`
        INNER JOIN `tmp_cat` AS `parents`
                ON `tmp_cat`.`cat_parent` = `parents`.`cat_id`
        WHERE      `tmp_cat`.`cat_left` IS NULL
          AND      `parents`.`cat_left`  IS NOT NULL
    ORDER BY `tmp_cat`.cat_left_old DESC
        LIMIT      1;

        # Finding the element's parent.
        SELECT  `cat_parent`
          INTO  currentParentId
        FROM    `tmp_cat`
        WHERE   `cat_id` = currentId;

        # Finding the parent's cat_left value.
        SELECT  `cat_left`
          INTO  currentLeft
        FROM    `tmp_cat`
        WHERE   `cat_id` = currentParentId;

        # Shifting all elements to the right of the current element 2 to the right.
        UPDATE `tmp_cat`
        SET    `cat_right` = `cat_right` + 2
        WHERE  `cat_right` > currentLeft;

        UPDATE `tmp_cat`
        SET    `cat_left` = `cat_left` + 2
        WHERE  `cat_left` > currentLeft;

        # Setting cat_left and rght values for current element.
        UPDATE `tmp_cat`
        SET    `cat_left`  = currentLeft + 1,
               `cat_right` = currentLeft + 2
        WHERE  `cat_id`   = currentId;

    END WHILE;

    # Writing calculated values back to physical table.
    UPDATE `catalog`, `tmp_cat`
    SET    `catalog`.`cat_left`  = `tmp_cat`.`cat_left`,
           `catalog`.`cat_right` = `tmp_cat`.`cat_right`
    WHERE  `catalog`.`cat_id`   = `tmp_cat`.`cat_id`;

    COMMIT;

    DROP TABLE IF EXISTS `tmp_cat`;

END|

#4

In all solutions provided, I was getting a problem where MySQL would prompt that it would be Running query for hours but nothing would happen.

在提供的所有解决方案中，我遇到了一个问题，MySQL会提示它将运行查询数小时，但什么也不会发生。

I then realized that if I set the lft and rght values to 1 and 2 in the first record of the tmp_tree table (the one with parent_id = 0), everything worked fine. Maybe the procedure needs updating to do this automatically.

然后我意识到，如果我在tmp_tree表的第一个记录(parent_id = 0的记录)中将lft和rght的值设置为1和2，那么一切都没问题。也许过程需要更新才能自动完成。