Tree in SQL database: The Nested Set Model

Tree structure are very commonly used by all the application who need to manage a large quantity of data. However once have in memory the tree data the application need a way to save them for future use. SQL database can be a good solution but, in this case, we need a way to store data in efficient mode. This post will propose a model called Nested Set.




IMPORTANT NOTE: Please note that I'm not the original author of this article. I found it some years ago in the MySQL web site but after some time the article disappeared and now it seem is very hard to find around. Since I saved it I want to repropose here mainly because I think is a very well written article explaining the model and proposing examples SQL query to use. The only example missing from this article is the query for move a branch with all the child nodes. I found a working query in a newsgroup and I added them at the end of this post. Again, since I think is very important, I repeat the content of this post are not mine. If you are the original author and you want I remove this post or want I add some copyright link or similar just ask.

The Nested Set Model

What I would like to focus on in this article is a different approach, commonly referred to as the Nested Set Model. In the Nested Set Model, we can look at our hierarchy in a new way, not as nodes and lines, but as nested containers. Try picturing our electronics categories this way:


Notice how our hierarchy is still maintained, as parent categories envelop their children.We represent this form of hierarchy in a table through the use of left and right values to represent the nesting of our nodes:

CREATE TABLE nested_category (
category_id INT AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(20) NOT NULL,
lft INT NOT NULL,
rgt INT NOT NULL
);


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


SELECT * FROM nested_category ORDER BY category_id;


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

We use lft and rgt because left and right are reserved words in MySQL, see http://dev.mysql.com/doc/mysql/en/reserved-words.html for the full list of reserved words.

So how do we determine left and right values? We start numbering at the leftmost side of the outer node and continue to the right:



This design can be applied to a typical tree as well:




When working with a tree, we work from left to right, one layer at a time, descending to each node's children before assigning a right-hand number and moving on to the right. This approach is called the modified preorder tree traversal algorithm.

Retrieving a Full Tree

We can retrieve the full tree through the use of a self-join that links parents with nodes on the basis that a node's lft value will always appear between its parent's lft and rgt values:

SELECT node.name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND parent.name = 'ELECTRONICS'
ORDER BY node.lft;

+----------------------+
| name                 |
+----------------------+
| ELECTRONICS          |
| TELEVISIONS          |
| TUBE                 |
| LCD                  |
| PLASMA               |
| PORTABLE ELECTRONICS |
| MP3 PLAYERS          |
| FLASH                |
| CD PLAYERS           |
| 2 WAY RADIOS         |
+----------------------+

Unlike our previous examples with the adjacency list model, this query will work regardless of the depth of the tree. We do not concern ourselves with the rgt value of the node in our BETWEEN clause because the rgt value will always fall within the same parent as the lft values.

Finding all the Leaf Nodes

Finding all leaf nodes in the nested set model even simpler than the LEFT JOIN method used in the adjacency list model. If you look at the nested_category table, you may notice that the lft and rgt values for leaf nodes are consecutive numbers. To find the leaf nodes, we look for nodes where rgt = lft + 1:

SELECT name
FROM nested_category
WHERE rgt = lft + 1;

+--------------+
| name         |
+--------------+
| TUBE         |
| LCD          |
| PLASMA       |
| FLASH        |
| CD PLAYERS   |
| 2 WAY RADIOS |
+--------------+

Retrieving a Single Path 

With the nested set model, we can retrieve a single path without having multiple self-joins:

SELECT parent.name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.name = 'FLASH'
ORDER BY parent.lft;

+----------------------+
| name                 |
+----------------------+
| ELECTRONICS          |
| PORTABLE ELECTRONICS |
| MP3 PLAYERS          |
| FLASH                |
+----------------------+

Finding the Depth of the Nodes

We have already looked at how to show the entire tree, but what if we want to also show the depth of each node in the tree, to better identify how each node fits in the hierarchy? This can be done by adding a COUNT function and a GROUP BY clause to our existing query for showing the entire tree:

SELECT node.name, (COUNT(parent.name) - 1) AS depth
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;

+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| ELECTRONICS          |     0 |
| TELEVISIONS          |     1 |
| TUBE                 |     2 |
| LCD                  |     2 |
| PLASMA               |     2 |
| PORTABLE ELECTRONICS |     1 |
| MP3 PLAYERS          |     2 |
| FLASH                |     3 |
| CD PLAYERS           |     2 |
| 2 WAY RADIOS         |     2 |
+----------------------+-------+


We can use the depth value to indent our category names with the CONCAT and REPEAT string functions:

SELECT CONCAT( REPEAT(' ', COUNT(parent.name) - 1), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;

+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  PORTABLE ELECTRONICS |
|   MP3 PLAYERS         |
|    FLASH              |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
+-----------------------+

Of course, in a client-side application you will be more likely to use the depth value directly to display your hierarchy. Web developers could loop through the tree, adding <li></li> and <ul></ul> tags as the depth number increases and decreases.

Depth of a Sub-Tree

When we need depth information for a sub-tree, we cannot limit either the node or parent tables in our self-join because it will corrupt our results. Instead, we add a third self-join, along with a sub-query to determine the depth that will be the new starting point for our sub-tree:

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth

FROM nested_category AS node,
     nested_category AS parent,
     nested_category AS sub_parent,
    (
        SELECT node.name, (COUNT(parent.name) - 1) AS depth
        FROM nested_category AS node,
        nested_category AS parent
        WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.name = 'PORTABLE ELECTRONICS'
        GROUP BY node.name
        ORDER BY node.lft
    )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.name = sub_tree.name
GROUP BY node.name
ORDER BY node.lft;
 

+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| PORTABLE ELECTRONICS |     0 |
| MP3 PLAYERS          |     1 |
| FLASH                |     2 |
| CD PLAYERS           |     1 |
| 2 WAY RADIOS         |     1 |
+----------------------+-------+

This function can be used with any node name, including the root node. The depth values are always relative to the named node.

Find the Immediate Subordinates of a Node

Imagine you are showing a category of electronics products on a retailer web site. When a user clicks on a category, you would want to show the products of that category, as well as list its immediate sub-categories, but not the entire tree of categories beneath it. For this, we need to show the node and its immediate sub-nodes, but no further down the tree. For example, when showing the PORTABLE ELECTRONICS category, we will want to show MP3 PLAYERS, CD PLAYERS, and 2 WAY RADIOS, but not FLASH.

This can be easily accomplished by adding a HAVING clause to our previous query:

SELECT node.name, (COUNT(parent.name) - (sub_tree.depth + 1)) AS depth
FROM nested_category AS node,
     nested_category AS parent,
     nested_category AS sub_parent,
    (
        SELECT node.name, (COUNT(parent.name) - 1) AS depth
        FROM nested_category AS node,
        nested_category AS parent
        WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.name = 'PORTABLE ELECTRONICS'
        GROUP BY node.name
        ORDER BY node.lft
    )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
    AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
    AND sub_parent.name = sub_tree.name
GROUP BY node.name
HAVING depth <= 1
ORDER BY node.lft;


+----------------------+-------+
| name                 | depth |
+----------------------+-------+
| PORTABLE ELECTRONICS |     0 |
| MP3 PLAYERS          |     1 |
| CD PLAYERS           |     1 |
| 2 WAY RADIOS         |     1 |
+----------------------+-------+

If you do not wish to show the parent node, change the HAVING depth <= 1 line to HAVING depth = 1.

Aggregate Functions in a Nested Set

Let's add a table of products that we can use to demonstrate aggregate functions with:

CREATE TABLE product(
product_id INT AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(40),
category_id INT NOT NULL
);


INSERT INTO product(name, category_id) VALUES('20" TV',3),('36" TV',3),
('Super-LCD 42"',4),('Ultra-Plasma 62"',5),('Value Plasma 38"',5),
('Power-MP3 5gb',7),('Super-Player 1gb',8),('Porta CD',9),('CD To go!',9),
('Family Talk 360',10);


SELECT * FROM product;


+------------+-------------------+-------------+
| product_id | name              | category_id |
+------------+-------------------+-------------+
|          1 | 20" TV            |           3 |
|          2 | 36" TV            |           3 |
|          3 | Super-LCD 42"     |           4 |
|          4 | Ultra-Plasma 62"  |           5 |
|          5 | Value Plasma 38"  |           5 |
|          6 | Power-MP3 128mb   |           7 |
|          7 | Super-Shuffle 1gb |           8 |
|          8 | Porta CD          |           9 |
|          9 | CD To go!         |           9 |
|         10 | Family Talk 360   |          10 |
+------------+-------------------+-------------+

Now let's produce a query that can retrieve our category tree, along with a product count for each category:

SELECT parent.name, COUNT(product.name)
FROM nested_category AS node ,
nested_category AS parent,
product
WHERE node.lft BETWEEN parent.lft AND parent.rgt
AND node.category_id = product.category_id
GROUP BY parent.name
ORDER BY node.lft;


+----------------------+---------------------+
| name                 | COUNT(product.name) |
+----------------------+---------------------+
| ELECTRONICS          |                  10 |
| TELEVISIONS          |                   5 |
| TUBE                 |                   2 |
| LCD                  |                   1 |
| PLASMA               |                   2 |
| PORTABLE ELECTRONICS |                   5 |
| MP3 PLAYERS          |                   2 |
| FLASH                |                   1 |
| CD PLAYERS           |                   2 |
| 2 WAY RADIOS         |                   1 |
+----------------------+---------------------+

This is our typical whole tree query with a COUNT and GROUP BY added, along with a reference to the product table and a join between the node and product table in the WHERE clause. As you can see, there is a count for each category and the count of subcategories is reflected in the parent categories.

Adding New Nodes

Now that we have learned how to query our tree, we should take a look at how to update our tree by adding a new node. Let's look at our nested set diagram again:


If we wanted to add a new node between the TELEVISIONS and PORTABLE ELECTRONICS nodes, the new node would have lft and rgt values of 10 and 11, and all nodes to its right would have their lft and rgt values increased by two. We would then add the new node with the appropriate lft and rgt values. While this can be done with a stored procedure in MySQL 5, I will assume for the moment that most readers are using 4.1, as it is the latest stable version, and I will isolate my queries with a LOCK TABLES statement instead:

LOCK TABLE nested_category WRITE;

SELECT @myRight := rgt FROM nested_category
WHERE name = 'TELEVISIONS';


UPDATE nested_category SET rgt = rgt + 2 WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft + 2 WHERE lft > @myRight;

INSERT INTO nested_category(name, lft, rgt) VALUES('GAME CONSOLES', @myRight + 1, @myRight + 2);

UNLOCK TABLES;

We can then check our nesting with our indented tree query:

SELECT CONCAT( REPEAT( ' ', (COUNT(parent.name) - 1) ), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;


+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  GAME CONSOLES        |
|  PORTABLE ELECTRONICS |
|   MP3 PLAYERS         |
|    FLASH              |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
+-----------------------+

If we instead want to add a node as a child of a node that has no existing children, we need to modify our procedure slightly. Let's add a new FRS node below the 2 WAY RADIOS node:

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft FROM nested_category

WHERE name = '2 WAY RADIOS';
UPDATE nested_category SET rgt = rgt + 2 WHERE rgt > @myLeft;
UPDATE nested_category SET lft = lft + 2 WHERE lft > @myLeft;

INSERT INTO nested_category(name, lft, rgt) VALUES('FRS', @myLeft + 1, @myLeft + 2);

UNLOCK TABLES;

In this example we expand everything to the right of the left-hand number of our proud new parent node, then place the node to the right of the left-hand value. As you can see, our new node is now properly nested:

SELECT CONCAT( REPEAT( ' ', (COUNT(parent.name) - 1) ), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;


+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  GAME CONSOLES        |
|  PORTABLE ELECTRONICS |
|   MP3 PLAYERS         |
|    FLASH              |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
|    FRS                |
+-----------------------+

Deleting Nodes

The last basic task involved in working with nested sets is the removal of nodes. The course of action you take when deleting a node depends on the node's position in the hierarchy; deleting leaf nodes is easier than deleting nodes with children because we have to handle the orphaned nodes.

When deleting a leaf node, the process if just the opposite of adding a new node, we delete the node and its width from every node to its right:

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1
FROM nested_category
WHERE name = 'GAME CONSOLES';


DELETE FROM nested_category WHERE lft BETWEEN @myLeft AND @myRight;


UPDATE nested_category SET rgt = rgt - @myWidth WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft - @myWidth WHERE lft > @myRight;

UNLOCK TABLES;

And once again, we execute our indented tree query to confirm that our node has been deleted without corrupting the hierarchy:

SELECT CONCAT( REPEAT( ' ', (COUNT(parent.name) - 1) ), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;



+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  PORTABLE ELECTRONICS |
|   MP3 PLAYERS         |
|    FLASH              |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
|    FRS                |
+-----------------------+


This approach works equally well to delete a node and all its children:

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1
FROM nested_category
WHERE name = 'MP3 PLAYERS';


DELETE FROM nested_category WHERE lft BETWEEN @myLeft AND @myRight;

UPDATE nested_category SET rgt = rgt - @myWidth WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft - @myWidth WHERE lft > @myRight;

UNLOCK TABLES;

And once again, we query to see that we have successfully deleted an entire sub-tree:

SELECT CONCAT( REPEAT( ' ', (COUNT(parent.name) - 1) ), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;



+-----------------------+
| name                  |
+-----------------------+
| ELECTRONICS           |
|  TELEVISIONS          |
|   TUBE                |
|   LCD                 |
|   PLASMA              |
|  PORTABLE ELECTRONICS |
|   CD PLAYERS          |
|   2 WAY RADIOS        |
|    FRS                |
+-----------------------+

The other scenario we have to deal with is the deletion of a parent node but not the children. In some cases you may wish to just change the name to a placeholder until a replacement is presented, such as when a supervisor is fired. In other cases, the child nodes should all be moved up to the level of the deleted parent:

LOCK TABLE nested_category WRITE;

SELECT @myLeft := lft, @myRight := rgt, @myWidth := rgt - lft + 1
FROM nested_category
WHERE name = 'PORTABLE ELECTRONICS';

DELETE FROM nested_category WHERE lft = @myLeft;

UPDATE nested_category SET rgt = rgt - 1, lft = lft - 1 WHERE lft BETWEEN @myLeft AND @myRight;
UPDATE nested_category SET rgt = rgt - 2 WHERE rgt > @myRight;
UPDATE nested_category SET lft = lft - 2 WHERE lft > @myRight;

UNLOCK TABLES;

In this case we subtract two from all elements to the right of the node (since without children it would have a width of two), and one from the nodes that are its children (to close the gap created by the loss of the parent's left value). Once again, we can confirm our elements have been promoted:

SELECT CONCAT( REPEAT( ' ', (COUNT(parent.name) - 1) ), node.name) AS name
FROM nested_category AS node,
nested_category AS parent
WHERE node.lft BETWEEN parent.lft AND parent.rgt
GROUP BY node.name
ORDER BY node.lft;


+---------------+
| name          |
+---------------+
| ELECTRONICS   |
|  TELEVISIONS  |
|   TUBE        |
|   LCD         |
|   PLASMA      |
|  CD PLAYERS   |
|  2 WAY RADIOS |
|   FRS         |
+---------------+

Other scenarios when deleting nodes would include promoting one of the children to the parent position and moving the child nodes under a sibling of the parent node, but for the sake of space these scenarios will not be covered in this article.

Move a branch with childs

For move a branch with all child nodes I found the following query in this newsgroup discussion:

CREATE PROCEDURE MoveSubtree @my_root CHAR(2), @new_parent CHAR(2)
AS
DECLARE
  @origin_lft INT,
  @origin_rgt INT,
  @new_parent_rgt INT

SELECT @new_parent_rgt = rgt

FROM Tree
WHERE node = @new_parent;

SELECT @origin_lft = lft, @origin_rgt = rgt

FROM Tree
WHERE node = @my_root;

IF @new_parent_rgt < @origin_lft BEGIN

  UPDATE Tree SET
    lft = lft + CASE
      WHEN lft BETWEEN @origin_lft AND @origin_rgt THEN
        @new_parent_rgt - @origin_lft
      WHEN lft BETWEEN @new_parent_rgt AND @origin_lft - 1 THEN
        @origin_rgt - @origin_lft + 1
      ELSE 0 END,
    rgt = rgt + CASE
      WHEN rgt BETWEEN @origin_lft AND @origin_rgt THEN
        @new_parent_rgt - @origin_lft
      WHEN rgt BETWEEN @new_parent_rgt AND @origin_lft - 1 THEN
        @origin_rgt - @origin_lft + 1
      ELSE 0 END
  WHERE lft BETWEEN @new_parent_rgt AND @origin_rgt
     OR rgt BETWEEN @new_parent_rgt AND @origin_rgt;

END
ELSE IF @new_parent_rgt > @origin_rgt BEGIN

  UPDATE Tree SET
    lft = lft + CASE
      WHEN lft BETWEEN @origin_lft AND @origin_rgt THEN
        @new_parent_rgt - @origin_rgt - 1
      WHEN lft BETWEEN @origin_rgt + 1 AND @new_parent_rgt - 1 THEN
        @origin_lft - @origin_rgt - 1
      ELSE 0 END,
    rgt = rgt + CASE
      WHEN rgt BETWEEN @origin_lft AND @origin_rgt THEN
        @new_parent_rgt - @origin_rgt - 1
      WHEN rgt BETWEEN @origin_rgt + 1 AND @new_parent_rgt - 1 THEN
        @origin_lft - @origin_rgt - 1
      ELSE 0 END
  WHERE lft BETWEEN @origin_lft AND @new_parent_rgt
     OR rgt BETWEEN @origin_lft AND @new_parent_rgt;

END
ELSE BEGIN

  PRINT 'Cannot move a subtree to itself, infinite recursion';
  RETURN;
END

It's a bit long but it work...

Comments

  1. http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/

    Plagiarism?

    ReplyDelete
  2. I explicitly written in the IMPORTANT NOTE on the top of this article that I'm not the original author of this article. I found it some years ago in the MySQL web site but after some time the article disappeared and now it seem is very hard to find around. I didn't know the link you posted. If the blog's administrator is the original author of this article and want I remove it he need to simply ask, as I wrote in the same note.

    ReplyDelete
  3. Looks difficult, but thanks for the great workout!

    ReplyDelete
  4. How to model this nested set method in UML class diagram

    ReplyDelete
  5. Does this queries apply also for adding a node as last child?

    I believe in that case it should be something like this...

    LOCK TABLE nested_category WRITE;

    SELECT @myRight:= rgt FROM nested_category

    WHERE name = '2 WAY RADIOS';
    UPDATE nested_category SET rgt = rgt + 2 WHERE rgt = @myRight;

    INSERT INTO nested_category(name, lft, rgt) VALUES('FRS', @myRight, @myRight+ 1);

    UNLOCK TABLES;

    Or I missed something?

    ReplyDelete
  6. Move a branch with childs
    store procedure in mysql not working,Please explain it, how to use it.
    And please explain these two var (@my_root CHAR(2), @new_parent CHAR(2)) with a example.
    What is meaning of rgt = rgt + CASE line.
    Any help please...

    ReplyDelete
  7. This is not standard SQL code but, I guess, is some specific DB format incuding some additional programming commands. my_root is the root node of the subtree you want to move and new_parent is the, as the name suggest, the parent node you want to "attach" your subtree. "rgt = rgt + CASE" is a command for use the value "returned" by the two following WHEN based to the current value processed. You can not use this code as is, this is only a snippet showing the alghoritm to convert in your programming language code.

    ReplyDelete
  8. What if i had other categories than ELECTRONICS? like i want to add SPORTS, GARDENING?

    ReplyDelete
    Replies
    1. If you mean add SPORTS and GARDENING as siblings of ELECTRONICS, following the numbers used in the example will be SPORTS lft=21, rgt=22 and GARDENING lft=23, rgt=24

      Delete
  9. Many Thanks for your reply.
    Lets say i have ELECTRONICS lft=1, rgt=20 and SPORTS lft=21,rgt=38 then how would i know these are siblings?

    ReplyDelete
    Replies
    1. Basically you have to understand how this method work, that is not so difficult. If you want to retrieve siblings of root items just look for the item with lft value of 1 (first root) and check if exist another item with lft value equal to first root item found rgt + 1. Following the example you have to look for an item like item.lft = ELECTRONICS.rgt + 1 and so on for next root items.

      Delete
  10. Great stuff! This is really gonna save our arses at work lol. Thanks!

    ReplyDelete

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