PHP tree menu, bottom-up

Question

I'm having some issues getting a tree menu to work from bottom-up. I already have a script to work from top-down, which works fine.

This is a very simplified version of my table:

+-----+-----------+--------------------+
| uid | parent_id | page_address       |
+-----+-----------+--------------------+
| 1   | 0         | index.php          |
| 2   | 0         | login.php          |
| 3   | 2         | dashboard.php      |
| 4   | 3         | bookings.php       |
| 5   | 3         | documents.php      |
| 6   | 4         | changebookings.php |
| 7   | 4         | activities.php     |
+-----+-----------+--------------------+

The page_address field is unique.

I can work out what page the user is currently on, for example changebookings.php

I would then like a menu to look like this:

login.php
    dashboard.php
    bookings.php
        changebookings.php
        activities.php
    documents.php

However, the closest I've got so far is the following tree:

login.php
    bookings.php
        changebookings.php

As you can see, my script currently only returns the actual parent, and not a list of links currently in the parent.

For those interested, the script I use in total is at the bottom of this post.

Is there any easier way to get the bottom-up tree as required? Many thanks

Phil

EDIT: I've finally got the code to work, for future users who stumble upon this post, I have added the functionality below:

$dataRows = $databaseQuery->fetchAll();     // Get all the tree menu records

$dataRows = $result->fetchAll(PDO::FETCH_ASSOC);

foreach($dataRows as $row)
{
    if($row['link_address']==substr($_SERVER['PHP_SELF'], 1, strlen($_SERVER['PHP_SELF'])-1))
    {
        $startingId = $row['parent_id'];
    }
}

$menuTree = $this->constructChildTree($dataRows, $startingId);


private function constructChildTree(array $rows, $parentId, $nesting = 0)
{
    $menu = array();

    if(!in_array($nesting, $this->nestingData))
    {
        $this->nestingData[] = $nesting;
    }

    foreach($rows as $row)
    {
        if($row['parent_id']==$parentId && $parentId!=0)
        {
            $menu[] = $row['link_address'];

            $newParentId = $this->getNextParent($rows, $row['parent_id']);

            $parentChildren = $this->constructChildTree($rows, $newParentId, ($nesting+1));

            if(count($parentChildren)>0)
            {
                foreach($parentChildren as $menuItem)
                {
                    $menu[] = 'NESTING' . $nesting . '::' . $menuItem;
                }
            }
        }
    }

    return $menu;
}


private function getNextParent($rows, $parentId)
{
    foreach($rows as $row)
    {
        if($row['uid']==$parentId)
        {
            return $row['parent_id'];
        }
    }
}

[Storing Hierarchical Data in a Database](http://www.sitepoint.com/hierarchical-data-database-2/) — DarkBee, Apr 22 '13 at 11:23

score 1 · Accepted Answer · answered Apr 22 '13 at 11:16

1

Without reading your code you should be doing:

1) Get current page, look at parent ID.

2) Load all with that parent ID.

3) Get next Parent ID using current Parent ID as ID.

4) If new parent ID != 0, goto step 2 passing in the new Parent ID.

Sounds like you just need to edit your script to include ALL pages with the given ID as their parent ID.

answered Apr 22 '13 at 11:16

MatthewMcGovern

3,466
1
19
19

I'll accept this answer as it was helpful towards re-writing the code which eventually made it work as expected! – Phil Cross Apr 22 '13 at 12:16
Cool :) I've done something similar in Javascript so I knew the steps but didn't have the time to go through your code and write it! – MatthewMcGovern Apr 22 '13 at 12:23

score 1 · Answer 2 · edited Jan 07 '14 at 13:52

1

<?PHP
$sql = "SELECT * FROM TABLE WHERE table parent_id=0";
$result = mysql_query($sql);
while($perant_menu = mysql_fetch_array($result))
{
    echo display_child($perant_menu["uid"],$perant_menu["page_address"]);
}

// Recursive function
function display_child($parent_id,$name)
{
    $sql= "SELECT * FROM table where parent_id = $parent_id";
    $result = mysql_query($sql);
    if(mysql_num_rows($result)>0)
    {
        while($menu = mysql_fetch_array($result))
        {
            echo display_child($menu["id"],$menu["page_address"]);
        }
    }
    else
    {
        echo $name;
    }
}
?>

edited Jan 07 '14 at 13:52

BenMorel

34,448
50
182
322

answered Apr 22 '13 at 11:32

Dharmesh Hariyani

394
1
14

Unfortunately I couldn't use this code as it uses deprecated mysql* functionality. Also, I'm trying to limit the number of SQL queries made to the server, which unfortunately with this answer, for each parent to retrieve another query would be sent though. I appreicate the time it took to response though :) – Phil Cross Apr 22 '13 at 12:21

PHP tree menu, bottom-up

2 Answers2