Help with PHP recursive navigation list menu

Question

I am trying to add a dynamic recursive navigation list menu to a site of am working on. The scenerio is that the menu has 2 levels related by a parentid(preid).

My issue is that I can display the 1st level list correctly, however I cannot get the second level to display properly. I am not sure where to add the UL and /UL tags for the second level.

This is what I am after

<ul>
<li>Item 1</li>
<li>item 2</li>
<li>item 3</li>
<ul>
  <li>sub item 1</li>
  <li>sub item 2</li>
</ul>
<li>Item 4</li>
<li>item 5</li>
<ul>
  <li>sub item 1</li>
  <li>sub item 2</li>
</ul>
<li>item 6</li>
</ul>

This is actually what i am getting with the below code:

    <ul>
  <li>item 1
    <ul>
    </ul>
  </li>
  <li>item 2
    <ul>
      <li>sub item 1</li>
      <ul>
      </ul>
      <li>sub item 2</li>
      <ul>
      </ul>
    </ul>
  </li>
  <li>Sports Injuries
    <ul>
    </ul>
  </li>
    </ul>
  </li>
</ul>

Below is the class file I am using to create the menu:

class Dynamic_Menu 
    {
        function getConfig()
        {
            $this->DB_SERVER = 'localhost';
            $this->DB_USER = '***';
            $this->DB_PASS = '***';
            $this->DB_NAME = '***';

        }

        function __construct()
        {
            $this->getConfig();
            $Conn = mysql_connect($this->DB_SERVER, $this->DB_USER, $this->DB_PASS);
            if (!$Conn)
                die("Error: ".mysql_errno($Conn).":- ".mysql_error($Conn));
            $DB_select = mysql_select_db($this->DB_NAME, $Conn);
            if (!$DB_select)
                die("Error: ".mysql_errno($Conn).":- ".mysql_error($Conn));
        }

        function select_row($sql)
        {
            //echo $sql . "<br />";
            if ($sql!="")
            {
                $result = mysql_query($sql) or die("Error: ".mysql_errno().":- ".mysql_error());
                if ($result)
                {
                    while($row = mysql_fetch_array($result))
                        $data[] = $row;
                }
                return $data;
            }
        }

        function recordCount($sql)
        {
            if ($sql!="")
            {
                $result = mysql_query($sql) or die("Error: ".mysql_errno().":- ".mysql_error());
                if ($result)
                {
                    $cnt = mysql_num_rows($result);
                    return $cnt;
                }
            }
        }

        function getChild($id)
        {
            $menu = "";
            $str = "";
            $s = "SELECT * FROM vcms_sys_explorer WHERE preid = '$id' ";
            $res = $this->select_row($s);
            $menu .= '<ul>';
            for ($i=0;$i<count($res);$i++)
            {
                $cnt_of_child = $this->recordCount("SELECT * FROM vcms_sys_explorer where preid = '".$res[$i][eid]."' ");
                //if ($cnt_of_child > 0)
                //  $str = '';
                //else
                //  $str = " (is sub menu item)";

                $menu .= '<li>'. $res[$i][name].$str.'</li>';   
                $menu .= $this->getChild($res[$i][eid]);
            }
            $menu .= '</ul>';       
            return $menu;
        }

        function getMenu($parentid)
        {
            $menu = "";
            $s = "SELECT * FROM vcms_sys_explorer WHERE preid = '$parentid'  ";
            $res = $this->select_row($s);

            $menu .= '<ul>';

            for ($i=0;$i<count($res);$i++)
            { 
                $menu .= '<li>'.$res[$i][name].$this->getChild($res[$i][eid]).'</li>';
                if ((count($res) - 1) > $i) {
                }
            } 

            $menu .= '</ul>';

            return $menu;
        }
    }

I call the menu with:

$menu = new Dynamic_Menu();
$menu->getMenu(1);

Could someone please help and explain where I need to place the level 2 UL and /UL tags. I have been banging my head with this for the last 2 days. Any help would be greatly appreciated, thanks...

Daniel · Accepted Answer · 2010-05-26 04:39:49Z

11

In a nested list, sub-lists will always be contained within a list element-- that's what makes them nested. You can print a full list in just one function using this format (in generic code, but you should get the basic idea):

function get_list($parent) {
    $children = query('SELECT * FROM table WHERE parent_id = '.$parent);
    $items = array();
    while($row = fetch_assoc($children)) {
        $items[] = '<li>'.$row['name'].get_list($row['id']).'</li>';
    }
    if(count($items)) {
        return '<ul>'.implode('', $items).'</ul>';
    } else {
        return '';
    }
}

And this will give you a list structured properly as:

<ul>
    <li>Item 1</li>
    <li>Item 2
        <ul>
            <li>Item 2.1</li>
            <li>Item 2.2</li>
        </ul>
    </li>
</ul>

answered May 26, 2010 at 4:39

Daniel

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1 Comment

Jason Over a year ago

Hi Daniel, you are a guru, thanks very much for your time and help on this. Your answer works like a charm...

Community · Accepted Answer · 2017-05-23 12:28:17Z

0

All though this question is not the exact same as the question I posted 2 days ago, here is the result of what I was attempting to do with folders rather than a DB. The following will traverse the directory and all sub directories of the specified $path and spits out the results in a nested un-ordered list upon completion of running the script. Hope it helps.

<?php
function readDirs($path){
    $dirHandle = opendir($path);
    echo "<ul>";
    while ($item = readdir($dirHandle)) {
        $newPath = $path . "/" . $item;

        if (is_dir($newPath) && $item != '.' && $item != '..') {
            echo "<li><a href='$newPath'>$item</a>";
            readDirs($newPath);
        }
    }
    echo "</li></ul>";
}
$path = "./galleries";
readDirs($path);

?>

edited May 23, 2017 at 12:28

CommunityBot

11 silver badge

answered Dec 30, 2012 at 14:03

Sam J. Dennis

681 silver badge6 bronze badges

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