使用PHP MYSQL和foreach语句处理多级菜单

Question

My Code and tables are given below

TABLES

    menu
    ====
    m_id    menu_name   parent_menu_id  menu_order  menu_url    status 
    1       M1          0               1           #           1
    2       M2          0               2           #           1
    3       S1          1               1           #           1
    4       S2          1               2           #           1

    user_rights
    ===========     
    user_id         rights
    1               1,2,3,4
    2               1,2,3

CODE

    <?php
    $get_user_id = "1"; //  user id will be 1 or 2 or any id from user table
    mysql_connect("localhost", "root", "password") or die(mysql_error());
    mysql_select_db("booksdb") or die(mysql_error());

    function getMenuRights($get_user_id) {
    $get_user_id = intval($get_user_id);
    $get_access_query = mysql_query("SELECT access FROM user_rights where user_id='".$get_user_id."'");
    $fetch_access_list = mysql_fetch_row($get_access_query);
    $get_access_list = $fetch_access_list[0];
    return $get_access_list;
    }

    function getUserMainMenus($get_user_id) {
    $get_user_id = intval($get_user_id);
    $rights = getMenuRights($get_user_id);
    $get_main_menu_query = mysql_query("SELECT * FROM menu where m_id IN ($rights) and parent_menu_id='0' order by menu_order");
    while ($row = mysql_fetch_assoc($get_main_menu_query)) {
    $results[] = $row;
    }
    return $results;
    }

    function getUserChildMenu($parent_menu_id, $get_user_id) {
    $parent_menu_id = intval($parent_menu_id);
    $get_user_id = intval($get_user_id);
    $rights = getMenuRights($get_user_id);
    $get_sub_menu = mysql_query("SELECT * FROM menu where parent_menu_id='".$parent_menu_id."' AND m_id IN ($rights)");
    while($row = mysql_fetch_assoc( $get_sub_menu )) {
    $results[] = $row;
    }
    return $results;
    }

    ?>
    <div id='cssmenu'>
    <ul>
    <?php foreach (getUserMainMenus($get_user_id) as $get_main_menu): ?>
    <li class='has-sub'><a href='#'><span><?=$get_main_menu['menu_name']; ?></span></a>
    <ul>
    <?php foreach (getUserChildMenu($get_main_menu['m_id'], $get_user_id) as $sub_menu): ?>
    <li class='has-sub'><a href='<?=$sub_menu['menu_url']; ?>'><span><?=$sub_menu['menu_name']; ?></span></a>
    </li>
    <?php endforeach; ?>   
    </ul>
    </li>
    <?php endforeach; ?>   
    </ul>
    </div>

The above code works fine, but it shows some error.

Undefined variable: results in phpfilename.php on line 21(and 32)   

Also, I have to check before each menu and sub-menu start whether there is value or not and then show the sub-menus.

How can I do that?

In other words, before the start of my menu and sub-menu I need to check for values and then show. If values, show the sub-menu. If not show the main menu alone.

Any help.

Thanks, Kimz

Answer 1

You are trying to append to the array in variable $results which isn't initiated yet.

Add the following:

$results = array();

Above each:

while() { ... }

---

For example:

function getUserMainMenus($get_user_id) {
    $get_user_id = intval($get_user_id);
    $rights = getMenuRights($get_user_id);
    $get_main_menu_query = mysql_query("SELECT * FROM menu where m_id IN ($rights) and parent_menu_id='0' order by menu_order");
    $results = array(); // add this here
    while ($row = mysql_fetch_assoc($get_main_menu_query)) {
    $results[] = $row;
    }
    return $results;
    }