创建树结构所需的 PHP 递归帮助

是的，有一个更好的方法。所谓修正的预序树遍历算法。您可以通过谷歌搜索找到大量信息，我相信堆栈溢出也是如此。

好处是您只需使用 1 个查询即可获取整个子树。选择会很快，但修改更重。

我认为您必须在函数调用中添加id作为参数。

$this->categoriesTree($row['menuid']) 

否则，每次

这是最好的例子。这是第一个答案的更正形式。

 function categoriesTree($id=0) {
    $s = "SELECT * FROM design_menu  WHERE menuparentId = '" . $id . "' 
    ORDER BY menuorder, menuid ";
    $rid = $this->db->query($s)->result_array();
    $treeArray = array();
    foreach ($rid as $row) {
        $treeArray[$row['menuid']] = $row;
        if ($row['menuhasChild'] == 1) {
            $treeArray[$row['menuname']] = $this->categoriesTree($row['menuid']); //results in Fatal error: Maximum function nesting level of '100' reached, aborting!
        }
    }
    return $treeArray;
}

行 ：

$treeArray[$row['menuid']] = $this->categoriesTree();

必须是 ：

$treeArray[$row['menuid']] = $this->categoriesTree($row['menuid']);

<?php
require_once ROOT_PATH . '/lib/dao/MySQLClass.php';
require_once ROOT_PATH . '/lib/confs/Conf.php';
/**
 * Generate HTML for multi-dimensional menu from MySQL database
 * with ONE QUERY and WITHOUT RECURSION
 * @author J. Bruni
 */
//print_r($_SESSION['symfony/user/sfUser/culture']);die;
class MenuBuilder
{
    /**
     * MySQL connection
     */
    var $conn;
    /**
     * Menu items
     */
    var $items = array();
    /**
     * HTML contents
     */
    var $html  = array();
        //var $culture = $_SESSION['symfony/user/sfUser/culture'];
        var $culture;
        var $columnName;
    /**
     * Create MySQL connection
     */
    function MenuBuilder()
    {
        $conf = new Conf();
            $db=new MySQLClass($conf);
           $this->conn = mysql_connect($db->myHost .':'.$db->myHostPort, $db->userName, $db->userPassword);
           $this->culture=$_SESSION['language'];
    }
    /**
     * Perform MySQL query and return all results
     */
    function fetch_assoc_all( $sql )
    {
                if($this->culture=="en"){
                $this->columnName='sm_mnuitem_name';
                }else{
                    $this->columnName='sm_mnuitem_name_'.$this->culture;
                }
                //die(print_r($_SESSION));
                if($_SESSION['user']=="USR001"){
                $query="SELECT sm_mnuitem_id, sm_mnuitem_parent, ".$this->columnName.", sm_mnuitem_webpage_url, sm_mnuitem_position FROM hs_hr_sm_mnuitem ORDER BY sm_mnuitem_parent, sm_mnuitem_position;";
                }
                else{
                    $query="select * from hs_hr_sm_mnuitem m left join hs_hr_sm_mnucapability c on m.sm_mnuitem_id=c.sm_mnuitem_id left join hs_hr_users u on u.sm_capability_id=c.sm_capability_id where u.id='".$_SESSION['user']."' ORDER BY m.sm_mnuitem_parent, m.sm_mnuitem_position;";
                }
        //$result = mysql_query("SELECT sm_mnuitem_id, sm_mnuitem_parent, ".$this->columnName.", sm_mnuitem_webpage_url, sm_mnuitem_position FROM hs_hr_sm_mnuitem ORDER BY sm_mnuitem_parent, sm_mnuitem_position;",$this->conn);
                $result = mysql_query($query,$this->conn);
        if ( !$result ){
            return false;
                }
        $assoc_all = array();
        while( $fetch = mysql_fetch_assoc( $result ) ){
            $assoc_all[] = $fetch;
                }
                //die(print_r($assoc_all));
        mysql_free_result( $result );
        return $assoc_all;
    }
    /**
     * Get all menu items from database
     */
    function get_menu_items()
    {
        // Change the field names and the table name in the query below to match tour needs
        $sql = 'SELECT sm_mnuitem_id, sm_mnuitem_parent, sm_mnuitem_name, sm_mnuitem_webpage_url, sm_mnuitem_position FROM hs_hr_sm_mnuitem ORDER BY s_mnuitem_parent, sm_mnuitem_position;';
        return $this->fetch_assoc_all( $sql );
    }
    /**
     * Build the HTML for the menu
     */
    function get_menu_html( $root_id = 0 )
    {
        $this->html  = array();
        $this->items = $this->get_menu_items();
                //print_r($this->items);die("");
        foreach ( $this->items as $item )
            $children[$item['sm_mnuitem_parent']][] = $item;
        // loop will be false if the root has no children (i.e., an empty menu!)
        $loop = !empty( $children[$root_id] );
        // initializing $parent as the root
        $parent = $root_id;
        $parent_stack = array();
        // HTML wrapper for the menu (open)
                //$this->html[] = '<div>';
        $this->html[] = '<ul id="qm0" class="qmmc">';
        while ( $loop && ( ( $option = each( $children[$parent] ) ) || ( $parent > $root_id ) ) )
        {
            if ( $option === false )
            {
                $parent = array_pop( $parent_stack );
                // HTML for menu item containing childrens (close)
                $this->html[] = str_repeat( "'t", ( count( $parent_stack ) + 1 ) * 2 ) . '</ul>';
                $this->html[] = str_repeat( "'t", ( count( $parent_stack ) + 1 ) * 2 - 1 ) . '</li>';
            }
            elseif ( !empty( $children[$option['value']['sm_mnuitem_id']] ) )
            {
                $tab = str_repeat( "'t", ( count( $parent_stack ) + 1 ) * 2 - 1 );
                // HTML for menu item containing childrens (open)
                $url="";
                if($option['value']['sm_mnuitem_webpage_url']=="#"){
                    $url="javascript:void(0);";
                }else{
                    $url=$option['value']['sm_mnuitem_webpage_url'];
                }        
                $this->html[] = sprintf(
                    '%1$s<li><a class="qmparent" href="%2$s">%3$s</a>',
                    $tab,   // %1$s = tabulation
                    //$option['value']['sm_mnuitem_webpage_url'],   // %2$s = link (URL)
                    $url,
                    $option['value'][$this->columnName]   // %3$s = title
                );
                $this->html[] = $tab . "'t" . '<ul>';
                array_push( $parent_stack, $option['value']['sm_mnuitem_parent'] );
                $parent = $option['value']['sm_mnuitem_id'];
            }
            else{
                // HTML for menu item with no children (aka "leaf")
                             if($_SESSION['user']!="USR001"){
                            if($option['value']['sm_mnuitem_webpage_url']!="#"){
                $this->html[] = sprintf(
                    '%1$s<li><a target="rightMenu" href="%2$s">%3$s</a></li>',
                    str_repeat( "'t", ( count( $parent_stack ) + 1 ) * 2 - 1 ),   // %1$s = tabulation
                    $option['value']['sm_mnuitem_webpage_url'],   // %2$s = link (URL)
                    $option['value'][$this->columnName]   // %3$s = title
                );
                            }
                             }else{
                                 $this->html[] = sprintf(
                    '%1$s<li><a target="rightMenu" href="%2$s">%3$s</a></li>',
                    str_repeat( "'t", ( count( $parent_stack ) + 1 ) * 2 - 1 ),   // %1$s = tabulation
                    $option['value']['sm_mnuitem_webpage_url'],   // %2$s = link (URL)
                    $option['value'][$this->columnName]   // %3$s = title
                );
                             }
                        }
        }
        // HTML wrapper for the menu (close)
        $this->html[] = '</ul>';
                //$this->html[] = '</div>';
        return implode( "'r'n", $this->html );
    }
}


?>

使用基于关系数据库的父/子关系的树结构构建菜单非常麻烦。关系数据库对于树结构来说是可怕的。它们要求您编写大量业务逻辑，只是为了以可读的格式表示数据。要使用附加功能更新菜单需要添加到该递归循环中......它可能会变得非常毛茸茸，具体取决于您希望菜单变得多么复杂。更不用说你最终会想要缓存整个东西，因为在重负载下循环在计算上变得非常昂贵。想一想，如果你有 5 个顶级菜单项，2 个子项，每个子项自己有 n 个子项，你将运行 16 个 SQL 语句。

我可以提供另一种解决方案：JSON。我曾经有这样的菜单表，现在我只是在SQL数据库中存储它的JSON表示形式（尽管即使这样也可以缓存在内存/文件系统中）。JSON菜单在空间方面要紧凑得多，简单阅读是合乎逻辑的，不需要摆弄父ID和子ID。它与PHP（json_encode/解码）将菜单转换为本机数组时效果更好。以及Javascript，这很重要，例如，如果你正在执行ajax调用来重新排序应用程序中的菜单。分层树结构是JSON擅长的。它还消除了跟踪"菜单顺序"的需要（因为数组顺序本质上是指定的）

菜单格式示例如下：

{
["en": "Home", "fr": "Accueil"],
["en": "Settings", "fr": "Paramètres", "child": 
    {
        ["en": "Email", "fr": "Email", "role": "EmailUser"]
    }
}

如您所见，它非常容易地提供了附加功能，例如绑定到菜单项的"角色"。添加此类功能不需要新的递归代码或更改 SQL 架构。它真的更加灵活。

所以，不是真正回答这个问题，但希望提供一些建议/见解，说明我认为是这个问题的更好解决方案。