具有递归功能的回显菜单树

怎么样：

function recurse($categories, $parent = null, $level = 0){    $ret = '<ul>';    foreach($categories as $index => $category)    {        if($category['root'] == $parent)        { $ret .= '<li><a href="#"><p >' . $category['name'] . '</p></a>'; $ret .= $this->recurse($categories, $category['id'], $level+1); $ret .= '</li>';        }    }    return $ret . '</ul>';}

此函数要求您首先在数据库中查询可用类别的整个列表，并假设您的根类别的值为null，但是可以根据当前模式的工作方式将该函数更改为接受-1或0。

$categories = { get from database into an multi-dimensional array };$Tree = $this->recurse($categories);echo $Tree;

您可以考虑执行以下操作，以防止在父级不存在任何子级时出现任何空的UL：

function recurse($categories, $parent = null, $level = 0){    $ret = '<ul>';    foreach($categories as $index => $category)    {        if($category['root'] == $parent)        { $ret .= '<li><a href="#"><p >' . $category['name'] . '</p></a>'; $sub = $this->recurse($categories, $category['id'], $level+1); if($sub != '<ul></ul>')     $ret .= $sub; $ret .= '</li>';        }    }    return $ret . '</ul>';}

但是，最好的解决方案是选择数据以包括一列，该列包含每个类别有多少个子类别。

select Category.*, (select count(distinct c1.id) from Category as c1 where c1.root = Category.id) as ChildCount from Category

您的功能将是：

function recurse($categories, $parent = null, $level = 0){    $ret = '<ul>';    foreach($categories as $index => $category)    {        if($category['root'] == $parent)        { $ret .= '<li><a href="#"><p >' . $category['name'] . '</p></a>'; if($category['ChildCount'] > 0)     $ret .= $this->recurse($categories, $category['id'], $level+1); $ret .= '</li>';        }    }    return $ret . '</ul>';}

希望有帮助吗？