Php Yii嵌套设置为下拉菜单

我使用的是Yii嵌套集行为，这有助于保持我的类别嵌套，如图所示（不管标题行如何，它们都是俄语）：

我想做的就是让引导嵌套菜单，应该是这样的：

    $criteria = new CDbCriteria;
    $criteria->order = 'root, lft';
    $categories = Category::model()->findAll($criteria);
    foreach($categories as $i => $category) {
      $items[$i]['label'] = $category->title;
      $items[$i]['url'] = $category->url;
      $items[$i]['active'] = false;
      $items[$i]['items'] = array(
        array('label'=>'123', 'url'=>'#'),
        array('label'=>'123', 'url'=>'#'),
        array('label'=>'123', 'url'=>'#', 'items'=>array(
            array('label'=>'1234', 'url'=>'#'),
            array('label'=>'1234', 'url'=>'#'),
            array('label'=>'1234', 'url'=>'#', 'items'=>array(
                array('label'=>'1234', 'url'=>'#'),
                array('label'=>'1234', 'url'=>'#'),
                array('label'=>'1234', 'url'=>'#'),
            )),
        )),
    ); 
}
$this->widget('bootstrap.widgets.TbMenu', array(
    'type'=>'pills',
    'stacked'=>false, // whether this is a stacked menu
    'items'=>$items
)); 

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我不知道该怎么做，顺便说一句，我读过，只是不知道如何实际应用这个函数来解决我的问题。感谢您的帮助。

这是我用来格式化为json对象的函数，您可以修改它来生成php数组

protected function formatJstree(){
        $categories = $this->descendants()->findAll();
        $level=0;
        $parent = 0;
        $data = array();
        foreach( $categories as $n => $category )
        {
            $node = array(
                'data'=> "{$category->title}",
                'attr'=>array('id'=>"category_id_{$category->category_id}")
            );
            if($category->level == $level){
                $data[$parent]["children"][] = $node;
            }
            else if($level != 0 && $category->level > $level){
                if(!isset($data[$n]["children"])){
                    $data[$n]["children"] = array();
                }
                $data[$parent]["children"][] = $node;
            }
            else
            {
                $data[] = $node;
                $parent = $n;
            }
            $level=$category->level;

        }
        return $data;

    }

---

最后，我自己的递归解决方案（适用于多个根）：

---

用PHP和Yii（MVC框架）创建多级分类系统，而不仅仅是一项简单的任务。 1.在模型中创建函数

`

` 现在在view/file.php中创建此代码

<?php
echo ‘<select id=”parent_id” class=”select” name=”Course[parent_id]” >’;
echo “<option value=’0′  >–Select Exam–</option>”;
echo Course::model()->getRootCategory($model->parent_id);
// ($model->parent_id) means selected text box
echo ‘</select>’; ?>

---

更多详细信息单击此url访问

您将此代码放在何处？如果我使用$categories=$this->subjections（），我会得到：“语法错误或访问冲突：1064您的SQL语法有错误；检查与您的MySQL服务器版本对应的手册，以了解使用文档中的“near”和

t

rgt

的正确语法，确保您正确设置了扩展，并且数据库中的信息一致

function getRootCategory($cur_cat='') {
$sql='select id, course_name, parent_id from course where parent_id="0" and status=0'; 
$command=Yii::app()->db->createCommand($sql);
$return =$command->queryAll();
foreach($return as $rootCat){
if ($rootCat['id']==$cur_cat){
$test= 'selected=selected'; 
}else{
$test='';
 }
$id=$rootCat['id'];
echo "".$rootCat['course_name'].'';
$this->sub_cat($rootCat['id'] , '', $cur_cat ); 
}
 }

   function sub_cat($parentID=0, $space='',$cur_cat ) {
$sql="select id, course_name, parent_id from course where parent_id='$parentID' and status=0";
$command=Yii::app()->db->createCommand($sql);
$return =$command->queryAll();
$count=count($return);
if($parentID==0){ $space=''; }else{ $space .=" - "; }

if($count > 0){
foreach($return as $subcat){
if ($subcat['id']==$cur_cat){$test='selected=selected';}else{$test='';}
$ids=$subcat['id'];
echo "".$space.$subcat['course_name'].'';
$this->sub_cat($subcat['id'],$space, $cur_cat ); 
}
 } 
}

<?php
echo ‘<select id=”parent_id” class=”select” name=”Course[parent_id]” >’;
echo “<option value=’0′  >–Select Exam–</option>”;
echo Course::model()->getRootCategory($model->parent_id);
// ($model->parent_id) means selected text box
echo ‘</select>’; ?>