Laravel 菜单自递归

Laravel Menu self recursion

提问by Hello

the model

该模型

 class Menu extends Eloquent {

        public static $table = 'menus';

        public function parent_menu()
        {
            return $this->belongs_to('Menu', 'parent_id');
        }

    }

how I get it in the controller:

我如何在控制器中获取它：

$menus = Menu::with('parent_menu')->get();

how do I render it in the view:

我如何在视图中呈现它：

foreach($menus as $m)
{
  echo $m->parent_menu->title;
}

looks like there is a problem when the relation is inside a table, i get an error

当关系在表中时，看起来有问题，我收到错误

`trying to get property of non object`

is there a solution for this?

有解决方案吗？

回答by Pevara

I have implemented a way to get endless depth in menu's in Laravel 4. It is not exactly what you ask, but the technique should be easily adaptable.

我已经在 Laravel 4 中实现了一种在菜单中获得无限深度的方法。这不是你所要求的，但该技术应该很容易适应。

For starters my menu is just an array (for now) that gets assigned to the master view and that looks something like this.

对于初学者来说，我的菜单只是一个数组（目前），它被分配给主视图，看起来像这样。

$menu = array(
   array(
     'name' => 'item1',
     'url' => '/'
   ),
    array(
     'name' => 'item2',
     'url' => '/',
     'items' => array(
        array(
          'name' => 'subitem1',
          'url' => '/'
        ),
        array(
          'name' => 'subitem2',
          'url' => '/'
        )
     )
   )
);

You could easily achieve this structure by using a Model as well. You will need function child_itemsor something as we will render the menu from the top down, and not from the bottom up.

您也可以通过使用模型轻松实现此结构。您将需要函数child_items或其他东西，因为我们将自上而下呈现菜单，而不是自下而上。

Now in my master blade template I do this:

现在在我的主刀片模板中，我这样做：

<ul>
    @foreach ($menu as $item)
       @include('layouts._menuItem', array('item' => $mainNavItem))
    @endforeach
</ul>

And then in the layouts._menuItemtemplate I do this:

然后在layouts._menuItem模板中我这样做：

<?php
$items = array_key_exists('items', $item) ? $item['items'] : false;
$name = array_key_exists('name', $item) ? $item['name'] : '';
$url = array_key_exists('url', $item) ? url($item['url']) : '#';
$active = array_key_exists('url', $item) ? Request::is($item['url'].'/*') : false;
?>

<li class="@if ($active) active @endif">
       <a href="{{ $url }}">{{ $name }}</a>
       @if ($items)
          <ul>
             @foreach ($items as $item)
               @include('layouts._menuItem', array('item' => $item))
             @endforeach
          </ul>
       @endif
</li>

As you can see this template calls itself recursively, but with a different $itemvariable. This means can go as deep as you want in your menu structure. (The php block is just there to prepare some variables so I could keep the actual template code clean and readable, technically it is not required).

如您所见，此模板以递归方式调用自身，但使用不同的$item变量。这意味着可以在菜单结构中尽可能深入。（php 块只是为了准备一些变量，所以我可以保持实际的模板代码干净和可读，从技术上讲，它不是必需的）。

I stripped the Twitter Bootstrap code in the snippets above to keep things simple (I actually have titles, dropdown toggles, icons, dividers, ... in my template / array), so the code is not tested. The full version is working fine for me though, so let me know if I made a mistake somewhere.

我删除了上面代码片段中的 Twitter Bootstrap 代码以保持简单（我的模板/数组中实际上有标题、下拉切换、图标、分隔符……），因此代码没有经过测试。不过，完整版对我来说很好用，所以如果我在某处犯了错误，请告诉我。

Hope this helps you (or anyone else, cause this is a rather old question) on the way. Let me know if you need any more pointers / help, or if you want my full code.

希望这对您（或其他任何人，因为这是一个相当古老的问题）有所帮助。如果您需要更多指示/帮助，或者您想要我的完整代码，请告诉我。

Happy coding!

快乐编码！

回答by ktsakas

I believe the following is the correct way to do recursion in laravel.

我相信以下是在 laravel 中进行递归的正确方法。

Supposing we have a children relation, you can add this to your model class:

假设我们有一个 children 关系，您可以将其添加到您的模型类中：

public function getDescendants ( $parent= false ) {
    $parent = $parent ?: $this;
    $children = $parent->children()->get();

    foreach ( $children as $child ) {
        $child->setRelation(
            'children', 
            getDescendants( $child )
        );
    }

    return $children;
}

The above will get all the children records recursively, and you can access them like this:

以上将递归获取所有子记录，您可以像这样访问它们：

$d = Category::find(1)->getDescendants();

foreach ( $d as $child_level_1 ) {
    foreach ( $child_level_1->children as $child_level_2 ) {
        foreach ( $child_level_2->children as $child_level_3 ) {
            // ...... this can go on for infinite levels
        }
    }
}

Although not tested, the following might be useful to flatten all the recursive relations into one collection of models (check the documentation on adding new methods to collections):

尽管未经测试，但以下内容可能有助于将所有递归关系平铺到一个模型集合中（查看有关向集合添加新方法的文档）：

// Add this to your model
public function newCollection ( array $models = array() ) {
    return new CustomCollection( $models );
}


// Create a new file that extends the orginal collection
// and add the flattenRelation method
class CustomCollection extends Illuminate\Database\Eloquent\Collection {
    // Flatten recursive model relations
    public static function flattenRelation ( $relation ) {
        $collection = $this;
        // Loop through the collection models
        foreach ( $collection as $model ) {

            // If the relation exists
            if ( isset($model->relations[$relation]) ) {
                // Get it
                $sub_collection = $model->relations[$relation];

                // And merge it's items with the original collection
                $collection = $collection->merge(
                    $sub_collection->flatten($relation)
                );

                // Them remove it from the relations
                unset( $model->relations[$relation] );
            }

        }

        // Return the flattened collection
        return $collection;
    }
}

That way you can do the following:

这样，您可以执行以下操作：

// This will get the descenands and flatten them recursively
$d = Category::find(1)->getDescendants()->flattenRelation( 'children' );

// This will give you a flat collection of all the descendants
foreach ( $d as $model ) {

}

回答by user3229469

My laravel menu with unlimited submenus (menu items from database)

我的 Laravel 菜单具有无限子菜单（来自数据库的菜单项）

public function CreateMenu( $parid, $menu, $level ) {

    $output = array();    
    $action= Route::current()->getUri();
    $uri_segments = explode('/', $action);
    $count=count($uri_segments);
    foreach( $menu as $item => $data ) {

            if ($data->parent_id == $parid) { 
                $uri='';
                $output[ $data->id ] = $data;
                for($i=0; $i<=$level; $i++) {
                    if($i < $count) {
                        $uri.="/".Request::segment($i+1);
                    }
                    if($uri == $data->link ) {
                        $output[ $data->id ]->activeClass = 'active';
                        $output[ $data->id ]->inClass = 'in';
                    }
                    else {
                        $output[ $data->id ]->activeClass = '';
                        $output[ $data->id ]->inClass = '';
                    }
                    $output[ $data->id ]->level = $level+2;
                }
                $output[ $data->id ]->submenu = self::CreateMenu( $data->id, $menu, $level+1 );
            }

    }
    return $output;

}

In the BaseController or where you want, put

在 BaseController 或你想要的地方，把

$navitems=DB::table('navigations')->get();
$menu=BaseController::CreateMenu(0,$navitems,0);
return View::share($menu);

After that, i put the menu html to the macro.php

之后，我把菜单 html 放到了 macro.php

HTML::macro('MakeNavigation', function($data) {

foreach ($data as $key => $value) {
    if($value->submenu) {
        echo '<li class="'.$value->activeClass.'">
        <a href="'.$value->link.'" class="'.$value->activeClass.'">"'
            .$value->name.' <span class="fa arrow"></span> 
        </a>';
        echo "<ul class='nav nav-".$value->level."-level ".$value->inClass." '>";
            HTML::MakeNavigation($value->submenu);
        echo "</ul>";
    }
    else {
        echo '<li class="'.$value->activeClass.'">
        <a href="'.$value->link.'" class="'.$value->activeClass.'">'
                      .$value->name.'
        </a>';
    }
    echo "</li>";
}});

And in the view (blade templating) just call the

并在视图（刀片模板）中调用

{{ HTML::MakeNavigation($menu) }}

回答by tuurbo

This may be of some help. Heres how i did it with product categories / sub categories

这可能会有所帮助。这是我如何处理产品类别/子类别

Model:

模型：

<?php

class Category extends Eloquent {

    protected $table = 'product_category';

    public function subcat()
    {
        return $this->hasMany('Category', 'node_id')->orderBy('position');
    }

Query (you can use conditions when using eager loading)

查询（使用急切加载时可以使用条件）

$categories = Category::with(['subcat' => function($query){
    $query->with(['subcat' => function($query){
        $query->orderBy('name');
    }])
}])->where('node_id', 0)->orderBy('position')->get(['id', 'name']);

foreach ($categories as $level1)
{
    echo $level1->name;

    foreach ($level1->subcat as $level2)
    {
        echo $level2->name;

        foreach ($level2->subcat as $level3)
        {
                echo $level3->name;
        }
    }
}