Php 获取引用对象的数据

我有一个配方模型和食品成分模型，属于配方

RecipeModel：

class Recipe extends Model
{
    protected $table = 'recipes';

    protected $fillable = [ 
        'details'
    ];

    public function foods(){
            return $this->hasMany('App\FoodComponent','id','food_id');
    }
}

FoodComponent模型：

class FoodComponent extends Model
{
    protected $table = 'food_components';

    protected $fillable = [ 'title',
        'menu_id',
        'image'
    ];

    public function recipes(){
        return $this->belongsTo('App\Recipe','id','food_id');
    }
}

RecipeController：

public function index()
    {  
        $recipes = Recipe::all();

        return view('recipes.index')->with('recipes',$recipes);
    }

在我看来

@foreach($recipes as $recipe)
       {{ $recipe->foods }}
 @endforeach

我得到这样的东西：

[{"id":1,"menu_id":1,"title":"Pollo Locco","image":"images\/uploads\/foodcomponents\/99254.jpg","created_at":"2016-08-12 10:01:38","updated_at":"2016-08-12 10:01:38"}]

但我只想要“头衔”

我试过$recipe->foods->title。但它不起作用

你能告诉我怎么才能只显示标题吗

谢谢大家!

试试这个

@foreach($recipes as $recipe)
  @foreach($recipe->foods as $food)
   {{ $food->title }}
  @endforeach
@endforeach

试试这个

@foreach($recipes as $recipe)
  @foreach($recipe->foods as $food)
   {{ $food->title }}
  @endforeach
@endforeach

---

试试这个$recipe->foods[0]->标题。让我知道这是否有效。它有效！非常感谢。很高兴听到，但这只适用于一张唱片。你可以创建一个计数器或使用@LinuxArtism提供的答案，试试这个$recipe->foods[0]->title。让我知道这是否有效。它有效！非常感谢。很高兴听到，但这只适用于一张唱片。您可以创建计数器或使用@linuxartisan提供的答案