laravel-使用自定义模型函数获取父类别平铺

Question

我已经用下面提到的“parent_id”在category类中定义了父关系，我想用父标题打印category title，比如mens

未定义的属性：Illumb\Database\Eloquent\Relations\BelongsTo:：$title

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Category extends Model
{
    //
    protected $table = 'categories';


    public function parent()
    {
        return $this->belongsTo('App\Category', 'parent_id');
    }   

    public function withParentTitle(): string
    {               
        if($this->parent()){            
            return $this->parent()->title.' > '. $this->title;
        }
        return $this->title;
    }   


}

类别控制器

................

namespace App\Http\Controllers;

use App\Category;

class CategoryController extends Controller
{

    public function index(){        
        $categories = Category::get();
        foreach($categories as $category){
            echo $category->withParentTitle();
        }
    }
}




Answer 1

public function withParentTitle(): string
    {               
        // Your $this->parent() would always be true, 
        // use $this->parent()->exists() instead
        if($this->parent()->exists()){            
            // $this->parent() refers to the relationship, 
            // you can't get title out of the relationship object, 
            // unless you retrieve it using $this->parent
            return $this->parent->title.' > '. $this->title;
        }
        return $this->title;
    }   

Answer 2

您必须在不使用（）的情况下进行呼叫，如下所示：

public function withParentTitle(): string
{               
    if($this->parent()){            
        return $this->parent->title.' > '. $this->title;
    }
    return $this->title;
}

因为如果您使用（），您将调用关系作为查询生成器，并且您可以使用其他闭包：

$this->parent()->where('title', 'foo')->first();

Answer 3

根据laravel文件https://laravel.com/docs/5.8/eloquent-mutators

您可以通过这种方式使用访问器

public function getParentTitleAttribute(){
  return $this->parent ? $this->parent->title . '>' . $this->title : $this->title;
}

然后调用$category-