Laravel-5.8：试图获取非对象的属性“display_name”

Question

当我尝试评估来自类别模型的数据时，我在 Laravel 中收到“试图获取非对象的属性‘display_name’”错误。我有 2 个模型：Post 和 Category；我有相应的表格：Posts('id', 'title', 'body', 'cover_image')和Categories('id', 'display_name','name', 'key).数据透视表是：category_post这是我的帖子模型：namespace App;use Illuminate\Database\Eloquent\Model;class Post extends Model{    //Table Name    protected $table = 'posts';    //Model Relationships    public function tag(){        return $this->belongsToMany('App\Tag', 'id');    }    public function category() {        return $this->belongsTo('App\Category','id');    }}这是我的类别模型：namespace App;use Illuminate\Database\Eloquent\Model;class Category extends Model{    protected $table = 'categories';//Model Relationships    public function post()     {        return $this->belongsTo('App\Post', 'id');    }}另外，这是我的 PagesController：namespace App\Http\Controllers;use Illuminate\Http\Request;use App\Post;class PagesController extends Controller{    public function index()    {        if ( !is_null(Post::class) ) {                $posts = Post::latest()->with('category')->paginate(5);            return view('pages.index')->with(['posts' => $posts]);        }        else {            return view('pages.empty');        }    }这是我的刀片文件：@if(count($posts) > 0)   @foreach($posts as $post)      <span> <i class="fa fa-folder-open"></i> {{ $post->category->display_name }} </span>   @endforeach@endif

Answer 1

问题是你没有在你的帖子表中保存CATEGORY_ID，保存好类别 ID 并尝试这样：

更新 POST 模型：

namespace App;

use Illuminate\Database\Eloquent\Model;

class Post extends Model

{

    //Table Name

    protected $table = 'posts';

    // Setup fields of table "posts"

    protected $fillable = ['id', 'title', 'body', 'cover_image','category_id'];

    //Model Relationships

    public function tag(){

        return $this->belongsToMany('App\Tag', 'id');

    }

    // Relation category to foreign key 

    public function category() {

        return $this->belongsTo('App\Category','category_id');

    }

}

类别模型

namespace App;

use Illuminate\Database\Eloquent\Model;

class Category extends Model

{

    protected $table = 'categories';

    // Setup fields of table "category"

    protected $fillable = ['id', 'display_name', 'name', 'key'];

    //Model Relationships

    public function posts() 

    {

        return $this->hasMany('App\Post');

    }

}

看法

@if(count($posts) > 0)

   @foreach($posts as $post)

      <span> <i class="fa fa-folder-open"></i> {{ $post->category->display_name }} </span>

   @endforeach

@endif

移民帖 - 外键 category_id

Schema::create('posts', function (Blueprint $table) {

            // Your other fields.....

            // You must setup category like this

            $table->integer('category_id')->nullable()->unsigned();

            $table->foreign('category_id')->references('id')->on('categories')->onUpdate('cascade')->onDelete('cascade');

        });

控制器

namespace App\Http\Controllers;

use Illuminate\Http\Request;

use App\Post;

class PagesController extends Controller

{

    public function index()

    {

        $posts = Post::all();

        if (!empty($posts)) {  

            // Please let me know this test

            dd($posts[0]->category->display_name);

            return view('pages.index', ['posts' => $posts]);

        }

        else {

            return view('pages.empty');

        }

    }

Answer 2

我认为您post()在Category模型中的方法有误。尝试像这样修改这个方法：

类别型号：

public function post() 
{    return $this->hasMany('App\Post', 'id');}

来源：Eloquent-relationship 一对多