我的一對多關系不起作用，當我嘗試在視圖中呼叫它時會出現錯誤“嘗試獲取屬性'senders_name'''

我希望能夠將相關表的值傳遞給我的視圖，我正在使用一對多關系。我也試圖在不使用外鍵的情況下進行連接，我被告知將它用于 mysql 資料庫效率不高，它可能會引發錯誤或不做任何事情。我實際上測驗了它，我的意思是外鍵，它不起作用。

崗位型號：

namespace App\Models;

use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;

class Post extends Model
{
   //Post model
    protected $table = 'posts';

    public function staff()
    {
        return $this->belongsTo('App\Models\Staff');
    }
}

員工模型：

namespace App\Models;

use Illuminate\Database\Eloquent\Factories\HasFactory;
use Illuminate\Database\Eloquent\Model;

class Staff extends Model
{
    protected $table = 'staff';

    public function posts()
    {
        return $this->hasMany('App\Models\Post');
    }
}

創建posts表遷移：

<?php

use Illuminate\Database\Migrations\Migration;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Support\Facades\Schema;

class CreatePostsTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {   //my post table
        Schema::create('posts', function (Blueprint $table) {
            $table->increments('id');
            $table->string('title');
            $table->string('name');
            $table->string('phone_number');
            $table->mediumText('fault_explained');
            $table->string('repair_type');
            $table->string('repair_part');
            $table->integer('repair_total');
            $table->mediumtext('justify');
            $table->string('vendor_name');
            $table->string('vendor_number');
            $table->string('status_option');
            $table->integer('status_value');
            $table->string('vehicle_name');
            $table->string('vehicle_number');
            $table->integer('staff_id')->nullable()->unsigned();
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::dropIfExists('posts');
    }
}

創建staff表遷移：

 <?php

use Illuminate\Database\Migrations\Migration;
use Illuminate\Database\Schema\Blueprint;
use Illuminate\Support\Facades\Schema;

class CreateStaffTable extends Migration
{
    /**
     * Run the migrations.
     *
     * @return void
     */
    public function up()
    {   //Staff table
        Schema::create('staff', function (Blueprint $table) {
            $table->increments('id');  
            $table->string('senders_name');
            $table->string('senders_number');
            $table->timestamps();
        });
    }

    /**
     * Reverse the migrations.
     *
     * @return void
     */
    public function down()
    {
        Schema::dropIfExists('staff');
    }
}

看法：

<div class="content-wrapper">
    <main class="st_viewport">
        <div class="border bg-warning">
            <div style="margin-left:14px; margin-top:5px;">
                <span class="h2 wrapper " > <em><b>SUBJECT:</b> {{ $posts->title }}</span></em>
            </div>
            <div class="" style="margin-left:14px; margin-bottom:5px;">
                <em><b>Sent on</b> {{ $posts->created_at }}</em> {{ $posts->staff->senders_name}}
            </div>

uj5u.com熱心網友回復：

在Post您應該使用的模型中 -

public function staff()
    {
        return $this->belongsTo(Staff::class);
    }

在Staff您應該使用的模型中 -

public function posts()
    {
        return $this->hasMany(Post::class);
    }

在 CreatePostsTable 遷移中，您需要再添加一行 -

$table->foreign('staff_id')->references('id')->on('staff');

uj5u.com熱心網友回復：

您必須更換App\Models\Staff與Staff::class在你的代碼。

...
return $this->belongsTo(Staff::class);
...

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