Question

我有两个桌子。 RTAList和RTARegCompany。 RTARegCompany包含引用RTAList的外键rta_id。

    Schema::create('rta_list', function (Blueprint $table) {
        $table->increments('rta_id');
        $table->string('rta_name');
        $table->string('rta_address');
        $table->string('rta_phone');
        $table->string('rta_email')->unique();
        $table->integer('count'); 
        $table->timestamps();
    });

 Schema::create('rta_reg_company', function (Blueprint $table) {
        $table->increments('company_id');
        $table->integer('rta_id')->unsigned();
        $table->foreign('rta_id')
            ->references('id')
            ->on('rta_lists')
            ->onDelete('cascade');
        $table->string('company_name');
        $table->string('company_isin');
        $table->string('company_script');
        $table->string('company_address');
        $table->string('company_phone');
        $table->string('company_email');
        $table->timestamps();
    });

我必须为RTAList和RTARegCompany建模。

protected $table = 'rta_list';
protected $fillable = ['rta_name', 'rta_address','rta_phone','rta_email'];



 protected $table = 'rta_reg_company';
protected $fillable = ['rta_id', 'company_name','company_isin','company_script','company_address','company_phone','company_email'];


public function rtaname(){
    return $this->belongsTo(RTAList::class);
 }

现在，我想在添加RTA注册公司时显示rta_name。但是我收到错误消息“试图获取标称属性；

在RTARegController中，我已经编写了这段代码。

 $rta = RTAList::where('rtaname')->get();

在add_company.blade.php中，我想显示rta_name

  <input type="text" name="rta_id" id="rta_id" value="{{ $rta->rtaname->rta_name }}" disabled>

帮我解决这个问题...

显示名称代替id Laravel 5.6

0 个答案: