Question

我有这个问题，并且急切加载＆＃39;在laravel。

我正在处理彼此相关的4个表。

我有这些模态：

<?php
class AgendaPersonalModel extends Model
{
    protected $table = 'agenda_personal';

    public function periods() {

        return $this->hasMany(AgendaPersonalPeriodModel::class, 'agenda_id');

    }
}
?>

class AgendaPersonalPeriodModel extends Model
{
    protected $table = 'agenda_personal_period';


        public function weekdays() {

        return $this->hasMany(AgendaPersonalWeekdaysModel::class, 'period_id');

    }

<?php
class AgendaPersonalWeekdaysModel extends Model
{
    protected $table = 'agenda_personal_weekdays';


    public function breaks() {

        return $this->hasMany(AgendaPersonalBreakModel::class, 'weekday_id');

    }    

}
?>

<?php
class AgendaPersonalBreakModel extends Model
{
    protected $table = 'agenda_personal_breaks';
}

?>

现在我希望得到＆＃39;一个对象中的所有数据。

当我这样做时：

$agendaTest = AgendaPersonalModel::with(array('periods', 'periods.weekdays'))->where('id', 1)->first();

完美无缺

但是当我这样做时：

$agendaTest = AgendaPersonalModel::with(array('periods', 'periods.weekdays', 'weekdays.breaks'))->where('id', 1)->first();

我收到以下错误：

(1/1) RelationNotFoundException
Call to undefined relationship [weekdays] on model [App\Models\AgendaPersonalModel].
in RelationNotFoundException.php (line 20)

Answer 1

你可以这样做

$agendaTest = AgendaPersonalModel::with(['periods', 'periods.weekdays.breaks'])->where('id', 1)->first();

Answer 2

这样做：

AgendaPersonalModel::with(['periods', 'periods.weekdays' => function ($q) {
        $q->with('breaks');
    }])->find(1);

Laravel渴望装载4张桌子

2 个答案: