laravel错误地没收了查询

Question

这是模型：

class TappaItinerario extends Model{
protected $table = "TappaItinerario";
protected $primaryKey = ['ref_idFermata', 'ref_codiceLinea', 'orarioTappa'];
public $incrementing = false;
public $timestamps = false;


public static function listStops($start, $end)
{
    $tester = TappaItinerario::select('ref_codiceLinea')
        ->distinct()
        ->whereExists(function ($query) use ($start, $end) {
            $query->select()
                ->from('' . (new static)->getTable() . ' AS T2')
                ->join('' . (new static)->getTable() . ' AS T3', 'T2.orarioTappa', '<', 'T3.orarioTappa')
                ->where([
                    ['T2.ref_codiceLineas', '=', '' . (new static)->getTable() . '.ref_codiceLinea'],
                    ['T3.ref_codiceLinea', '=', '' . (new static)->getTable() . '.ref_codiceLinea'],
                    ['T2.tipoTappa', '=', 'A'],
                    ['T3.tipoTappa', '=', 'A'],
                    ['T2.ref_idFermata', '=', $start],
                    ['T3.ref_idFermata', '=', $end]
                ]);
        })
        ->orWhereExists(function ($query) use ($start, $end) {
            $query->select()
                ->from('' . (new static)->getTable() . ' AS T2')
                ->join('' . (new static)->getTable() . ' AS T3', 'T2.orarioTappa', '<', 'T3.orarioTappa')
                ->where([
                    ['T2.ref_codiceLinea', '=', '' . (new static)->getTable() . '.ref_codiceLinea'],
                    ['T3.ref_codiceLinea', '=', '' . (new static)->getTable() . '.ref_codiceLinea'],
                    ['T2.tipoTappa', '=', 'R'],
                    ['T3.tipoTappa', '=', 'R'],
                    ['T2.ref_idFermata', '=', $start],
                    ['T3.ref_idFermata', '=', $end]
                ]);
        })
        ->get();
    return $tester;
}}

如果我尝试启动它，Laravel会给我完整的停靠点清单。

但是，如果使用相同的查询，则从sql正常启动，并使我仅显示一站式。

我指定从数据库启动的查询是从laravel生成的，然后将其复制以启动它。这样就不会有参数错误或其他任何事情。

我不知道如何解决问题。你有什么想法吗？