Question

我在Laravel项目中使用存储库模式。问题是我看到它正在第二个调用查询中复制绑定！

这是我的代码：

class AirController extends Controller
{
    private $airportService;

    public function __construct(AirportService $airportService) {
        $this->airportService = $airportService;
    }

    public function getAirports(){
        $departure_airport = $this->airportService->getCityFromAirport("6");
        $destiny_airport = $this->airportService->getCityFromAirport("10");
    }
}

调试时，$departure_airport获得记录，但是$destiny_airport失败。您会认为id: 10有问题。不。如果我交换并先放置$destiny_airport，它会得到一条记录，但随后$departure_airport会失败。然后，我考虑按照建议的here打印原始SQL查询。

这是结果：

INFO: "select * from `airports` where `airports`.`id` = ? limit 1"  
INFO: ["6"]  
INFO: "select * from `cities` where `cities`.`id` = ? limit 1"  
INFO: [441]  
*****************************************************************************************
INFO: "select * from `airports` where `airports`.`id` = ? and `airports`.`id` = ? limit 1"  
INFO: ["6","10"]

当我在第二个查询中仅将参数10传递为参数时，为什么在第三个查询中（星号之后）使用参数6和10复制列“ id”？！我想要这样：

INFO: "select * from `airports` where `airports`.`id` = ? limit 1"  
INFO: ["10"]

这是实现：

AirportService.php ：

use App\Repositories\AirportRepository as Airport;

class AirportService {

    private $airport;

    public function __construct(Airport $airport){
        $this->airport = $airport;
    }

    public function getCityFromAirport($airportId){
        $airport = $this->airport->find($airportId);
        return $airport->City;
    }
}

Repository.php

...

public function find($id, $columns = array('*')) {
    return $this->model->find($id, $columns);
}

...

IRepository.php

...

public function find($id, $columns = array('*'));

...

给出的错误是：

local.ERROR: Trying to get property of non-object {"userId":41,"email":"...@...","exception":"[object] (ErrorException(code: 0): Trying to get property of non-object at C:\\wamp\\www\\project\\API\\app\\Services\\AirportService.php:21)

Answer 1

尝试在存储库的find方法中调用newModelInstance。

return $this->model->newModelInstance()->find($id, $columns);

在第二个调用中雄辩地重复绑定

1 个答案: