Question

我有一张人员表

CREATE TABLE `people` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `name` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL,
  `job_title` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL,
  `profile_pic` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL,
  `departments` varchar(500) COLLATE utf8mb4_unicode_ci DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=95 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

部门表

CREATE TABLE `departments` (
  `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `people_id` int(11) NOT NULL,
  `department_id` varchar(255) COLLATE utf8mb4_unicode_ci NOT NULL,
  `created_at` timestamp NULL DEFAULT NULL,
  `updated_at` timestamp NULL DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=50 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_unicode_ci;

一个人可以在多个部门
在departments表中，foreign_key是people_id

在我的人物模型中

public function departments()
    {
 return $this->hasMany('App\Models\Departments', 'id', 'people_id');

    }

最后我尝试返回一些结果

 return $people::find(1)->departments;

我收到了错误，

**ErrorException in web.php line 129:
Trying to get property of non-object**

请原谅我，如果这是一个愚蠢的错误，我已经使用了laravel一段时间，但从未在之前设置过具有关系的模型。

Answer 1

试试这段代码： -

Get All records
$people = People::with('departments')->get();
$people = json_decode(json_encode($people),true) //Convert in to array
echo "<pre>"; print_r($people); die; //print it

Get 1 record
$people = People::with('departments')->first();
$people = json_decode(json_encode($people),true) //Convert in to array
echo "<pre>"; print_r($people); die; //print it

希望它有所帮助！

Answer 2

所以最后这是一个愚蠢的问题。

$people::find(1)->departments;

Find需要一个id，但id不存在。

Laravel模型非对象的Poperty

2 个答案: