Laravel响应内容必须是实现__toString()的字符串或对象，给定的" object“

Question

我想运行技能函数，但我做不到。

Route.php

Route::get('setting',function(){
    return \App\User::first()->skills();
});

User.php

protected $casts = [
    'skills' => 'json'
];

public function skills(){
    return new Skills($this , $this->skills);
}

Skills.php

namespace App;
use App\User;
use Mockery\Exception;

class Skills
{
    protected $user;
    protected $skills = [];

    public function __construct(User $user,array $skills){

        $this->user=$user;
        $this->skills=$skills;
    }
}

我想进入/settings页面，我有"The Response content must be a string or object implementing __toString(), "object" given.“错误。

我尝试在路由中添加dd()函数的返回值，我看到了所有的JSON数据，但$skills->get()，$skill->set()当时不起作用。

编辑：

Skills.php

<?php
    /**
     * Created by PhpStorm.
     * User: root
     * Date: 01.08.2015
     * Time: 11:45
     */

    namespace App;
    use App\User;
    use Mockery\Exception;

    class Skills
    {
        protected $user;
        protected $skills = [];

        public function __construct(User $user,array $skills){
            $this->user=$user;
            $this->skills=$skills;
        }

        public function get($key){
            return array_get($this->skills,$key);
        }

        public function set($key,$value){
            $this->skills[$key]=$value;
            return $this->duration();
        }

        public function has($key){
            return array_key_exists($key,$this->skills);
        }

        public function all(){
           return $this->skills;
        }

        public function merge(array $attributes){
            $this->skills = array_merge(
                $this->skills,
                array_only(
                    $attributes,
                    array_keys($this->skills)
                )
            );
            return $this->duration();
        }

        public function duration(){
            return $this->user->update(['skills' => $this->skills]);
        }

        public function __get($key){
            if ($this->has($key)) {
                return $this->get($key);
            }
            throw new Exception("{$key} adlı Yetenek bulunamadı");
        }
    }

Answer 1

当你这样做的时候

return \App\User::first()->skills();

您返回的是关系定义对象，该对象不实现__toString()方法。返回相关对象所需的内容是

return \App\User::first()->skills;

这将返回一个包含相关技能的集合对象-它将被正确序列化。

Answer 2

也许你可以通过对象变量使用get_object_vars($skills)和后来的loop。示例：

foreach(get_object_vars($skills) as $prop => $val){
}

Answer 3

在Route.php中，返回如下所示的集合

Route::get('setting',function(){
     return \App\User::first()->skills;
});

或

Route::get('setting',function(){
    $skills = \App\User::first(['id','skills']);
    return $skills->skills;
});

在你的User.php模型上，你可以这样分解你的代码

protected $casts = [
    'skills' => 'json'
];

public function skills()
{
    return $this->skills;
}

我希望这对某些人有帮助。