php Laravel 的 Response 内容必须是字符串或对象，实现了 __toString()，给定的“对象”

Laravel The Response content must be a string or object implementing __toString(), "object" given

提问by Alameddin ?elik

I want to run Skills function but I cant it.

我想运行技能功能，但我不能。

Route.php

路由.php

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

User.php

用户名

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

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

Skills.php

技能.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;
    }
}

I want to enter /settings page I have "The Response content must be a string or object implementing __toString(), "object" given." error.

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

I tried to add dd()function's return in route, I see all JSON data but $skills->get(), $skill->set()didn't working at the time.

我尝试添加dd()函数的返回路线中，我看到所有的JSON数据，但是$skills->get()，$skill->set()在当时没有工作。

Edit:

编辑：

Skills.php

技能.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} adl? Yetenek bulunamad?");
        }
    }

采纳答案by jedrzej.kurylo

When you do

当你做

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

you are returning the Relation definition object, which doesn't implement __toString()method. What you need in order to return the related object is

您正在返回未实现__toString()方法的 Relation 定义对象。为了返回相关对象，您需要的是

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

This will return a Collectionobject cotaining related skills - this will be properly serialized.

这将返回一个包含相关技能的Collection对象 - 这将被正确序列化。

回答by itsazzad

May be you could go with get_object_vars($skills)and later loopthrough the object variables. Example:

可能您可以使用get_object_vars($skills)并稍后loop通过对象变量。例子：

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

回答by nensamuel

in the Route.php, return a collection like so

在Route.php 中，像这样返回一个集合

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

OR

或者

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

And on you User.php model, you can factor your code say so

在你的 User.php 模型上，你可以考虑你的代码这样说

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

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

I hope this helps someone.

我希望这可以帮助别人。