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Laravel无法获取试图获取非对象属性的访问器

laravel php

BARNOWL · 技术社区 · 6 年前

我正在尝试获取map对象的值,以检查用户是否进行了评级。

{{ $book->rated }} 不是假的,也不是真的。

修补匠

$book->getTypeAttribute();PHP注意:正在尝试获取在线/Applications/MAMP/htdocs/eliating/app/Book.php中的非对象四十

此方法需要获取评级的类型,默认值设置为false

public function getTypeAttribute(){        
    return Rate::where('type', $this->attributes['id'])->where('user_id',auth()->user()->id) === self::RATED; 
}

Book.php文件

use Illuminate\Database\Eloquent\Model;
use willvincent\Rateable\Rateable;
use App\User;
use App\Rate;

class Book extends Model
{

    use Rateable;

    const RATED = "true";
    const NOT_RATED = "false";

    protected $fillable = [ 'user_id', 'title', 'description'];


    public function scopeGetBook($query, $book_name )
    {
        return $query->where('slug',  $book_name );
    }

    public function setTitleAttribute($value)
    {
        $this->attributes['title'] = $value;
        $this->attributes['slug'] = str_slug($value);

    }

    public function scopeGetBookUser($query, $user_id)
    {
        return $query->where('user_id',  $user_id )->first();
    }

    public function getTypeAttribute(){        
        return Rate::where('type', $this->attributes['id'])->where('user_id',Auth()->user()->id) === self::RATED; 
    }

率.php

<?php

namespace App;


use App\User;

use Illuminate\Database\Eloquent\Model;
use willvincent\Rateable\Rateable;

class Rate extends Model
{
    protected $fillable = [
        'user_id',
        'type',
        'rating'
    ];


    public $timestamps = false;

    protected $table = 'ratings';





}

BookController.php文件 (这是设置类型的方式,也是我尝试检索值的方式)

public function rate(Request $request, $book_id)
{

    $book = Book::find($book_id);
    $rating = $book->ratings()->where('user_id', auth()->user()->id)->first();

    if(is_null($rating)){
        $ratings = new Rating();
        $ratings->rating =  $request['rating'];
        $ratings->user_id = auth()->user()->id;
        $ratings->type = Book::RATED;
        $book->ratings()->save($ratings);




        return json_encode($book);

    }
    else{
       return response()->json(['status' => 'You already left a review']);
    }
}

  public function show($book_name)
{
    $books = Book::with('ratings')->GetBook($book_name)->get();


    $data = $books->map(function(Book $book){

        $book['rated'] = $book->getTypeAttribute();

        return $book;

    });

    return view('books.show', compact('data', $data));
}

HTML格式

  <div id="rateYo" data-rateyo-rating="{{  $book->userSumRating or 0}}" data-rateyo-read-only="{{ $book->rated }}" > ></div>

图像(只读属性必须有true或false)两者都不显示。

1 回复 | 直到 6 年前

Brian Lee 6 年前

你需要打电话 first 或 get 在访问器中的查询上:

return Rate::where(['type' => $this->getKey(), 'user_id' => auth()->id()])->first()

另外,在使用访问器时,不调用完整函数,如 getTypeAttribute ,只使用属性名, type .

// wrong
$book['rated'] = $book->getTypeAttribute();

// right
$book['rated'] = $book->type;

但是,我认为如果用户留下了评分,您应该使用 exists 或 doesntExist 查询函数。例如:

// this will return true/false
public function getRatedAttribute()
{
    return Rate::where(['type' => $this->getKey(), 'user_id' => auth()->id()])->exists();
}

将属性附加到 Book 型号:

// this will add the attribute to each book in the query result.
protected $appends = ['rated']; // or type

然后您可以简单地使用:

$book->rated; // true if a rating exists, otherwise false.