试图在 Laravel 中获取非对象的属性，即使它存在答案

【问题标题】：Trying to get property of non-object in Laravel even when it is present试图在 Laravel 中获取非对象的属性，即使它存在
【发布时间】：2016-05-25 16:48:58
【问题描述】：

我正在测试的数据在feed变量中是这样的

{
    "id": 32,
    "user_id": 1,
    "target_id": 3,
    "feedable_id": 7,
    "feedable_type": "review",
    "created_at": "2016-05-23 14:18:22",
    "updated_at": "2016-05-23 14:18:22",
    "feedable": {
        "id": 7,
        "user_id": 1,
        "slug_id": 3,
        "review": "y this product is so quite?",
        "stars": 9,
        "created_at": "2016-05-23 14:18:22",
        "updated_at": "2016-05-23 14:18:22",
        "slug": {
            "id": 3,
            "value": "Quite",
            "views": 5,
            "user_id": 1,
            "category_id": 1,
            "created_at": "2016-05-23 14:18:03",
            "updated_at": "2016-05-24 12:47:29"
        }
    }
}

但我无法访问该值：在我看来相当

{{ $feed->feedable->slug->value }}

Laravel 抛出

Trying to get property of non-object (View: ...

【问题讨论】：

模型上的关系是否正确设置？
是的，关系很好，而且回报很好

标签： php laravel view eloquent

【解决方案1】：

这是一个 json 变量，因此您需要先对其进行解码，然后再访问该值。使用json_decode可以解码变量并得到期望值。

Check online.

$json = '{
    "id": 32,
    "user_id": 1,
    "target_id": 3,
    "feedable_id": 7,
    "feedable_type": "review",
    "created_at": "2016-05-23 14:18:22",
    "updated_at": "2016-05-23 14:18:22",
    "feedable": {
        "id": 7,
        "user_id": 1,
        "slug_id": 3,
        "review": "y this product is so quite?",
        "stars": 9,
        "created_at": "2016-05-23 14:18:22",
        "updated_at": "2016-05-23 14:18:22",
        "slug": {
            "id": 3,
            "value": "Quite",
            "views": 5,
            "user_id": 1,
            "category_id": 1,
            "created_at": "2016-05-23 14:18:03",
            "updated_at": "2016-05-24 12:47:29"
        }
    }
}';
$feed = json_decode ($json);

echo $feed->feedable->slug->value; //Quite

【讨论】：