试图获得Laravel中非对象的属性,即使它存在

时间:2016-05-25 16:48:59

标签: php laravel view eloquent

我正在测试的数据在{{ $feed->feedable->slug->value }} 变量

中是这样的
Trying to get property of non-object (View: ...

但我无法访问该值:完全在我看来

{{1}}

Laravel抛出

{{1}}

1 个答案:

答案 0 :(得分:1)

这是一个json变量,因此您需要先解码它然后访问该值。使用json_decode,您可以解码变量并获得所需的值。

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