【问题标题】:Laravel: JSON data in view bladeLaravel:视图刀片中的 JSON 数据
【发布时间】:2017-09-04 14:18:59
【问题描述】:

控制器:

$commission = Commission::findOrFail($id);

    $json_calculation = (array) json_decode($commission->price_calculation, true);

    return view('tradesman.commissions-list-detail')
        ->with('commission', $commission)
        ->with('calculations', $json_calculation);

JSON:

{"price_per_hour":"180","hours":"4","material_type_1":"Something","material_price_1":"1200","material_type_2":"Something","material_price_2":"800"}

查看:

@forelse($calculations as $calculation)
                <p>{!! $calculation->price_per_hour !!}</p>

错误:

ErrorException: Trying to get property of non-object

如果我只使用 {!! $calculation !!} 它有效,但我需要像 {!! $calculation->material_type1 !!}

【问题讨论】:

    标签: php json laravel view blade


    【解决方案1】:

    您正在将您的 Json 转换为数组,这就是您无法访问您的密钥之类的对象。所以改变你的这一行

    $json_calculation = (array) json_decode($commission->price_calculation, true);
    

    喜欢这样,然后尝试。它会像对象一样解码你的json

    $json_calculation = json_decode($commission->price_calculation);
    

    【讨论】:

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