Php Laravel：如何在blade视图中拆分/获取json嵌套元素？

当我通过下面的视图时，blade显示了从mongo获取的整个Json结构

{!! $fees !!}

输出：

[{"_id":"5e04a06ca445f78401a16d0a","University":[{"fees":"$200"},{"month":"June"}]}]

Array ([0] => Array ([_id] => 5e04a06ca445f78401a16d0a [University] 
=> Array ([0] => Array ( [fees] => $200 ) [1] => Array ( [month] => June ) )))

 fees $200month June

现在如何获取嵌套内容。我想在blade中显示Fees=$200和Month=June

像这样尝试，但结果在刀片上为“空白”，没有任何错误。在上面的JSOn输入中，嵌套的open/closing:[，就在'University:'之后，是否存在任何问题。请建议

@foreach ($fees as $value)   
    {{$content[] = $value->['University'] }}  
    @foreach ($content as $key)
        <h1>{{ $key['fees'] }}</h1>
        <h1>{{ $key['month'] }}</h1>
    @endforeach
@endforeach

而且

我还从控制器中尝试了以下操作：

..
public function getFees()
{

  # database fetch test (with any one of the db query)
  $fees = Finance::all();

  $fees = Finance::where(['University.month' => 'June'])->get()->toArray();

  $fees = Finance::where('University.month', 'June')->first();

  $fees = Finance::where('University.month', 'June')->get();


  # Return test (with any one of the return)
  return view('tables')->with('fees', json_encode($fees, true));

  return view('tables', compact('fees'));

  return view('tables')->with(compact('fees'));

  return view('tables')->with('fees', $fees);

 }
  ..

编辑2：

在视图中，我尝试了如下操作，但得到的例外是：尝试获取非对象的属性“费用”

<?php
$fees = json_decode($fees, true); 
#echo "</br>";
#print_r($fees)
?>

    @foreach ($fees[0] as $value)   
    @php $content = $value->University @endphp  // or without @
    @foreach ($content as $key)
        <h1>{{ $key['fees'] }}</h1>
        <h1>{{ $key['month'] }}</h1>
    @endforeach
@endforeach

唯一的输出是合并，没有逗号分隔。 我可以如下展示它们吗

费用=200美元 月份=六月

或者作为html

<td>{{$k}}</td><td>{{$v}}</td>

---

其工作原理如下：-

@foreach ($fees as $value)   
    {{$content[] = $value->['University'] }} 

    @if($content->['fees']!==null)
        <h1>{{ $content['fees'] }}</h1>
    @else
        <h1>-</h1>
    @endif

    @if($content->['fees']!==null)
        <h1>{{ $content['fees'] }}</h1>
    @else
        <h1>-</h1>
    @endif

@endforeach

您可以尝试以下方法：

$fees = json_decode($fees);
$univ = $fees[0]->University;
//print_r($univ);
foreach ($univ as $key => $value) {
  foreach($univ[$key] as $k =>$v){
     echo $k .'= '.$v.' ';
  }
}

---

那么，用$value->University而不是“$value->['University']”来做什么呢？你想做什么sense@AlbertoSinigaglia是的，这段代码是错误的。它无法找到带有$value->University的elementstry，因此这应该可以工作。$value['University']您是否尝试过json_decode$fees，对吗？如果它以输出中显示的方式在呈现页面中打印出来：那么它似乎还不是一个PHP数组，而是一个包含json的字符串。这不起作用。有关更多信息，请参阅我的编辑。使用json_decode for$fees=json_decode$fees，对；以及foreach$fees[0]作为$value尝试thisphp$content=$value->University endphp，因为$fees[0]是objectshow试图获取非对象的属性'fees'。我像这样尝试Edit2$fees=json_decode$fees；$univ=$fees[0]->University；//print_r$univ；foreach$univ as$key=>$value{foreach$univ[$key]as$k=>$v{echo$k..$v；}这将起作用。这不起作用。有关详细信息，请参阅我的编辑。

 fees $200month June

<td>{{$k}}</td><td>{{$v}}</td>

@foreach ($fees as $value)   
    {{$content[] = $value->['University'] }} 

    @if($content->['fees']!==null)
        <h1>{{ $content['fees'] }}</h1>
    @else
        <h1>-</h1>
    @endif

    @if($content->['fees']!==null)
        <h1>{{ $content['fees'] }}</h1>
    @else
        <h1>-</h1>
    @endif

@endforeach

$fees = json_decode($fees);
$univ = $fees[0]->University;
//print_r($univ);
foreach ($univ as $key => $value) {
  foreach($univ[$key] as $k =>$v){
     echo $k .'= '.$v.' ';
  }
}