preg_replace（）：参数不匹配，pattern是一个字符串，而replacement是laravel 5.1中的一个数组

Question

我是laravel的新人。我尝试将复选框的所有选中值插入数据库表中一个colume，例如fixed =＆＃34;＆gt; Table / Chair Arrangemen，Mike / Speaker Testing，...＆＃34; 。 如果有人能解决这个问题，我感到非常感激和感激 当我尝试下面的代码时，我得到上面提到的错误

<div class="form-group">
  <label class="control-label col-sm-2" for="lead">List of Stationaries</label>
  <div class="col-sm-4">
        <input type="checkbox" name="list_of_stationary[]" value="Table/Chair Arrangement">Table/Chair Arrangement<br>
        <input type="checkbox" name="list_of_stationary[]" value="Projector Testing">Projector Testing<br>
        <input type="checkbox" name="list_of_stationary[]" value="Mike/Speaker Testing">Mike/Speaker Testing<br>
        <input type="checkbox" name="list_of_stationary[]" value="Laptop/Desktop">Laptop/Desktop<br>
        <input type="checkbox" name="list_of_stationary[]" value="Handsout Ready">Handsout Ready
  </div>
  <label class="control-label col-sm-2" for="stationaries">Stationaries</label>
  <div class="col-sm-4">
    <input type="text" class="form-control datepicker" name="stationary_done_date" value="" placeholder="Stationaries Done Date">
  </div>
  </div>

我将数据存储到数据库的控制器功能是： -

  public function store(StationaryRequest $request)
   {   

    $st = new Stationary();


     $st -> stationary_done_date = $request->stationary_done_date;

    $st->list_of_stationary=$request->list_of_stationary


      $st -> save();


  return redirect()->back()->with('message','Stationary Added');
 }