在Laravel中，如何在name属性中显示php值？

Question

我在刀片文件中有以下html代码：@foreach ($engagements as $engagement)    {!! Form::checkbox('engagements[]', $engagement->id, in_array($engagement->id, $user->privileges->pluck(engagement_id)->toArray()) ? true : false) !!} {{ $engagement->name }}    {!! Form::select('roles[]', $userRoles, $user->privileges->where(engagement_id, $engagement->id)->first()[role]) !!}@endforeach`我想通过$engagement->id在roles[]如roles[{{@engegement_id}}]但是，{{@engegement_id}}与其显示的值，不如将其显示为[{{@engegement_id}}]html视图源中的角色。我在这里做错了什么？

Answer 1

我们不能这样，它给的错误是“未定义变量角色” @php echo $roleWithId = roles[$engagement->id] @endphp。但是，我尝试选择本身作为

{!! Form::select('roles[@php echo $engagement->id @endphp]', $userRoles, $user->privileges->where(\App\Privilege::COLUMN_ENGAGEMENT_ID, $engagement->   {\App\Engagement::COLUMN_ID})->first()[\App\Privilege::COLUMN_ROLE]) !!}

但是，它显示为

<select name="roles[<?php echo $engagement->id ?>]"> 在查看源中

Answer 2

试试这个

@foreach ($engagements as $engagement)

@php echo $roleWithId = roles[$engagement->id] @endphp

{!! Form::checkbox('engagements[]', $engagement->id, in_array($engagement->id, $user->privileges->pluck(engagement_id)->toArray()) ? true : false) !!} {{ $engagement->name }}

{!! Form::select($roleWithId , $userRoles, $user->privileges->where(engagement_id, $engagement->id)->first()[role]) !!}

@endforeach

生成要传递的字符串，并将其存储在变量中，然后在Formselect中传递该变量。