如果条件是真实的显示块按钮，则条件false show unblock按钮:laravel，javaScript

Question

如果 $block->user_username数据空显示 block按钮否则会在其他条件下运行另一个条件。但是在foreach循环中，其他不起作用的@else <a href="#0" class="cd-popup-trigger">Block2</a>请帮助我。任何人都知道其他任何条件可以做这件事。

@else <a href="#0" class="cd-popup-trigger">Block2</a> @endif仅当条件正在工作时，这种其他条件才会粘住。如果此@if(Auth::user()->name == $block->user_username && $user->username == $block->block_username)条件是错误的代码，则没有显示其他条件。

 @foreach($blocked as $block)    
                   @if(empty($block->user_username ))
                     <a href="#0" class="cd-popup-trigger">Block1</a>
                   @else
                        @if(Auth::user()->username == $block->user_username and $user->username == $block->block_username)
                         <a href="#0" class="cd-popup-trigger">Unblock</a>
                         @else
                            <a href="#0" class="cd-popup-trigger">Block2</a>
                        @endif
                    @endif
            @endforeach  

Answer 1

尝试此代码

@foreach($blocked as $block)    
  @if(!empty($block->user_username ))
    @if(Auth::user()->name == $block->user_username && $user->username == $block->block_username)
      <a href="#0" class="cd-popup-trigger">Unblock</a>
    @else
      <a href="#0" class="cd-popup-trigger">Block2</a>
    @endif
  @else
     <a href="#0" class="cd-popup-trigger">Block1</a> 
  @endif
@endforeach 

我希望它会有所帮助