所以我使用dropzone.js,我想在dropzone成功事件后重新加载特定图像。dropzone成功事件后的Ajax图像加载
我控制器
public function edit($id)
{
$offer = Offer::find($id);
if(!is_object($offer->getMedia('featimgs')->first())){
$offerfeatimg = '/assets/images/offerfeatimg.jpg';
} else {
$offerfeatimg = $offer->getMedia('featimgs')->first()->getUrl('medium');
}
return view('admin.offers.edit')->with(compact('offer', 'offerfeatimg'));
}
所以这是我的形象传递给视图:
<div class="panel-body">
<img src="{{ $offerfeatimg }}" class="img-responsive">
@if($offerfeatimg != '/assets/images/offerfeatimg.jpg')
<div class="removebutton">
<a href="/admin/offer/featimg/delete/{{ $offer->id }}" class="btn btn-danger" role="button">Izbrisi sliku</a>
</div>
@endif
<form action="/admin/offer/featimg/{{ $offer->id }}" class="dropzone" id="my-awesome-dropzone">
{!! csrf_field() !!}
<div class="dz-message">Prebacite glavnu sliku za ovu ponudu</div>
</form>
</div>
的观点:
所以我希望以后重新加载通过AJAX这部分成功dropzone事件:
<img src="{{ $offerfeatimg }}" class="img-responsive">
@if($offerfeatimg != '/assets/images/offerfeatimg.jpg')
<div class="removebutton">
<a href="/admin/offer/featimg/delete/{{ $offer->id }}" class="btn btn-danger" role="button">Izbrisi sliku</a>
</div>
@endif
有何想法?
当我通过dropzone发送图像时,我的回应是实际的图像路径。 但我如何使用该响应在var imageSrc = response? –
好吧,我得到它的工作。如果(imageSrc!='/assets/images/offerfeatimg.jpg') –
也许你想使用'$(“。removebutton”)。hide()'和'$( “.removebutton”)。show()'基于'imageSrc'的值?如果您需要,我会编辑我的答案。 – csum