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我是laravel framework.please的新手,你能帮我找到这个好的解决方案吗?可能这是一件小事情,但我并没有对它进行不幸的处理。Laravel 5.1 - @include标记不起作用
<!DOCTYPE html>
<html lang="en">
<head>
<title>Site</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"> </script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"> </script>
</head>
<body>
<div class="container">
<div class="panel panel-default">
<div class="panel-heading">
<button type="button" class="btn btn-info" id="add">New Site</button>
</div>
<div class="panel-body">
@include('newSite')
<table class="table table-hover">
<caption>Site Info</caption>
<thead>
<th>Site ID</th>
<th>Site Name</th>
<th>Date</th>
<th>Description</th>
</thead>
<tbody>
@foreach($sites as $key => $site)
<tr id="site{{$site->site_id}}">
<td>{{$site->site_id}}</td>
<td>{{$site->site_name}}</td>
<td>{{$site->site_description}}</td>
<td>
<button class="btn btn-success btn-edit">Edit</button>
<button value="btn btn-danger btn-delete">Delete</button>
</td>
</tr>
@endforeach
</tbody>
</table>
</div>
</div>
<script type="text/javascript">
$("#add").on('click',function(){
$('#site').modal('show');
})
</script>
</div>
</body>
</html>
这是我的看法site.blade.php需要file.i包括newSite.blade.php文件,这file.i使用@include( 'newSite')line.but它无法正常工作。还当我把这个命令不工作都低于codes.please帮助我尽快
当你调用@include(“newSite”)发生什么事? –
我的出门只显示“新网站”Button.table不显示在我的out.i要点击新网站按钮时,打开newSite.blade.php作为弹出 – Harshan
好吧,意味着你想显示newSite.blade.php作为一个弹出窗口? –