Php 如何在laravel中调用引导库?

Php 如何在laravel中调用引导库?,php,twitter-bootstrap-3,laravel-5.2,Php,Twitter Bootstrap 3,Laravel 5.2,我刚刚通过composercomposer require twbs/bootstrap安装了bootstrap 3,我正在考虑如何在母版页default.blade.php中调用bootstrap库,而不是像这样通过href调用:,这样我可以在导航中应用一些样式 <!doctype html> <html> <head> <title>Home</title> </head> <body> @inclu

我刚刚通过composer
composer require twbs/bootstrap
安装了bootstrap 3,我正在考虑如何在母版页default.blade.php中调用bootstrap库,而不是像这样通过href调用:
,这样我可以在导航中应用一些样式

<!doctype html>
<html>
<head>
<title>Home</title>
</head>
<body>
    @include ('templates.partials.navigation')
<div class = "container">
    @yield('content') {{-- @yield is always used to get content from a child page into master page. So this page will be master page. --}}
</div>
</body>
</html>

家
@包括('templates.partials.navigation')
@yield('content'){{--@yield始终用于将子页面的内容获取到母版页。因此此页面将成为母版页。-}}
navigation.blade.php

<div class = "navbar navbar-inverse">
<ul>
    <li>
        <a href = "#">Login</a>
        <a href = "#">Register</a>
    </li>
</ul>
</div>


如何在导航中应用样式而不导入母版页中的样式链接库?

刚刚找到了解决问题的方法

<link rel = "stylesheet" href= "{{ URL::asset('css/bootstrap.min.css')}}">
<div class = "navbar navbar-default">
<ul>
    <li>
        <a href = "#">Login</a>
        <a href = "#">Register</a>
    </li>
</ul>
</div>

您只需使用url()



阅读此链接上的说明:
<link rel = "stylesheet" href= "{{ url('css/bootstrap.min.css')}}">
<div class = "navbar navbar-default">
<ul>
    <li>
        <a href = "#">Login</a>
        <a href = "#">Register</a>
    </li>
</ul>
</div>