Laravel 5.1，在视图文件中使用变量名

用户名

有人可以解释为什么下一个问题在laravel 5.1中不起作用：我已经查看了文件login1.blade.php

@extends('ajaxforms.form1')
@section('form')
<input type="text" name="email" /><input type="password" name="password" />
@endsection
@section('script')
<script type="text/javascript">var modalName = '#{{$modalname}}';</script>
@endsection
@section('link')
<li ng-controller="login_controller" id="login_controller">
<a href="#" data-toggle="modal" data-target="#{{$modalname}}">Sign in</a></li>
@endsection

register1.blade.php

@extends('ajaxforms.form1')
@section('form')
<input type="text" name="name" /><input type="text" name="email" />
<input type="password" name="password" /><input type="password2" name="password2" />
@endsection
@section('script')
<script type="text/javascript">var modalName = '#{{$modalname}}';</script>
@endsection
@section('link')
<li ng-controller="register_controller" id="register_controller">
<a href="#" data-toggle="modal" data-target="#{{$modalname}}">Sign in</a></li>
@endsection

..和form1.blade.php

<div id="{{ $modalname }}">
@yield('form')
</div>
@yield('script')
@yield('link')

并且还链接了master.blade.php：

@include('ajaxforms.login1',['modalname' => 'modalLogin'])
@include('ajaxforms.register1', ['modalname' => 'modalRegister'])

这将生成一个代码：

<div id="modalLogin">
<input type="text" name="email" /><input type="password" name="password" />
</div>
<script type="text/javascript">var modalName = '#modalLogin';</script>
<li ng-controller="login_controller" id="login_controller">
<a href="#" data-toggle="modal" data-target="#modalLogin">Sign in</a></li>

<div id="modalRegister">
<input type="text" name="email" /><input type="password" name="password" />
</div>
<!--Here I want to '#modalRegister' instead of '#modalLogin'; And register_controller 
instead of login_controller. Why doesn't work?????????????????????????-->
<script type="text/javascript">var modalName = '#modalLogin';</script>
<li ng-controller="login_controller" id="login_controller">
<a href="#" data-toggle="modal" data-target="#modalLogin">Sign in</a></li>

谢谢

Anwerj

我相信问题在于节定义，您正在为相同的请求使用两个具有相同名称的不同节。如果部分形式被定义，则将其缓存并稍后与相同代码一起使用。该链接可以提供其他方式。

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