绑定View :: composer以使用通配符匹配所有视图吗? [英] Binding View::composer to match all views using wildcards?

问题描述

我有一个这样的导航栏.

I have a navigation bar like this.

<li>Account</li>
   <ul>
      <li>Register</li>
      <li>Login/li>
      ...

我想根据Auth::check()动态地更新它.例如，如果用户已登录，则帐户"将更改为我的个人资料页面"，子兄弟姐妹将替换为适当的数组.

I want to update this dynamically depending on Auth::check(). For example, if the user is logged in, "Account" will be changed with "My Profile Page" and child siblings will be replaced with an appropriate array.

我需要在控制器中不编辑View::make calls的情况下执行此操作.看起来很糟糕.

I need to do this without editing View::make calls in my controllers. It looks pretty bad.

我正在寻找这样的解决方案；

A solution like this is what I'm looking for;

View::composer('home.*', function($view) {
    if(Auth::check())
       return $view->nest('accountArea', 'home.navigation-loggedIn', null);
    else
       return $view->nest('accountArea', 'home.navigation-visitor', null);
});

如果有更好的选择，我也想听听他们！

If there are better alternatives, I would like to hear them too!

推荐答案

似乎类似于Laravel中的通配符.到目前为止，它们还没有记录.

Seems like the wildcards in Laravel works. They're just undocumented as of now.

View::composer('admin.layouts.*', function($view)
{
     if (Sentry::check()) $view->with('navigation', View::make('admin._partials.navigation'));
     else                 $view->with('navigation', null);
});

这就是我想要的.

更新:这是替代解决方案

您还可以将其绑定到布局，因此扩展该布局的所有子视图都将从composer中受益.

You can also bind it to the layout, so all the subviews that extend that layout will benefit from composer.

View::composer('admin.layouts.main_layout', function($view)
{
     if (Sentry::check()) $view->with('navigation', View::make('admin._partials.navigation'));
     else                 $view->with('navigation', null);
});

它将作曲家绑定到每个执行@extend('admin.layouts.main_layout')的视图.

It will bind composers to every view that does @extend('admin.layouts.main_layout').

这篇关于绑定View :: composer以使用通配符匹配所有视图吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！