Laravel 4 - 包括一个“部分"视图中的视图(不使用 Blade 模板) [英] Laravel 4 - including a "partial" view within a view (without using Blade template)
问题描述
在 Laravel 3 中，我曾经这样做过.
In Laravel 3, I used to do this.
<?php render('partials.header'); ?>
这是在PHP"视图中完成的，没有使用 Laravel 的 Blade 模板.
This was done in "PHP" views, without using Laravel's Blade templates.
在版本 4 中相当于什么?
What's the equivalent of this in version 4?
我试过了
<?php @include('partials.header'); ?>
这不起作用.
如果我这样做
@include('partials.header')
我必须将我的文件保存为.blade.php"
I have to save my file as ".blade.php"
如何在不使用刀片模板的情况下包含子视图"?
How do I include a "subview" without using the blade template?
推荐答案
在 Laravel 4 中，有多种方法可以在视图中包含视图.您的选择将取决于下面列出的任何一种结果...
There are different ways to include a view within a view in Laravel 4. Your choice will depend on any one of the outcomes outlined below...
您可以在适当的控制器中编译(渲染)部分视图，并使用 $data[''] 数组将这些视图传递给主视图.
You can compile (render) the partial views in the appropriate Controller, and pass these views to the Main View using the $data[''] array.
随着观看次数的增加，这可能会变得乏味，但是嘿，至少有很大的灵活性:)
This may become tedious as the number of views increase, but hey, at least there's a lot of flexibility :)
示例见下面的代码:
...
public function showMyView()
{
/* Header partial view */
$data['header'] = View::make('partials.header');
/* Flexible enough for any kind of partial views you require (e.g. a Header Menu partial view) */
$data['header_menu'] = View::make('partials.header_menu');
/* Footer partial view */
$data['footer'] = View::make('partials.footer');
return View::make('myView', $data);
}
...
查看
您可以按如下方式包含上述部分内容(在您的视图代码中的任何位置):
View
You can include the partials above as follows (at any position in your View code):
<html>
<head></head>
<body>
<!-- include partial views -->
<?php echo ($header) ?>
<?php echo ($header_menu) ?>
<div id="main-content-area"></div>
<?php echo ($footer) ?>
</body>
</html>
您的部分视图现在将添加到您的主视图中.
Your partial views will now be added to your main View.
其实有一个比使用上面的方法更简单的方法:只需在视图的 html 中包含这个...
There's actually a much easier way than using the method above: Simply include this in the html of the view...
<html>
<head></head>
<body>
<!-- include partial view: header -->
<?php echo View::make('partials.header') ?>
<div id="main-content-area">
</div>
<!-- include partial view: footer -->
<?php echo View::make('partials.footer') ?>
</body>
</html>
确保部分的文件夹结构是 [views/partials/header.php] 以便为 Laravel 的 View::make() 函数提供正确的文件路径.
Make sure that the folder structure for the partials is [views/partials/header.php] in order to provide the correct file-path to the View::make() function of Laravel.
如果您尝试在控制器中传递 $data['page_title']，嵌套视图将不会接收数据.
要将数据传递给这些嵌套视图，您需要这样做:
If you try to pass the $data['page_title'] in a controller, the nested views wont receive the data.
To pass data to these nested views you need to do it like this:
<html>
<head></head>
<body>
<?php
/* Pass page title to header partial view */
$data ['page_title'] = "My awesome website";
echo View::make('partials.header', $data);
?>
<div id="main-content-area"></div>
<?php echo View::make('partials.footer') ?>
</body>
</html>
注意
问题明确指出:不使用 Blade 模板"，所以我确保给出一个不包含任何 Blade 模板代码的解决方案.
NOTE
The question clearly stated: "Without using Blade template", so I have made sure to give a solution that does not include any Blade templating code.
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