CodeIgniter / PHP - 从视图中调用视图 [英] CodeIgniter/PHP - Calling a view from within a view

问题描述

基本上对于我的webapp我试图组织它好一点。因为它目前，每次我想加载一个页面，我必须这样从我的控制器，如：

Basically for my webapp I'm trying to organise it a bit better. As it at the moment, every time I want to load a page, I have to do it from my controller like so:

        $this->load->view('subviews/template/headerview');
    $this->load->view('subviews/template/menuview');
    $this->load->view('The-View-I-Want-To-Load');
    $this->load->view('subviews/template/sidebar');
    $this->load->view('subviews/template/footerview'); 

因为你可以说它不是非常高效。

As you can tell it's not really very efficient.

所以我想我会创建一个主视图 - 这叫做template.php。这是模板视图的内容：

So I thought I'd create one 'master' view - It's called template.php. This is the contents of the template view:

<?php
    $view = $data['view'];

        $this->load->view('subviews/template/headerview');
        $this->load->view('subviews/template/menuview');
        $this->load->view($view);
        $this->load->view('subviews/template/sidebar');
        $this->load->view('subviews/template/footerview');
?>

然后我想我可以从这样的控制器调用它：

And then I thought I'd be able to call it from a controller like this:

    $data['view'] = 'homecontent';
    $this->load->view('template',$data);

不幸的是，我根本无法使这项工作。有谁有任何方式围绕这个或修复我可以放到位？我试着在template.php的$视图中放置和，但是没有什么区别。通常的错误是未定义的变量：数据或无法加载视图：$ view.php等。

Unfortunately I simply cannot make this work. Does anyone have any ways around this or fixes I can put into place? I've tried putting ""s and ''s around $view in template.php but that makes no difference. The usual error is "Undefined variable: data" or "Cannot load view: $view.php" etc.

谢谢大家！

杰克

推荐答案

我相信你有：

$view = $data['view'];

$this->load->view('subviews/template/headerview');
$this->load->view('subviews/template/menuview');
$this->load->view($view);
$this->load->view('subviews/template/sidebar');
$this->load->view('subviews/template/footerview');

你需要摆脱这一行：

$view = $data['view'];

这是因为当数组从控制器传递时，视图上的变量可以被简单地访问由$ view而不是$ data ['view']。

This is because when the array is passed from the controller, the variable on the view can be accessed simply by $view rather than $data['view'].

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