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codeigniter /杂货店CRUD发送数据以查看不起作用

[英]codeigniter / groceryCRUD send data to view not working

I have a script that works, this one: 我有一个有效的脚本,这个:

    $output = $crud->render();

    $data = array();
    $data['title'] = 'Invoice details';
    $output->data = $data;

    $this->output($output); /* crud render and extra fields */

the thing is that this script is opening a function in the controler: 问题是该脚本正在控制器中打开一个函数:

    public function output($output = null) {
    $this->load->view('template.php', $output);
}

and the template file as this: (and more) 和模板文件如下:(还有更多)

<title><?php echo ucfirst($data['title']); ?></title>

works like a charm. 奇迹般有效。

the template file is loading the header and footer and a bunch of css 模板文件正在加载页眉和页脚以及一堆CSS

when I want to open the invoice, I don't want to open the template, just invoice.php 当我想打开发票时,我不想打开模板,只需打开invoice.php

so I created this: 所以我创建了这个:

    public function invoice_print() {
    $data = array();
    $data['company'] = 'test';

    $this->load->view('invoice.php', $data);
}

when I open the page I get a error: 当我打开页面时出现错误:

A PHP Error was encountered Severity: Notice Message: Undefined variable: data Filename: views/invoice.php Line Number: 16 遇到PHP错误严重性:注意消息:未定义的变量:数据文件名:views / invoice.php行号:16

invoice.php line 16 = this: invoice.php第16行=此:

<p class="bold"> <?php echo $data['company']; ?> </p>

how do I fix this? 我该如何解决?

regards, 问候,

Ralph 拉尔夫

When you pass the variable to the view, you call the variable by the index: 将变量传递给视图时,可以通过索引调用变量:

$data['company'] = 'test';
$this->load->view('invoice.php', $data);

view: 视图:

<p class="bold"> <?php echo $company; ?> </p>

http://ellislab.com/codeigniter%20/user-guide/general/views.html http://ellislab.com/codeigniter%20/user-guide/general/views.html

$output = $crud->render();
$misDatos =  'Invoice details';
$output->misDatos = $misDatos;
$this->_example_output($output);

And you print $misDatos in your VIEW. 然后在视图中打印$misDatos

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