Php 已传递变量，但仍引发异常？

我创建并传递一个数据数组：

public function home()
{


    $data['page'] = 'home';
    $data['table'] = 'pageData';
    $data['temp'] = 'temp_1';

    $this->template($data);


}


    public function template($data)
{

    $this->load->model("model_get");


    $data['results'] = $this->model_get->getData($data);

    $this->load->view('template', $data);


}

这是模板视图：

<?php

$this->load->view('header');

$this->load->view('nav', $data);

$data['results'] = $results;

$this->load->view($temp, $data);

$this->load->view('footer');

?>

但是仍然加载视图并完成其中的所有if语句，并从存储在

$temp

中的名称加载视图

---

为什么会抛出异常？

您没有在该视图中填充$data以加载第二个视图。您的代码名称太模糊，无法真正理解它应该做什么，但让我们将其分解

public function home()
{
$data['page'] = 'home';
$data['table'] = 'pageData';
$data['temp'] = 'temp_1';
$this->template($data);
}


public function template($data)
{
$this->load->model("model_get");
$data['results'] = $this->model_get->getData($data);
//I assume getData takes a table name and returns all the data
//This only works because you're taking all the data passed from the first function
//in the function above.
$this->load->view('template', $data);
}


<?php
$this->load->view('header');
$this->load->view('nav', $data);
//At this point $data is empty. You have available $results, $page, $table and $temp
//because they were passed from the template function above.
$this->load->view('nav', $results);
//The line above is what it appears you need, rather than $data.
$data['results'] = $results;
//You've repopulated $data here but now all it contains is $results.
$this->load->view($temp, $data);
$this->load->view('footer');
?>

公共功能主页（）
{
$data['page']='home'；
$data['table']='pageData'；
$data['temp']='temp_1'；
$this->template（$data）；
}
公共功能模板（$data）
{
$this->load->model（“model_get”）；
$data['results']=$this->model\u get->getData（$data）；
//我假设getData使用一个表名并返回所有数据
//这只适用于获取从第一个函数传递的所有数据
//在上面的函数中。
$this->load->view（'template'，$data）；
}
显然，您需要检查数据变量并查看它返回什么。使用print\r该豁免声明其未定义。它所做的只是回显它未定义，但它使用了变量，在xamp的一侧，它工作正常，我单独添加了导航菜单，它在错误后不精确，但直到错误发生，这已关闭服务器，就像我说的，它仍然加载内容，我认为这是变量的作用域问题。我如何纠正作用域问题，重新定义它？
public function home()
{
$data['page'] = 'home';
$data['table'] = 'pageData';
$data['temp'] = 'temp_1';
$this->template($data);
}


public function template($data)
{
$this->load->model("model_get");
$data['results'] = $this->model_get->getData($data);
//I assume getData takes a table name and returns all the data
//This only works because you're taking all the data passed from the first function
//in the function above.
$this->load->view('template', $data);
}


<?php
$this->load->view('header');
$this->load->view('nav', $data);
//At this point $data is empty. You have available $results, $page, $table and $temp
//because they were passed from the template function above.
$this->load->view('nav', $results);
//The line above is what it appears you need, rather than $data.
$data['results'] = $results;
//You've repopulated $data here but now all it contains is $results.
$this->load->view($temp, $data);
$this->load->view('footer');
?>