variables - 代码点火器 View 记住以前的变量!

Question

我在 Controller 中有以下代码:

$data['what'] = 'test';
$this->load->view('test_view', $data);
$this->load->view('test_view');

看法:

<?php
    echo $what;
?>

运行此代码时的结果是:

testtest

不应该只是“测试”，因为我第二次没有传递变量 $data 吗？
如何让 CodeIgniter 以这种方式运行？

编辑1:

我想出了一个临时解决这个问题的方法:

在 Loader.php 中替换:

/*
* Flush the buffer... or buff the flusher?
*
* In order to permit views to be nested within
* other views, we need to flush the content back out whenever
* we are beyond the first level of output buffering so that
* it can be seen and included properly by the first included
* template and any subsequent ones. Oy!
*
*/ 

和:

 /*
 * Flush the buffer... or buff the flusher?
 *
 * In order to permit views to be nested within
 * other views, we need to flush the content back out whenever
 * we are beyond the first level of output buffering so that
 * it can be seen and included properly by the first included
 * template and any subsequent ones. Oy!
 *
 */ 

 if (is_array($_ci_vars)){
   foreach ($_ci_vars as $key12 => $value12) {
      unset($this->_ci_cached_vars[$key12]);
   }
 }

这应该在使用完毕后从缓存中删除变量。

错误报告: http://bitbucket.org/ellislab/codeigniter/issue/189/code-igniter-views-remember-previous

Answer 1

这很有趣，我从来没有像这样使用它，但你是对的，它不应该这样做，也许这是一些缓存选项。在最坏的情况下，您必须这样称呼它:

$this->load->view('test_view', '');

编辑:

我刚刚检查了他们存储库中的 Code Igniter 代码。这样做的原因是他们确实在缓存变量:

    /*
     * Extract and cache variables
     *
     * You can either set variables using the dedicated $this->load_vars()
     * function or via the second parameter of this function. We'll merge
     * the two types and cache them so that views that are embedded within
     * other views can have access to these variables.
     */ 
    if (is_array($_ci_vars))
    {
        $this->_ci_cached_vars = array_merge($this->_ci_cached_vars, $_ci_vars);
    }
    extract($this->_ci_cached_vars)

如果我理解正确，您必须不幸地这样做:

$this->load->view('test_view', array('what' => ''));