CodeIgniter; $ this在非对象上下文中的用法 [英] CodeIgniter; $this usage in non object context

问题描述

我想在我的CodeIgniter项目中编辑我的 config / database.php ，所以我不必每次推新的时候不断更改数据库信息版本到服务器。为了做后者，我创建了一个 $ debug 变量，如下所示，它检查uri以查看该网站是否在我的本地主机上运行项目上传之前）或在实际服务器上。

I am trying to edit my config/database.php within my CodeIgniter project so that I don't have to keep changing the database information every time I push a new version to the server. To do the latter, I have created a $debug variable, shown below, which checks against the uri to see if the site is being run on my localhost machine (where I build the project before upload) or on the actual server.

不幸的是，在database.php文件中，我没有访问 $ this 变量，我个人是PHP的新OOP，我不完全确定一个方法。

Unfortunately in the database.php file, I don't have access to the $this variable, and as I am personally new to OOP in PHP, I am not entirely sure of a way around this. Please can you tell me how I can do so?

感谢，

最多。

致命错误：在第51行应用程序/ config / database.php中不在对象上下文中时使用$ this

Fatal error: Using $this when not in object context in application/config/database.php on line 51

$debug = strpos($this->uri->config->item('base_url'), 'localhost'); //line 51

$db['default']['hostname'] = $debug == TRUE ? 'x' : 'y';
$db['default']['username'] = $debug == TRUE ? 'x' : 'y';
$db['default']['password'] = $debug == TRUE ? 'x' : 'y';
$db['default']['database'] = $debug == TRUE ? 'x' : 'y';

推荐答案

您需要获取codeigniter对象的实例 -

You need to a get an instance of the codeigniter object -

$CI = get_instance();

对于旧版本的php，请使用=&

For older versions of php, use =&

然后，使用 $ CI ，无论您通常使用 $ this

Then, use $CI wherever you would normally use $this

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