Php 5.2的新static()或get_调用_类
我在本地主机上创建了一个数据类,它使用了PHP5.3中的一个新特性,Php 5.2的新static()或get_调用_类,php,Php,我在本地主机上创建了一个数据类,它使用了PHP5.3中的一个新特性,static()或get\u,称为\u class() 现在我的Web主机支持php 5.2,我不得不尝试其他方法,所以我尝试了以下方法: public static function get_called_class() { $objects = array(); $traces = debug_backtrace(); foreach ($traces as $t
static()
或get\u,称为\u class()
现在我的Web主机支持php 5.2,我不得不尝试其他方法,所以我尝试了以下方法:
public static function get_called_class()
{
$objects = array();
$traces = debug_backtrace();
foreach ($traces as $trace)
{
if (isset($trace['object']))
{
if (is_object($trace['object']))
{
$objects[] = $trace['object'];
}
}
}
if (count($objects))
{
return get_class($objects[0]);
}
}
结果大致相同,只是我得到了这个对象名:
[__PHP_Incomplete_Class_Name] => user
[table] => user
[data:private] =>
[className:private] => user
[arFields:private] => Array
(
[id] => id
[firstname] => firstname
[initials] => initials
[lastname] => lastname
[email] => email
[password] => password
)
[id] => 15
[firstname] => ...
[initials] => ...
[lastname] =>...
[email] => ...
[password] => ...
)
如何使名称为用户对象而不是此对象?duplicate