Php 使用自静态功能时出错
我有一个示例代码:Php 使用自静态功能时出错,php,Php,我有一个示例代码: class Assets { public static $my_static = ''; public static function custom_js() { return self::add_custom_js(); } public static function add_custom_js($str) { return self::$my_static = $str;
class Assets {
public static $my_static = '';
public static function custom_js()
{
return self::add_custom_js();
}
public static function add_custom_js($str)
{
return self::$my_static = $str;
}
}
和php
<?php
Assets::add_custom_js("ABC");
?>
<html>
...
<?php echo Assets::custom_js(); ?>
...
</html>
...
...
错误无法显示数据字符串为“ABC”我想您希望返回
$my_static
值以在echo Assets::custom_js()中获取ABC
代码>。考虑如下:
class Assets {
public static $my_static = '';
public static function custom_js()
{
// return self::add_custom_js();
return self::$my_static; //This is what you want i believe
}
public static function add_custom_js($str)
{
return self::$my_static = $str;
}
}
您必须从custom\u js()
返回$my\u static
,因此您的类应如下所示:
<?php
class Assets {
public static $my_static = '';
public static function custom_js()
{
return self::$my_static; //<-------------change this line
}
public static function add_custom_js($str="")
{
return self::$my_static = $str;
}
}
?>
在custom_js()中调用add_custom_js()时没有传递参数