Question

我的页脚上的示例链接

$powered = 'Powered by <a href="htp://stackoverflow.com">Stackoverflow</a>';

主题文件

<?php echo $powered; ?>

如果从我的页脚中删除$powered并且错误/提醒通知它，如何制作。

示例die('Do not remove the powered link')

Answer 1

简短回答：不可能。

Answer 2

可能......您可以使用Zend Guard来保护您的代码不被更改。

Answer 3

答案很长：

<?php

$powered = 'Powered by <a href="htp://stackoverflow.com">Stackoverflow</a>';

ob_start();
echo $powered;
$html = ob_get_clean();

/*
 * If you remove the "echo $powered;" part, please don't remove the following lines
 * so I can detect the license violation. Thank you for your cooperation!
 */
if( strpos($html, $powered)===FALSE ){
    echo die('Do not remove the powered link');
}else{
    echo $html;
}

?>

当有人移除我的主题页脚的电源时，如何发出通知提醒

3 个答案: