返回后Php代码不执行：false；

下面的Php代码（最后一页）检查用户是否按下了“回复”按钮，然后执行一些操作。一切正常，但如果在下面继续的巨大脚本中，我在某个点触发了来自Php的javascript警报，返回false，那么它绝对不会做任何事情。。。如何让Php记住并执行这一部分

if (cmtx_setting('show_reply')) { //if reply field is enabled
        $cmtx_reply_id = trim($_POST['cmtx_reply_id']); //remove any space at beginning and end
        cmtx_is_injected($cmtx_reply_id); //check for injection attempt
        cmtx_validate_reply($cmtx_reply_id, $cmtx_page_id); //validate reply
        $cmtx_reply_to = cmtx_sanitize($cmtx_reply_id, true, true); //sanitize reply
    }

即使在脚本中的某个点，我有

返回false

完整的代码片段：

/* Reply To */
    if (!isset($_POST['cmtx_reply_id'])) { //if reply ID not submitted
        $_POST['cmtx_reply_id'] = 0; //set it with a zero value
    }
    if (cmtx_setting('show_reply')) { //if reply field is enabled
        $cmtx_reply_id = trim($_POST['cmtx_reply_id']); //remove any space at beginning and end
        cmtx_is_injected($cmtx_reply_id); //check for injection attempt
        cmtx_validate_reply($cmtx_reply_id, $cmtx_page_id); //validate reply
        $cmtx_reply_to = cmtx_sanitize($cmtx_reply_id, true, true); //sanitize reply
    } else {
        $cmtx_reply_to = 0;
}

有
返回false充当杀手-它
存在脚本运行。比如,

<?php

echo "Hello";
return false;
echo "World";

很抱歉，打字错误（我脑子里有太多CSS）编辑了这篇文章。但是我输入的代码放在第一次返回false之前；在我的档案里。这是我不理解的…确实如此，但表单验证也会停止（表单已提交，但即使是空的）。