返回后不执行PHP代码：false;

Question

下面的PHP代码（最后一页）检查用户是否按下了“回复”按钮，然后执行操作。一切OK，但如果在某个点下方延续了巨大的脚本触发我从PHP JavaScript警告，它返回假的，那绝对没有......我怎样才能使PHP记住并执行此部分

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return false;？

的代码全部片段：

/* 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; 
} 

Answer 1

有return false;作为一个杀手锏 - 它exists;脚本运行。例如;

<?php 

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

https://eval.in/201188

输出只是：Hello

通过阅读the docs

[...] 回报也结束的一个eval（）语句的执行或脚本文件。如果从全球范围内调用，那么执行当前的脚本文件的结束