ajax 获取后 Php 变量将无法工作

Question

我对 PHP 变量有疑问。我显示的用户名是使用 $_SESSION['uid']; 登录的用户 ID .

赞$uid = $_SESSION['uid'];

我必须用

获取用户名

$data = $Details->UserDetails($uid);
$data_username = $data['username'];

所以在索引页中显示我<?php echo $data_username;?>但是当我使用 $data_username在下面的代码中，它给我错误。

    <?php 

    include_once ("includes/datas.php"); 
    //session_start(); and $data_username will be come within datas.php
    if ($_GET['action'] == "get_all_posts") { get_all_posts($db,$config); }
    function get_all_posts($db,$config) {
     $sql = "select * from `posts` where ((
     to_uname = '".mysqli_real_escape_string($db, $data_username)."' AND 

     from_uname = '".mysqli_real_escape_string($db,$_GET['client'])."' ) OR (

     to_uname = '".mysqli_real_escape_string($db,$_GET['client'])."' AND 

     from_uname = '".mysqli_real_escape_string($db,$data_username)."' )) order by post_id DESC ";
    }
    ?>

我收到通知: undefined variable :data_username我在这里遗漏了什么有人可以告诉我吗？

Answer 1

将 $data_username 传递给函数即，

if ($_GET['action'] == "get_all_posts") { get_all_posts($db,$config, $data_username); }