Php 传递变量值

如果$query中有值，我可以知道如何传递它的值吗。如果它为空，我仍然需要传递变量。虽然sql数据库中确实存在未定义的变量，但我不断收到未定义变量的错误

<?php
include("dbconnect.php");
include("header.php");

if (isset($_POST['btn'])) {
    $uname        = $MySQLi_CON->real_escape_string(trim($_POST['user_name']));
    $email        = $MySQLi_CON->real_escape_string(trim($_POST['user_email']));
    $upass        = $MySQLi_CON->real_escape_string(trim($_POST['password']));
    $enroller_id_n = $MySQLi_CON->real_escape_string(trim($_POST['enroller_id_n']));        
    $enrolled_id_n=  $MySQLi_CON->real_escape_string(trim($_POST['enrolled_id_n'])); 
    $direction = $MySQLi_CON->real_escape_string(trim($_POST['direction'])) ;
    $new_password = password_hash($upass, PASSWORD_DEFAULT);
    $query = $MySQLi_CON->query("select * from personal where enroller_id='".$enroller_id_n."'");
    if($query){
        while ($row = $query->fetch_array()) {
            $enroller_id3 = $row['enroller_id'];
            $left_mem     = $row['left_mem'];
            $right_mem    = $row['right_mem'];
            $test         = "left_mem";
            $test2        = "right_mem";
            $direc        = $direction;
        }
    }
}
?>

---

您需要在此处尝试一些调试。比如说,

var_dump($enroller_id_n); // Check if the variable is not empty

$query = $MySQLi_CON->query("select * from personal where enroller_id='".$enroller_id_n."'") or die($MySQLi_CON->error);

$row = $query->fetch_array(MYSQLI_ASSOC);

echo "<pre>";
print_r($row);  // Check your result array

将值从
$query
传递到何处？是否尝试在$query上打印？结果是什么？传递$query的变量，例如$left mem和$right mem。这里的问题是我得到了一个未定义的变量，因为我的注册者id在personal中为emptylearn about prepares语句。每次在meow代码中使用数据库扩展时，它都会被弃用，多年来一直在PHP7中消失。如果你只是在学习PHP，那么就把精力花在学习
PDO
或
mysqli
数据库扩展上。但是里面有空数据。该页面将不会显示：（您的意思是$enroller\u id\n为空？我的enroller\u id数据在个人表中为空。我只想允许以下查询在以下情况下运行：$left\u mem！=$test&&$right\u mem！=$test2&&$enroller\u id3！=$enroller\u id\n&&$direc=$test）{}即使没有数据，您也需要从查询中删除where条件，并将其更改为：$query=$MySQLi\u CON->query（“select*from personal”）；
$query = $MySQLi_CON->query("select * from personal");