传递变量值

Question

我可以知道如何从$ query中传递值，如果其中有值的话。如果它是空的，我仍然需要传递变量。虽然变量确实存在于sql数据库中，但我仍然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; 
     } 
    } 
} 
?> 

Answer 1

您需要在这里尝试一些调试。例如，根据该意见

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 

，你需要删除WHERE从查询条件，将其更改为简单：

$query = $MySQLi_CON->query("select * from personal");