php - 如何防止 SQL 注入(inject)更改 URL 参数(DELETE 语句)PHP

Question

我有一个如下代码，用于通过 URL 参数删除条目

<td><a href="deletecar.php?car_id=<?php echo $row_cars['car_id']; ?>" onclick=" if ( !confirm('Are you sure to DELETE?') ) return false; ">Delete</a></td>

这是 URL 参数输出

http://localhost/html/deletecar.php?car_id=17

但是如果我将 car_id=17 更改为 car_id=23(位于其他用户的汽车列表中)，它将删除

如何防止这种情况

deletecar.php如下所示

<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  if (PHP_VERSION < 6) {
    $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;
  }

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? doubleval($theValue) : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

if ((isset($_GET['car_id'])) && ($_GET['car_id'] != "") && (isset($_SESSION['MM_Username']))) {
  $deleteSQL = sprintf("DELETE FROM cars WHERE car_id=%s",
                       GetSQLValueString($_GET['car_id'], "int"));

  mysql_select_db($database_conn, $conn);
  $Result1 = mysql_query($deleteSQL, $conn) or die(mysql_error());

  $deleteGoTo = "myaccount.php";
  if (isset($_SERVER['QUERY_STRING'])) {
    $deleteGoTo .= (strpos($deleteGoTo, '?')) ? "&" : "?";
    $deleteGoTo .= $_SERVER['QUERY_STRING'];
  }
  header(sprintf("Location: %s", $deleteGoTo));
}
?>

这是我在数据库中的表

INSERT INTO `car` (`car_id`, `c_id`, `c_brand`, `c_model`, `c_model_nd`, `c_model_year`, `c_color`, `c_capacity`, `c_owner`, `c_statu`, `c_show`) VALUES
(16, '34DA1593', 'Volkswagen', 'Volt', '313 CDI', 2006, 'Beyaz', '', 18, 'yakamozturizm', 'Boş', 0),
(17, '34BC5897', 'Mercedes', 'Sprinter', '313CDI', 2006, 'Gri', '', 14, 'PcRestorer', 'Boş', 0),
(18, '34DBC145', 'Volkswagen', 'Volt', '213 CDI', 2013, 'Beyaz', '', 16, 'PcRestorer', 'Boş', 0);

编辑....

我已经这样改变了我的代码

$colname_delete = "-1";
if (isset($_GET['car_id'])) {
  $colname_delete = $_GET['car_id'];
}
$owner_delete = "-1";
if (isset($_SESSION['MM_Username'])) {
  $owner_delete = $_SESSION['MM_Username'];
}

if ((isset($_GET['car_id'])) && ($_GET['car_id'] != "")) {
  $deleteSQL = sprintf("DELETE FROM minibusler  WHERE car_id = %s AND c_owner =%s", 

GetSQLValueString($colname_delete, "int"),
GetSQLValueString($owner_delete, "text"));

  mysql_select_db($database_conn, $conn);
  $Result1 = mysql_query($deleteSQL, $conn) or die(mysql_error());

  $deleteGoTo = "myaccount.php";
  if (isset($_SERVER['QUERY_STRING'])) {
    $deleteGoTo .= (strpos($deleteGoTo, '?')) ? "&" : "?";
    $deleteGoTo .= $_SERVER['QUERY_STRING'];
  }
  header(sprintf("Location: %s", $deleteGoTo));
}

看起来可行，您认为这是安全的方法吗

感谢您的帮助

Answer 1

无论如何，在删除汽车之前，您应该检查它是否属于当前用户。如果没有显示合适的消息。