"Cant process the request",处理基本的参数化查询
"Cant process the request", dealing with basic parameterized queries
我正在尝试我在网上找到的东西(对此非常陌生)并且 none 有效。这是一个随机的科学项目,我决定了解更多,但我仍停留在 "procedures" 的第 2 部分。 https://www.sciencebuddies.org/science-fair-projects/project-ideas/Cyber_p008/cybersecurity/sql-injection#procedure
我看过视频,但它们只包含 user_ID 而不是用户名和密码。注意:只有处理 login.php 的代码才会导致问题。
<?php
include("global.php");
include("db.php");
session_start();
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password are sent in the form
$username = $_POST['username'];
$password = $_POST['password'];
// Check if the username and password exist in the database
$sql = "SELECT username FROM users WHERE username = '$username' AND password = '$password'";
$stmt = msqli_stmt_init($db);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL Statement Failed";
} else {
mysqli_stmt_bind_param($stmt, "ss", $username, $password );
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);
$count = mysqli_num_rows($result);}
// If username and password matched then there is one row in the result
if ($count != 0) {
$_SESSION['login_user'] = strtolower($username);
header("location: search.php");
}
else {
$error = "Your Username or Password is invalid";
}
}
?>
它本应阻止基本的“'or''='”注入攻击,但它决定不完全起作用。
如果您使用查询参数——这绝对是个好主意——您必须在查询中保留占位符。使用 ?
作为占位符。
像这样:
$sql = "SELECT username FROM users WHERE username = ? AND password = ?";
您稍后将变量绑定到这些参数。您必须绑定与参数占位符数量相同的变量。
您遇到了您描述的错误,因为您试图将变量绑定到没有参数占位符的查询。
我正在尝试我在网上找到的东西(对此非常陌生)并且 none 有效。这是一个随机的科学项目,我决定了解更多,但我仍停留在 "procedures" 的第 2 部分。 https://www.sciencebuddies.org/science-fair-projects/project-ideas/Cyber_p008/cybersecurity/sql-injection#procedure
我看过视频,但它们只包含 user_ID 而不是用户名和密码。注意:只有处理 login.php 的代码才会导致问题。
<?php
include("global.php");
include("db.php");
session_start();
if ($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password are sent in the form
$username = $_POST['username'];
$password = $_POST['password'];
// Check if the username and password exist in the database
$sql = "SELECT username FROM users WHERE username = '$username' AND password = '$password'";
$stmt = msqli_stmt_init($db);
if (!mysqli_stmt_prepare($stmt, $sql)) {
echo "SQL Statement Failed";
} else {
mysqli_stmt_bind_param($stmt, "ss", $username, $password );
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
$row = mysqli_fetch_array($result, MYSQLI_ASSOC);
$count = mysqli_num_rows($result);}
// If username and password matched then there is one row in the result
if ($count != 0) {
$_SESSION['login_user'] = strtolower($username);
header("location: search.php");
}
else {
$error = "Your Username or Password is invalid";
}
} ?>
它本应阻止基本的“'or''='”注入攻击,但它决定不完全起作用。
如果您使用查询参数——这绝对是个好主意——您必须在查询中保留占位符。使用 ?
作为占位符。
像这样:
$sql = "SELECT username FROM users WHERE username = ? AND password = ?";
您稍后将变量绑定到这些参数。您必须绑定与参数占位符数量相同的变量。
您遇到了您描述的错误,因为您试图将变量绑定到没有参数占位符的查询。