[英]MySQL Prepared statement confusion
好的,所以我在Prepared語句上遇到了很多麻煩。 我做了幾個小時的研究,似乎還不能完全理解一切......
我真的覺得我需要了解Prepared語句,因為我正准備在我的網站上發布一些新的免費API(需要API Key來執行API)但我最近意識到一切都不安全....我可以簡單地使用SQL注入繞過API密鑰檢查,例如'OR'1'='1
以下是我驗證API密鑰的方法:
$apikey = $_GET['key'];
$sql = "SELECT * FROM `table` WHERE `key` = '$apikey'";
$query = mysqli_query($con, $sql);
if($query)
{
$fetchrow = mysqli_fetch_row($query);
if(isset($fetchrow[0]))
{
echo "API Key is valid!";
}
else
{
echo "API KEY is invalid";
}
}
如上所述,通過像這樣執行我的API可以很容易地繞過這個
http://website.com/api.php?key='OR'1'='1
這一開始真的嚇到了我,但后來我做了一些研究,並且學會了一種防止任何形式的SQL注入的好方法是使用預處理語句,所以我做了很多研究,這對我來說似乎很復雜:/
所以我想我的問題是,如何使用上面的代碼,並使用預處理語句使其功能相同?
可能你需要的一切:
class Database {
private static $mysqli;
連接到DB:
public static function connect(){
if (isset(self::$mysqli)){
return self::$mysqli;
}
self::$mysqli = new mysqli("DB_HOST", "DB_USER", "DB_PASS", "DB_NAME");
if (mysqli_connect_errno()) {
/*Log error here, return 500 code (db connection error) or something... Details in $mysqli->error*/
}
self::$mysqli->query("SET NAMES utf8");
return self::$mysqli;
}
執行語句並獲得結果:
public static function execute($stmt){
$stmt->execute();
if ($mysqli->error) {
/*Log it or throw 500 code (sql error)*/
}
return self::getResults($stmt);
}
將結果綁定到純數組:
private static function getResults($stmt){
$stmt->store_result();
$meta = $stmt->result_metadata();
if (is_object($meta)){
$variables = array();
$data = array();
while($field = $meta->fetch_field()) {
$variables[] = &$data[$field->name];
}
call_user_func_array(array($stmt, "bind_result"), $variables);
$i = 0;
while($stmt->fetch()) {
$array[$i] = array();
foreach($data as $k=>$v)
$array[$i][$k] = $v;
$i++;
}
$stmt->close();
return $array;
} else {
return $meta;
}
}
課程結束:)
}
用法示例:
public function getSomething($something, $somethingOther){
$mysqli = Database::connect();
$stmt = $mysqli->prepare("SELECT * FROM table WHERE something = ? AND somethingOther = ?");
$stmt->bind_param("si", $something, $somethingOther); // s means string, i means number
$resultsArray = Database::execute($stmt);
$someData = $resultsArray[0]["someColumn"];
}
解決您的問題:
public function isKeyValid($key){
$mysqli = Database::connect();
$stmt = $mysqli->prepare("SELECT * FROM table WHERE key = ? LIMIT 1");
$stmt->bind_param("s", $key);
$results = Database::execute($stmt);
return count($results > 0);
}
PHP會自動關閉數據庫連接,因此不必擔心它。
$sql = "SELECT * FROM `table` WHERE `key` = ?";
if(stmt = $mysqli->prepare($sql)) {
$stmt->bind_param("i", $apikey);
$stmt->execute();
$stmt->bind_result($res);
$stmt->fetch();
$stmt->close();
}
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