Question

在我的应用程序中，我使用2个php文件，其中一个包含应用程序所需的所有功能，另一个是检索并将数据发送回用户

我的第一个php文件是DB_Functions.php 其中包含以下代码

<?php

class DB_Functions {

     private $conn;

    // constructor
    function __construct() {
        require_once 'DB_Connect.php';
        // connecting to database
        $db = new Db_Connect();
        $this->conn = $db->connect();
    }

/**
     * Storing new user
     * returns user details
     */
     public function storeUser($name,$sex,$dob,$email,$college,$password,$latitude,$longitude,$pass) {
    $uuid = uniqid('', true);
    $hash = $this->hashSSHA($password);
    $encrypted_password = $hash["encrypted"]; // encrypted password
    $salt = $hash["salt"]; // salt

    $stmt = $this->conn->prepare("INSERT INTO CFLASH_USERS(Name, sex, dob, mail, college, password, latitude, longitude, pass, salt, created_at) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?, ?, NOW())");
  $stmt->bind_param("ssssssssss", $name, $sex, $dob, $email, $college, $password, $latitude, $longitude, $pass, $salt);

    $result = $stmt->execute();
    $stmt->close();


    // check for successful store
    if ($result) {
        $stmt = $this->conn->prepare("SELECT * FROM CFLASH_USERS WHERE mail = ?");
        $stmt->bind_param("s", $email);
        $stmt->execute();
        $user = $stmt->get_result()->fetch_assoc();
        $stmt->close();

        return $user;
    } else {
        return false;
    }
}

 /**
 * Encrypting password
 * @param password
 * returns salt and encrypted password
 */
public function hashSSHA($password) {

    $salt = sha1(rand());
    $salt = substr($salt, 0, 10);
    $encrypted = base64_encode(sha1($password . $salt, true) . $salt);
    $hash = array("salt" => $salt, "encrypted" => $encrypted);
    return $hash;
}

/**
 * Decrypting password
 * @param salt, password
 * returns hash string
 */
public function checkhashSSHA($salt, $password) {

    $hash = base64_encode(sha1($password . $salt, true) . $salt);

    return $hash;
}

}

?>

我的第二个php文件Register.php包含

<?php

require_once 'include/DB_Functions.php';
$db = new DB_Functions();

// json response array
$response = array("error" => FALSE);

if (isset($_POST['name']) && isset($_POST['email']) && isset($_POST['password'])) {

// receiving the post params
    $name = $_POST['name'];
$sex = $_POST['sex'];
$dob = $_POST['dob'];
$email = $_POST['email'];
$college = $_POST['college'];
$password = $_POST['password'];
$latitude = $_POST['latitude']; 
$longitude  = $_POST['longitude']; 
$pass  = $_POST['pass'];

// create a new user
    $user = $db->storeUser($name, $sex, $dob, $email, $college ,$password, $latitude, $longitude, $pass);
    if ($user) {
        // user stored successfully
        $response["error"] = FALSE;
        $response["uid"] = $user["mail"];
        $response["user"]["name"] = $user["Name"];
        $response["user"]["sex"] = $user["sex"];
        $response["user"]["dob"] = $user["dob"];
        $response["user"]["email"] = $user["mail"];
        $response["user"]["college"] = $user["college"];
        $response["user"]["created_at"] = $user["created_at"];
        $response["user"]["updated_at"] = $user["updated_at"];
        echo json_encode($response);
    } else {
        // user failed to store
        $response["error"] = TRUE;
        $response["error_msg"] = "Unknown error occurred in registration!";
        echo json_encode($response);
    }

} else {

$response["error"] = TRUE;
$response["error_msg"] = "Required parameters (name, email or password) is missing!";
echo json_encode($response);
}
?>

我的表结构是

我无法将数据插入其中。我已经检查过meathod是错的，但没错。所以我认为bind_param（）会出错。请告诉我错误的位置以及纠正错误的解决方案。

请解释一下“调用未定义的方法mysqli_stmt :: get_result（）”

Answer 1

您可能缺少MySQLND。该方法仅适用于该驱动程序。

http://php.net/manual/en/mysqli-stmt.get-result.php

无法使用绑定参数

1 个答案: