php - 登录时需要传递其中一个字段值

Question

这是我的用户登录代码。

<?php

require("Conn.php");
require("MySQLDao.php");
$email = htmlentities($_POST["email"]);
$password = htmlentities($_POST["password"]);
$returnValue = array();

if(empty($email) || empty($password))
{
    $returnValue["status"] = "error";
    $returnValue["message"] = "Missing required field";
    echo json_encode($returnValue);
    return;
}

$secure_password = md5($password);

$dao = new MySQLDao();
$dao->openConnection();
$userDetails = $dao->getUserDetailsWithPassword($email,$secure_password);

if(!empty($userDetails))
{
    $returnValue["status"] = "Success";
    $returnValue["message"] = "User is Logged in";
    echo json_encode($returnValue);
} else {

    $returnValue["status"] = "error";
    $returnValue["message"] = "User is not found";
    echo json_encode($returnValue);
}

$dao->closeConnection();

?>

我的sql代码如下:

<?php
class MySQLDao {
    var $dbhost = null;
    var $dbuser = null;
    var $dbpass = null;
    var $conn = null;
    var $dbname = null;
    var $result = null;

    function __construct() {
        $this->dbhost = Conn::$dbhost;
        $this->dbuser = Conn::$dbuser;
        $this->dbpass = Conn::$dbpass;
        $this->dbname = Conn::$dbname;
    }


    // function to open connection

    public function openConnection() {
        $this->conn = new mysqli($this->dbhost, $this->dbuser, $this->dbpass, $this->dbname);
        if (mysqli_connect_errno())
        echo new Exception("Could not establish connection with database");
    }

    // function to return the connection

    public function getConnection() {
        return $this->conn;
    }

    // function to close the connection

    public function closeConnection() {
        if ($this->conn != null)
        $this->conn->close();
    }

    // function to get user email

    public function getUserDetails($email)
    {
        $returnValue = array();
        $sql = "select * from ap_users where user_email='" . $email . "'";

        $result = $this->conn->query($sql);
        if ($result != null && (mysqli_num_rows($result) >= 1)) {
            $row = $result->fetch_array(MYSQLI_ASSOC);
            if (!empty($row)) {
                $returnValue = $row;
            }       
        }
        return $returnValue;
    }

    // get user details using email and password

    public function getUserDetailsWithPassword($email, $userPassword)
    {
        $returnValue = array();
        $sql = "select id,user_email from ap_users where user_email='" . $email . "' and user_password='" .$userPassword . "'";

        $result = $this->conn->query($sql);
        if ($result != null && (mysqli_num_rows($result) >= 1)) {
            $row = $result->fetch_array(MYSQLI_ASSOC);
            if (!empty($row)) {
                $returnValue = $row;
            }
        }
        return $returnValue;
    }

    // register user with all fields

    public function registerUser($email, $password, $username, $fname, $lname, $mobile, $roleid)
    {
        $sql = "insert into ap_users set user_email=?, user_password=?, user_username=?, user_fname=?, user_lname=?, user_mobile=?, user_roleid=?";
        $statement = $this->conn->prepare($sql);

        if (!$statement)
            throw new Exception($statement->error);

        $statement->bind_param("sssssss", $email, $password, $username, $fname, $lname, $mobile, $roleid);
        $returnValue = $statement->execute();

        return $returnValue;
    }

}
?>

当前在登录时，我根据状态和消息获得“成功”和“用户已登录”值。但我想用登录消息推送用户的“roleid”，请帮忙!

Answer 1

roleid is "user_roleid" column in same table.

更改 getUserDetailsWithPassword-Method 中的代码

$sql = "select id,user_email, user_roleid from ap_users where user_email..."

请查看评论中的其他建议(密码和哈希)!