使用用户名登录应用程序时，未使用电子邮件，因此会引发以下错误，反之亦然：“注意：未定义的索引：/Applications/XAMPP/xamppfiles/htdocs/menthor/login.php中的电子邮件16英寸
我尝试过将产生错误的条件放在条件中，但是这些努力被证明是徒劳的 Login.php文件（与错误有关的部分）
// STEP 1. Receive data / info paassed to current file
if ((empty($_REQUEST['username']) || empty($_REQUEST['password'])) && (empty($_REQUEST['email']) || empty($_REQUEST['password']))) {
$return['status'] = '400';
$return['message'] = 'Missing requird information';
echo json_encode($return);
return;
}
// securing received info / data from hackers or injections
$email = htmlentities($_REQUEST['email']);
$username = htmlentities($_REQUEST['username']);
$password = htmlentities($_REQUEST['password']);
// STEP 2. Establish connection with the server
require('secure/access.php');
$access = new access('localhost' , 'root', '' , 'menthor');
$access->connect();
// STEP 3. Check existence of the user. Try to fetch the user with the same email address
// STEP 3. Check availability of the login/user information
$username_aval = $access->selectUser_Username($username);
$email_aval = $access->selectUser_Email($email);
//$return = array();
// user is found
if ($username_aval) {
// Get encrypted password and salt from the server for validation
$encryptedPassword = $username_aval['password'];
$salt = $username_aval['salt'];
if ($encryptedPassword == sha1($password . $salt)) {
$return['status'] = '200';
$return['message'] = 'Successfully Logged In';
$return['id'] = $username_aval['id'];
$return['username'] = $username_aval['username'];
$return['email'] = $username_aval['email'];
$return['fullName'] = $username_aval['fullName'];
$return['lastName'] = $username_aval['lastName'];
$return['birthday'] = $username_aval['birthday'];
$return['gender'] = $username_aval['gender'];
$return['cover'] = $username_aval['cover'];
$return['ava'] = $username_aval['ava'];
$return['bio'] = $username_aval['bio'];
$return['allow_follow'] = $username_aval['allow_follow'];
} else {
// In event that encrypted password and salt does not match
$return['status'] = '201';
$return['message'] = 'Password do not match';
}
} else if ($email_aval) {
// Get encrypted password and salt from the server for validation
$encryptedPassword = $email_aval['password'];
$salt = $email_aval['salt'];
if ($encryptedPassword == sha1($password . $salt)) {
$return['status'] = '200';
$return['message'] = 'Successfully Logged In';
$return['id'] = $email_aval['id'];
$return['username'] = $email_aval['username'];
$return['email'] = $email_aval['email'];
$return['fullName'] = $email_aval['fullName'];
$return['lastName'] = $email_aval['lastName'];
$return['birthday'] = $email_aval['birthday'];
$return['gender'] = $email_aval['gender'];
$return['cover'] = $email_aval['cover'];
$return['ava'] = $email_aval['ava'];
$return['bio'] = $email_aval['bio'];
$return['allow_follow'] = $email_aval['allow_follow'];
} else {
// In event that encrypted password and salt does not match
$return['status'] = '202';
$return['message'] = 'Password do not match';
}
}else {
// In event that user is not found
$return['status'] = '403';
$return['message'] = 'User was not found';
}
// stop connection with server
$access->disconnect();
// pass info as JSON
echo json_encode($return);
Access.php文件（与错误有关的部分） 会根据收到的电子邮件尝试在数据库中选择任何值
public function selectUser_Email($email) {
// array to store full user related information with the logic: key=>Value
$returnArray = array();
// SQL Language / Commande to be sent to the server
// SELECT * FROM users WHERE email='rondell@gmail.com'
$sql = "SELECT * FROM users WHERE email='" . $email . "'";
// executing query via already established connection with the server
$result = $this->conn->query($sql);
// result isn't zero and it has least 1 row / value / result
if ($result != null && (mysqli_num_rows($result)) >= 1) {
// converting to JSON
$row = $result->fetch_array(MYSQLI_ASSOC);
// assign fetched row to ReturnArray
if (!empty($row)) {
$returnArray = $row;
}
}
// throw back returnArray
return $returnArray;
}
// Will try to select any value in the database based on received Email
public function selectUser_Username($username) {
// array to store full user related information with the logic: key=>Value
$returnArray = array();
// SQL Language / Commande to be sent to the server
// SELECT * FROM users WHERE username='rondell'
$sql = "SELECT * FROM users WHERE username='" . $username . "'";
// executing query via already established connection with the server
$result = $this->conn->query($sql);
// result isn't zero and it has least 1 row / value / result
if ($result != null && (mysqli_num_rows($result)) >= 1) {
// converting to JSON
$row = $result->fetch_array(MYSQLI_ASSOC);
// assign fetched row to ReturnArray
if (!empty($row)) {
$returnArray = $row;
}
}
// throw back returnArray
return $returnArray;
}
通过网络服务器登录时的当前结果
Notice: Undefined index: email in /Applications/XAMPP/xamppfiles/htdocs/menthor/login.php on line 16
{"status":"200","message":"Successfully Logged In","id":"44","username":"rondell","email":"rondell@gmail.com","fullName":"rondell","lastName":"","birthday":"","gender":"","cover":"","ava":"","bio":"","allow_follow":"1"}
预期结果
{"status":"200","message":"Successfully Logged In","id":"44","username":"rondell","email":"rondell@gmail.com","fullName":"rondell","lastName":"","birthday":"","gender":"","cover":"","ava":"","bio":"","allow_follow":"1"}
答案 0 :(得分：0)
使用电子邮件作为对象，或者您可以转储请求并查看发生的情况
答案 1 :(得分：0)
结果是，解决方案非常简单，经过一番刻苦的思考...我只需要简单地创建两个login.php文件...一个专用于使用用户名和密码登录的用户，另一个对于使用电子邮件和密码登录的用户...欢呼