我想使用 PHP 将我的 ios 应用程序(使用 swift 3)中的用户注册详细信息保存在 MySQL 中。但这不起作用。 这是我的代码:
registeruser.php
$response = array ();
if ($_SERVER ['REQUEST_METHOD'] == 'POST') {
// getting values
$username = $_POST ["username"];
$password = $_POST ['password'];
// including the db operation file
require_once '../include/DbOperation.php';
// inserting values
if ($db->createUser($username, $password)) {
$response ['error'] = false;
$response ['message'] = 'user added successfully';
} else {
$response ['error'] = true;
$response ['message'] = ' user not added';
}
echo json_encode ( $response );
public function createUser($username, $password)
{
$stmt = $this->conn->prepare("INSERT INTO registerUser(username, password) values(?, ?)");
$stmt->bind_param("register", $username, $password);
$result = $stmt->execute();
$stmt->close();
echo $result;
if ($result) {
return true;
} else {
return false;
}
}
不知道为什么不起作用?
显示的错误是
Undefined index: Username and password.
最佳答案
public function actionXXX()
{
$response = array ();
if ($_SERVER ['REQUEST_METHOD'] == 'POST') {
// getting values
$username = $_POST ["username"];
$password = $_POST ['password'];
// including the db operation file
require_once '../include/DbOperation.php';
// inserting values
// $db = new ***
if ($db->createUser($username, $password)) {
$response ['error'] = false;
$response ['message'] = 'user added successfully';
} else {
$response ['error'] = true;
$response ['message'] = 'user not added';
}
return json_encode($response);
}
}
public function createUser($username, $password)
{
$stmt = $this->conn->prepare("INSERT INTO registerUser(username, password) values(?, ?)");
$stmt->bind_param("register", $username, $password);
$result = $stmt->execute();
$stmt->close();
if ($result) {
return true;
} else {
return false;
}
}
关于php - 如何使用php从ios应用程序将数据存储在mysql中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41583018/