问题描述
在Android Studio中处理登录脚本。它将api调用发送到在WAMP中运行的PHP脚本。一切正常,除非会话变量设置不正确。疯狂的事情是,它完全适合邮递员使用。它以应有的方式返回用户标识。当我从仿真器运行登录名时,它工作正常并且没有显示错误。问题是我无法将userid设置为会话变量,以便可以在其他PHP页面(程序的其余部分)中使用它。我设置了一个测试会话变量PHP文件,它将会话变量设置得很好。我可以从其他PHP文件访问它们。我什至在脚本的顶部放置了一个测试脚本-$ _SESSION ['mystery'] =“问题”; -设置得很好,我可以从其他PHP文件访问它。我从来没有这么难过。它可能与WAMP不允许从和android调用设置会话变量有关吗?我没有尝试过将它放在实时服务器上。我不知道它的行为是否会有所不同。有什么想法吗?
这是我的登录API:
<?PHP
session_start();
$response = array();
// Check if the user is already logged in,if yes then redirect him to welcome page
if(isset($_SESSION["loggedin"]) && $_SESSION["loggedin"] === true){
echo "already logged in";
//header("location: welcome.PHP");
//exit;
}
// Include config file
require_once "configuration.PHP";
$_SESSION['mystery'] = "problem";
// Define variables and initialize with empty values
$username = "";
$password = "";
$username_err = "";
$password_err = "";
// Processing form data when form is submitted
if($_SERVER["REQUEST_METHOD"] == "POST"){
// Check if username is empty
if(empty(trim($_POST["username"]))){
$username_err = "Please enter username.";
} else{
$username = trim($_POST["username"]);
}
// Check if password is empty
if(empty(trim($_POST["password"]))){
$password_err = "Please enter your password.";
} else{
$password = trim($_POST["password"]);
}
// Validate credentials
if(empty($username_err) && empty($password_err)){
// Prepare a select statement
$sql = "SELECT id,username,password FROM users WHERE username = ?";
if($stmt = MysqLi_prepare($conn,$sql)){
// Bind variables to the prepared statement as parameters
MysqLi_stmt_bind_param($stmt,"s",$param_username);
// Set parameters
$param_username = $username;
// Attempt to execute the prepared statement
if(MysqLi_stmt_execute($stmt)){
// Store result
MysqLi_stmt_store_result($stmt);
// Check if username exists,if yes then verify password
if(MysqLi_stmt_num_rows($stmt) == 1){
// Bind result variables
MysqLi_stmt_bind_result($stmt,$id,$username,$hashed_password);
if(MysqLi_stmt_fetch($stmt)){
if(password_verify($password,$hashed_password)){
// Password is correct,so start a new session
session_start();
$user = array(
'id'=>$id,'username'=>$username
);
$response['error'] = false;
$response['message'] = 'Login successfull';
$response['user'] = $user;
// Store data in session variables
// $_SESSION["loggedin"] = true;
$_SESSION['id'] = $user['id'];
// Redirect user to welcome page
//header("location: welcome.PHP");
} else{
// display an error message if password is not valid
$password_err = "The password you entered was not valid.";
}
}
} else{
// display an error message if username doesn't exist
$username_err = "No account found with that username.";
}
} else{
echo "Oops! Something went wrong. Please try again later.";
}
// Close statement
MysqLi_stmt_close($stmt);
}
}
// Close connection
MysqLi_close($conn);
echo json_encode($response).'<br><br>';
print_r ($_SESSION).'<br><br>';
}
?>
解决方法
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