php - 无法为对象输入值。导致 JSON 对象响应

Question

我有一个有效的 php 代码，它从我的 Android 应用程序的 JsonObjectRequest 中获取 id 和 deviceID。问题是什么时候在“函数 authenticateUser”中。我似乎无法执行此操作 operation $response->auth = "1";。

当我评论此代码时，它不会产生任何错误并为我提供正确的输出，除了我为对象身份验证获取 null。我可以在对象 (response->$isAuthenticated 和 response->$isSameUser) 中为这两个变量输入值，但是当我尝试输入 (response->$auth)

class Response{
    public $isAuthenticated;
    public $isSameUser;
    public $auth;
}

$response = new Response();
$error = array();
$log= array();

if(isset($decoded['id']) && isset($decoded['deviceID'])){
    $conn = mysqli_connect($servername,$username,$password,$dbname);
    $id = $decoded['id'];
    $deviceID = $decoded['deviceID'];


    if (mysqli_connect_errno())
    {
        array_push($error,"Failed to connect to MySQL: " . mysqli_connect_error());
    }
    else
    {
        $response -> isAuthenticated = checkIfAlreadyAuthenticated($conn, $id);
        if($response -> isAuthenticated ==0){
            array_push($log, $response -> isAuthenticated);
            authenticateUser($response, $conn, $id, $deviceID);

        }
        elseif($response -> isAuthenticated ==1){
            array_push($log, $response -> isAuthenticated);
            $response -> isSameUser = checkIfSameUser($conn, $id, $deviceID);
        }

    }

}
else{
    //echo 'POST ERROR';
}

function checkIfSameUser($conn, $id, $deviceID){
    $sql = "SELECT pin, deviceID FROM nextrack_userauthentication";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        while($row = $result->fetch_assoc()) {
            $id_fromDB = $row["pin"]; 
            $deviceID_fromDB = $row["deviceID"];
        }
    } else {
        //echo "checkifSameUser Method SQL ERROR";
    }

    if((($id_fromDB == $id) == TRUE) AND (($deviceID_fromDB == $deviceID)== TRUE)){
        return 1;
    }

}

function authenticateUser($conn, $id, $deviceID){
    $authenticate = "1";
    $sql = "INSERT INTO nextrack_userauthentication(pin, activated, deviceID) VALUES ('".$id."','".$authenticate."','".$deviceID."')";
    mysqli_query($conn, $sql);

    if(mysqli_affected_rows($conn)>0)
    {
        $response->auth = "1";
    } 
    else 
    {
        $response->auth = "0";
    }

}

function checkIfAlreadyAuthenticated($conn, $id){
    $sql = "SELECT EXISTS(SELECT'". $id ."' FROM nextrack_userauthentication WHERE pin='" . $id ."'";

    $result = $conn->query("SELECT '". $id ."' FROM nextrack_userauthentication WHERE pin='" . $id ."'");

    if($result->num_rows == 0) {
        return 0;
    } else {
        return 1;
    }
}

echo json_encode($response, JSON_FORCE_OBJECT);

Answer 1

让 authenticateUser() 返回 1 或 0 可能会更好，具体取决于登录的成功并在调用部分分配它。这意味着 authenticateUser() 不直接链接到响应，而只是操作是否正常的情况...

   function authenticateUser($conn, $id, $deviceID){
        $authenticate = "1";
        $sql = "INSERT INTO nextrack_userauthentication(pin, activated, deviceID) VALUES ('".$id."','".$authenticate."','".$deviceID."')";

        mysqli_query($conn, $sql);

        if(mysqli_affected_rows($conn)>0)
        {
            return "1";
        } 
        else 
        {
            return "0";
        }

    }

然后……

    if($response -> isAuthenticated ==0){
        $response->auth = authenticateUser($conn, $id, $deviceID);
    }    

你也可以使用 auth 作为 true 或 false 然后你的函数变成......

   function authenticateUser($conn, $id, $deviceID){
        $authenticate = "1";
        $sql = "INSERT INTO nextrack_userauthentication(pin, activated, deviceID) VALUES ('".$id."','".$authenticate."','".$deviceID."')";

        mysqli_query($conn, $sql);

        return (mysqli_affected_rows($conn)>0);
    }