Php 分析数据json时出错。。。

我正在开发一个android应用程序，并试图允许用户通过php/mysql Web服务登录。 然而，该应用程序在输入电子邮件和密码后崩溃。检查logcat，我看到以下内容

JSON显示为--nnn{“success”：“1”，“uid”：“111”，“name”：“test”，“email”：“test”}n

然后错误——解析数据org.json.JSONException时出错：java.lang.String类型的值nnn无法转换为JSONObject

这个“nnn”是从哪里来的？我在我的代码中看不到这一点（可能是打字错误）

这是php-->

---

非常感谢。

原来是我的JSONParser中的一个输入错误

而（（line=reader.readLine（））！=null）{ sb.附加（第+行“n”）

已更改为\n，错误消失

谢谢

// get tag
if (isset($_POST['tag']) && $_POST['tag'] != '') {
$tag = $_POST['tag'];

//json response array
$response = array();

// check for tag type
if ($tag == 'login') {

$email = $_POST['email'];
$password = $_POST['password'];

//check user exists
$qry = "SELECT * FROM users WHERE email = ? AND password = ? ";
$stmt = $mysqli->prepare($qry);
$stmt->bind_param('ss', $email, $password);

if($stmt->execute()){

$result = $stmt->get_result();

$user = $result->fetch_assoc();

        // if user exists, set json values
        $response["success"] = "1";
        $response["uid"] = $user["unique_id"];
        $response["name"] = $user["name"];
        $response["email"] = $user["email"];
        echo json_encode($response);
    } else {
        // user not found
        $response["error"] = 1;
        $response["error_msg"] = "Incorrect email or password!";
        echo json_encode($response);
        exit;
    }
    // else register new user
} else if ($tag == 'register') {

    $name = $_POST['name'];
    $email = $_POST['email'];
    $password = $_POST['password'];

    // first check if user already exists

    $query = "SELECT email FROM users WHERE email=?";
    $stmt = $mysqli->prepare($query);
    $stmt->bind_param('s',$email);
    $stmt->execute();
    /* store result */
    $stmt->store_result();
    if ($stmt->num_rows > 0){
       //if result greater than 0 , email exists
        $response["error"] = 2;
        $response["error_msg"] = "Email address already registered.";
        echo json_encode($response);
        exit;
    } else {
        // create user
        $qry = "INSERT INTO users (name, email, password) VALUES(?, ?, ?)";
        $stmt = $mysqli->prepare($qry);
        $stmt->bind_param('sss', $name, $email, $password);

        if($stmt->execute())
        {
            $uid = mysql_insert_id();
            // user stored successfully
            $response["success"] = 1;
            $response["uid"] = $uid;
            $response["name"] = $name;
            $response["email"] = $email;
            echo json_encode($response);
        } else {
            // user failed to store
            $response["error"] = 1;
            $response["error_msg"] = "Error with Registartion";
            echo json_encode($response);
        }
    }
} else {
    echo "Invalid Request";
}
 } else {
 echo "Access Denied";
}
?>