在 android 设备上显示我的 php echo

Question

当我的数据库中已经存在用户名或电子邮件时，我想显示我的 php 回显

$sql = "SELECT * FROM user WHERE username='$username' OR email='$email'";

    $check = mysqli_fetch_array(mysqli_query($con,$sql));

        if(isset($check)){
            echo "username or email already exist";
        }else {

            $statement = mysqli_prepare($con, "INSERT INTO User (username, email, age, password) VALUES (?, ?, ?, ?)");
            mysqli_stmt_bind_param($statement, "ssis", $username, $email, $age, $password);
        }

mysqli_stmt_execute($statement);

mysqli_stmt_close($statement);
mysqli_close($con);

当电子邮件或用户名已经存在时，它不会创建帐户，但它不会显示我的回显并退出我的注册活动。

这是我的 ServerRequests.java

 @Override
    protected Void doInBackground(Void... params) {
        ArrayList<NameValuePair> dataToSend = new ArrayList<>();
        dataToSend.add(new BasicNameValuePair("username", user.username));
        dataToSend.add(new BasicNameValuePair("email", user.email));
        dataToSend.add(new BasicNameValuePair("password", user.password));
        dataToSend.add(new BasicNameValuePair("age", user.age + ""));

        HttpParams httpRequestParams = getHttpRequestParams();

        HttpClient client = new DefaultHttpClient(httpRequestParams);
        HttpPost post = new HttpPost(SERVER_ADDRESS
                + "Register.php");


        try {
            post.setEntity(new UrlEncodedFormEntity(dataToSend));
            client.execute(post);

        } catch (Exception e) {
            e.printStackTrace();
        }

        return null;

    }

    private HttpParams getHttpRequestParams() {
        HttpParams httpRequestParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpRequestParams,
                CONNECTION_TIMEOUT);
        HttpConnectionParams.setSoTimeout(httpRequestParams,
                CONNECTION_TIMEOUT);
        return httpRequestParams;
    }

    @Override
    protected void onPostExecute(Void result) {
        super.onPostExecute(result);
        progressDialog.dismiss();
        userCallBack.done(null);
    }

Answer 1

在 PHP 上用于打印的 echo 方法 如果您想向您的应用发送消息 只需将变量用作 $msg 而不是 echo 并在您的应用中请求 $msg

$sql = "SELECT * FROM user WHERE username='$username' OR email='$email'";

    $check = mysqli_fetch_array(mysqli_query($con,$sql));

        if(isset($check)){
            $msg = "username or email already exist";
        }else {

            $statement = mysqli_prepare($con, "INSERT INTO User (username, email, age, password) VALUES (?, ?, ?, ?)");
            mysqli_stmt_bind_param($statement, "ssis", $username, $email, $age, $password);
            $msg = "Created";
        }
mysqli_stmt_execute($statement);
mysqli_stmt_close($statement);
mysqli_close($con);