Question

我试图从MySQL表中获取一些数据并将此数据插入到JSONArray中。

在我的代码下面：

public static JSONArray get_inactive_bookings(String username) {
        JSONArray array = new JSONArray();
        JSONObject obj = new JSONObject();
        Connection conn = null;
        ResultSet rs = null;
        int index = 0;
        int user_id = Users.get_user_id(username);
        String sql = "select * from bookings where user_id =" + user_id + 
                " and Effettiva_Restituzione is not null";
        try {
            conn = Utilities.connect();
            Statement st = conn.createStatement();
            rs = st.executeQuery(sql);
            //This is where I think problem shows up
            while (rs.next()) {

                obj.put("Booking_ID", rs.getInt("booking_id"));
                obj.put("Book_Title", Books.get_title(rs.getInt("book_id")));
                obj.put("Booking_Date", rs.getString("Data_Prenotazione"));
                obj.put("Return_Date", rs.getString("Data_Restituzione"));
                array.add(index, obj);
                index++;
            }
        } catch (SQLException ecc) {
            System.out.println("ERROR: " + ecc.getMessage());
        } finally {
            Utilities.disconnect(conn);
        }
        return array;
    }

在我的表中，一旦执行sql查询，我会发生两次，但在结果数组中，我得到表中最后一次出现的两次。我想这是错的。你们知道如何在JSONArray中同时出现这两种情况吗？谢谢

Answer 1

在第一次迭代中，首先添加键值。但是在第一次迭代之后，您将覆盖对象中每个键的映射值。因为，您多次添加相同的对象会覆盖其条目。您必须在每次迭代时创建一个新的jsonObject。所以

JSONObject obj = new JSONObject（）;

必须在while循环中。像

public static JSONArray get_inactive_bookings(String username) {
    JSONArray array = new JSONArray();
    //JSONObject obj = new JSONObject(); -> this line must be in while loop
    Connection conn = null;
    ResultSet rs = null;
    int index = 0;
    int user_id = Users.get_user_id(username);
    String sql = "select * from bookings where user_id =" + user_id + 
            " and Effettiva_Restituzione is not null";
    try {
        conn = Utilities.connect();
        Statement st = conn.createStatement();
        rs = st.executeQuery(sql);
        //This is where I think problem shows up
        while (rs.next()) {
            JSONObject obj = new JSONObject(); // moved to here
            obj.put("Booking_ID", rs.getInt("booking_id"));
            obj.put("Book_Title", Books.get_title(rs.getInt("book_id")));
            obj.put("Booking_Date", rs.getString("Data_Prenotazione"));
            obj.put("Return_Date", rs.getString("Data_Restituzione"));
            array.add(index, obj);
            index++;
        }
    } catch (SQLException ecc) {
        System.out.println("ERROR: " + ecc.getMessage());
    } finally {
        Utilities.disconnect(conn);
    }
    return array;
}

此外，您不需要单独保留'index'来指定要添加的元素的位置。它已默认添加到数组的末尾。因此，如果将新元素添加到数组的末尾，最好添加如array.add(obj);之类的元素。

Answer 2

您需要将对象实例化移动到循环中，如下所示：

public static JSONArray get_inactive_bookings(String username) {
        JSONArray array = new JSONArray();
        Connection conn = null;
        ResultSet rs = null;
        int index = 0;
        int user_id = Users.get_user_id(username);
        String sql = "select * from bookings where user_id =" + user_id + 
                " and Effettiva_Restituzione is not null";
        try {
            conn = Utilities.connect();
            Statement st = conn.createStatement();
            rs = st.executeQuery(sql);
            //This is where I think problem shows up
            while (rs.next()) {
                JSONObject obj = new JSONObject();

                obj.put("Booking_ID", rs.getInt("booking_id"));
                obj.put("Book_Title", Books.get_title(rs.getInt("book_id")));
                obj.put("Booking_Date", rs.getString("Data_Prenotazione"));
                obj.put("Return_Date", rs.getString("Data_Restituzione"));
                array.add(index, obj);
                index++;
            }
        } catch (SQLException ecc) {
            System.out.println("ERROR: " + ecc.getMessage());
        } finally {
            Utilities.disconnect(conn);
        }
        return array;
    }

Answer 3

您正在尝试将每个元素放入数组中使用相同的JSONObject（相同的引用），因此您只需在循环中覆盖其值。

<script src="https://code.jquery.com/jquery-3.2.1.min.js"></script>
<div class="col-lg-12">
    <label>Username</label>
    <input type="text" class="form-control">
    <div class="field-help"> Choose a username that contains only letters and numbers, or use your email address. </div>
</div>
<div class="col-lg-12">
    <label>Password</label>
    <input type="password" class="form-control">
    <div class="field-help"> It's allways good to have one. </div>
</div>

Answer 4

您需要在每次迭代中创建一个新对象，因为您要覆盖该对象 -

while (rs.next()) {
    JSONObject obj = new JSONObject();
    obj.put("Booking_ID", rs.getInt("booking_id"));
    obj.put("Book_Title", Books.get_title(rs.getInt("book_id")));
    obj.put("Booking_Date", rs.getString("Data_Prenotazione"));
    obj.put("Return_Date", rs.getString("Data_Restituzione"));
    array.add(index, obj);
    index++;
}

如何将MySQL表数据插入JSONArray？

4 个答案: