Question

我可以获得有关使用外键插入两个表的帮助吗？我已经有了这个，当我点击“第一次”它可以工作，但当我尝试第二次或第一次后，外键不会改变，它保持不变。

我的数据库表：

passenger (name, p_id) // p_id is auto-increment
luggage (lug_id, brand, type, p_id)

到目前为止，这是我的代码：

private void InsertLuggageInfo() {    
conn = jdbcDBconnection.ConnectDB();
        String query_luggage, query_passenger;    
query_passenger = "INSERT INTO passenger (name) VALUES (?)";
stmt1 = conn.prepareStatement(query_passenger);
stmt1.setString(1, txtFirstname.getText());    
stmt1.execute();
            stmt1.close();
} catch (SQLException ex) {
            System.err.println(ex);
        }  
try {
String p_id = null;    
            String p_id_query = "SELECT p_id FROM passenger WHERE name = '" + txtFirstname.getText() + "'";
            stmt2 = conn.prepareStatement(p_id_query);
            rs2 = stmt2.executeQuery();
            if (rs2.next()) {
                p_id = rs2.getString("p_id");
            }    
query_luggage = "INSERT INTO luggage (lug_id, brand, type, p_id) VALUES (?,?,?,?)";    
stmt3 = conn.prepareStatement(query_luggage);    
stmt2.setString(1, txtLugNr.getText());
            stmt2.setString(2, txtbrand.getText());
            stmt2.setString(3, txtType.getText());
            stmt2.setString(4, p_id);
stmt2.execute();
            stmt2.close();
}catch (SQLException ex) {
            System.err.println(ex);
        }

Answer 1

首先，您应该更改获取生成ID的方式。有很多约翰史密斯，它可以给你带来问题。您可以使用stmt1.getGeneratedKeys()，如下所示：

int affectedRows = stmt1.executeUpdate();

if (affectedRows == 0) {
    throw new SQLException("Insert Passenger error: no rows were affected.");
}

try (ResultSet generatedKeys = stmt1.getGeneratedKeys()) {
    if (generatedKeys.next()) {
        user.setId(generatedKeys.getLong(1));
    }
    else {
        throw new SQLException("Insert Passenger error: no ID obtained.");
    }
}

这将保证您获得新插入的数据ID（您的乘客FK）。

第二：您的stmt3是 INSERT行李。但是您正在设置值并再次执行stmt2。

单击一次按钮，插入两个带外键的表

1 个答案: