java - ResultSet 方法 previous() 不起作用

Question

   public void actionPerformed(ActionEvent E)
   {
       int id;
       String name,address,phone;
       Connection conn = null;
       PreparedStatement stmt = null;
       try {
           //STEP 2: Register JDBC driver
           Class.forName("com.mysql.jdbc.Driver");

           //STEP 3: Open a connection
           System.out.println("Connecting to database...");
           conn = DriverManager.getConnection(DB_URL,USER,PASS);

           //STEP 4: Execute a query
           String sql;
           sql = "SELECT * FROM person";
           System.out.println("Creating statement...");
           stmt = conn.prepareStatement (sql,ResultSet.TYPE_SCROLL_INSENSITIVE , 
                                     ResultSet.CONCUR_UPDATABLE );


            ResultSet rs = stmt.executeQuery(sql);

            if(E.getSource()== bNext) {
                rs.next();
                id  = rs.getInt("id");
                name = rs.getString("name");
                address = rs.getString("address");
                phone = rs.getString("phone");

                //Display values
                System.out.print("ID: " + id);
                System.out.print(", Name: " + name);
                System.out.print(", Address: " + address);
                System.out.println(", Phone: " + phone);
            }
            if(E.getSource()== bPrevious) {
                rs.previous();
                id  = rs.getInt("id");
                name = rs.getString("name");
                address = rs.getString("address");
                phone = rs.getString("phone");

                //Display values
                System.out.print("ID: " + id);
                System.out.print(", Name: " + name);
                System.out.print(", Address: " + address);
                System.out.println(", Phone: " + phone);
            }
            if(E.getSource()==bLast) {
                rs.last();
                id  = rs.getInt("id");
                name = rs.getString("name");
                address = rs.getString("address");
                phone = rs.getString("phone");

               //Display values
               System.out.print("ID: " + id);
               System.out.print(", Name: " + name);
               System.out.print(", Address: " + address);
               System.out.println(", Phone: " + phone);
           }
           if(E.getSource()==bFirst) {
               rs.first();
               id  = rs.getInt("id");
               name = rs.getString("name");
               address = rs.getString("address");
               phone = rs.getString("phone");

               //Display values
               System.out.print("ID: " + id);
               System.out.print(", Name: " + name);
               System.out.print(", Address: " + address);
               System.out.println(", Phone: " + phone);
          }
          //STEP 6: Clean-up environment
          rs.close();
          stmt.close();
          conn.close();
    }

我不知道为什么，但是 previous() 方法不起作用，每当用户按下 next 按钮时，输出都保持不变，并且不会向前移动。有人可以帮忙吗？

Answer 1

每次调用 actionPerformed 方法时(我假设按下某个按钮时会调用它)，您都会再次执行查询。因此 next() 将始终为您提供第一个元素，而 previous 将返回 false。

您可能应该将所有行读入某些数据结构，例如ArrayList(作为初始化步骤，而不是在单击按钮时)，并维护该List<的当前索引，这将允许您在单击按钮时向前或向后移动。

附注在不检查返回值的情况下，不应使用 rs.next() 或 rs.previous() 。如果这些方法返回 false，则不应调用 rs.getInt("id") 等方法，因为它们会引发异常。

P.P.S 一些 JDBC 驱动程序不支持 previous()、first 和 last 方法，这意味着它们可能会抛出 SQLFeatureNotSupportedException。但是，如果您按照我的建议在初始化步骤中读取了所有数据，则只需要 next()。