Warning: file_get_contents(/data/phpspider/zhask/data//catemap/9/java/374.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
检查MySQL Java中是否存在用户名_Java_Mysql_Database - Fatal编程技术网

检查MySQL Java中是否存在用户名

检查MySQL Java中是否存在用户名,java,mysql,database,Java,Mysql,Database,我有一个寄存器对话框,它实现了一个动作侦听器。如果用户输入的名称已经存在,我想在控制台上打印一条消息。如果用户名不存在,MySQL应该将其添加到数据库中。不幸的是,此代码无法工作: private void regButtonActionPerformed(java.awt.event.ActionEvent evt) { if(!userBox.getText().equals("")) {

我有一个寄存器对话框,它实现了一个动作侦听器。如果用户输入的名称已经存在,我想在控制台上打印一条消息。如果用户名不存在,MySQL应该将其添加到数据库中。不幸的是,此代码无法工作:

private void regButtonActionPerformed(java.awt.event.ActionEvent evt) {                                          
    if(!userBox.getText().equals(""))
    {
        try
        {
            Class.forName("com.mysql.jdbc.Driver");
            Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/genx", "root", "Warlock1989");
            Statement stmt = con.createStatement();
            String query = "SELECT name FROM accounts";
            ResultSet rs = stmt.executeQuery(query);
            while(rs.next())
            {
                String uname = rs.getString("name");
                if(!uname.equals(userBox.getText()))
                {
                    PreparedStatement pstmt = con.prepareStatement("INSERT INTO accounts(name) VALUES(?)");
                    pstmt.setString(1, userBox.getText());
                    pstmt.executeUpdate();
                    System.out.println("Username " + userBox.getText() + " has been registered.");
                }
                else
                {
                    System.out.println("Username " + userBox.getText() + " already exists.");
                }
            }
        }
        catch(Exception ex)
        {
            ex.printStackTrace();
        }
    } 

您当前的方法从数据库加载所有记录,并尝试查找用户,如果数据库包含大量记录,则会导致内存异常。所以 不要从数据库中获取所有记录,只需使用
where name=?
运行查询,以检查数据库中是否已存在用户,如下所示:

PreparedStatement pstmt1 = null;
PreparedStatement pstmt2 = null;
ResultSet rs = null;
try {

    String userInput = userBox.getText();

    String query = "SELECT name FROM accounts where name=?";
    pstmt1 = con.preparedStatement(query);
    pstmt1.setString(1, userInput);

    rs = pstmt1.executeQuery();

    if(rs.next()) {
            pstmt2 = con.prepareStatement("INSERT INTO accounts(name) VALUES(?)");
            pstmt2.setString(1, userInput);
            pstmt2.executeUpdate();
            System.out.println("Username " + userInput + " has been registered.");
    }  else {
            System.out.println("Username " + userInput + " already exists.");
    }

} catch(SQLException sqlexe) {
    //add logging
} finally {
    if(pstmt1 != null)
       pstmt1.close();
    if(pstmt2 != null)
       pstmt2.close();
     if(rs != null)
       rs.close();
}

您当前的代码不会释放
resultset
&
preparedstatement
对象,因此请确保您正在释放finally块中的资源

“不工作”是什么意思?打印堆栈跟踪没有堆栈跟踪。什么也没发生。它不叫if、else或catch。你的桌子空了吗?因为这样select将不返回任何行,
rs.next()
将返回
false
。。。