如何解决一个keyPressed在java swing中延迟?
问题描述:
我想开发自动建议Textfield。当我在文本字段我打字这是一个延时键和提示一个键晚**如何解决一个keyPressed在java swing中延迟?
当我按在list.When我打字AAA它查询只为“AA”,并建议按AAA后,然后将查询为aaa
private void initialize() {
frame = new JFrame();
frame.setBounds(100, 100, 450, 300);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.getContentPane().setLayout(null);
JPanel panel = new JPanel();
panel.setBounds(10, 11, 414, 239);
frame.getContentPane().add(panel);
panel.setLayout(null);
textField = new JTextField();
textField.addKeyListener(new KeyAdapter() {
@Override
public void keyPressed(KeyEvent arg0) {
try {
addListItem();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
});
textField.setBounds(71, 24, 86, 20);
panel.add(textField);
textField.setColumns(10);
list = new JList<String>();
list.setBounds(71, 55, 86, 97);
list.setVisible(false);
list.getScrollableTracksViewportHeight();
panel.add(list);
}
public void addListItem() throws Exception {
list.setVisible(true);
Class.forName("com.mysql.jdbc.Driver");
c = DriverManager.getConnection("jdbc:mysql://localhost:3306/studentdemo", "root", "root");
ps = c.prepareStatement(
"SELECT stud_name FROM student_info where stud_name like '" + textField.getText().trim() + "%' ");
ResultSet s = ps.executeQuery();
model = new DefaultListModel<String>();
while (s.next()) {
model.addElement(s.getString("stud_name"));
System.out.println(s.getString("stud_name"));
}
list.setModel(model);
}
答
使用keyReleased
,而不是KeyAdapter
类的keyPressed
方法的任意键(额外的按键)。
使用'keyReleased'? – Berger
我应该在keyReleased中写什么? @Berger – Nilay
只需使用'keyPressed'的'keyReleased' INSTEAD。 – Berger