【发布时间】:2015-03-23 10:34:16
【问题描述】:
当我第一次输入错误密码时,showPopUpPassword 工作正常,但当我第二次或更多次重复密码时。它仍然在 JOptionPane.showMessageDialog(null, "incorrect password"); 之后弹出。
table.addMouseListener(new MouseAdapter() {
public void mousePressed(MouseEvent e) {
if (e.getButton() == MouseEvent.BUTTON3)
{
popup.show(table, e.getX(), e.getY());
EditProfile.addActionListener(new ActionListener() {
public void actionPerformed(ActionEvent e) {
int row = table.getSelectedRow();
String name = (String) table.getModel().getValueAt(row, 0);
if (!namelist.contains(name)) {
String pass = ctr.getNamebyPassword(name, password); // get password on database
password = showPopUpPassword(); // get the user input password
if(!pass.equals(password)) {
JOptionPane.showMessageDialog(null, "incorrect password");
}else if (pass.equals(password)){
edit = new editProfileFrame(ctr.getData(name), ctr.getAccount(name));
namelist.add(name);
}
}else {
JOptionPane.showConfirmDialog(null, "Cannot Duplicate Profile Window");
}
}
});
}
}
});
【问题讨论】:
-
为了获得更好的帮助,请尽快发布 SSCCE/MCVE,简短、可运行、可编译,在局部变量中包含 JTable/XxxTableModel 的硬编码值
-
不要使用
if (e.getButton() == MouseEvent.BUTTON3)。人们可能不知道 button3 是什么。相反,您可以使用:if (SwingUtilities.isRightMouseButton(e).
标签: java swing jtable joptionpane jpopup