我想要的只是登录一个简单的用户密码。
我已经放了一个ActionListener
,当我按下登录时,只需弹出密码并检查它是否正常。
import java.awt.BorderLayout;
import java.awt.GridBagConstraints;
import java.awt.GridBagLayout;
import java.awt.GridLayout;
import java.awt.Insets;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JOptionPane;
import javax.swing.JPanel;
import javax.swing.JPasswordField;
import javax.swing.JTextField;
public class PasswordForm
{
private static String password = "mypass";
public static void main(String[] args)
{
// Basic form create
JFrame frame = new JFrame("Form 1");
frame.setSize(300,300);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
// Creating the grid
JPanel panel = new JPanel(new GridBagLayout());
frame.getContentPane().add(panel, BorderLayout.NORTH);
GridBagConstraints c = new GridBagConstraints();
// Create some elements
JTextField usernameInput = new JTextField(10);
c.gridx = 0;
c.gridy = 1;
panel.add(usernameInput,c);
JPasswordField passwordInput = new JPasswordField(10);
c.gridx = 0;
c.gridy = 2;
panel.add(passwordInput,c);
JButton loginInput = new JButton("Login");
c.gridx = 0;
c.gridy = 3;
loginInput.addActionListener(new LoginButton());
panel.add(loginInput,c);
frame.setVisible(true);
}
static class LoginButton implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
JTextField usernameInput = (JTextField)e.getSource();
JOptionPane.showMessageDialog(null,"Text is:");
}
}
}
任何有帮助的人?
错误
Exception in thread "AWT-EventQueue-0" java.lang.ClassCastException: javax.swing.JButton
at PasswordForm$LoginButton.actionPerformed(PasswordForm.java:56)
at javax.swing.AbstractButton.fireActionPerformed(Unknown Source)
at javax.swing.AbstractButton$Handler.actionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.fireActionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.setPressed(Unknown Source)
at javax.swing.plaf.basic.BasicButtonListener.mouseReleased(Unknown Source)
at java.awt.Component.processMouseEvent(Unknown Source)
at javax.swing.JComponent.processMouseEvent(Unknown Source)
at java.awt.Component.processEvent(Unknown Source)
at java.awt.Container.processEvent(Unknown Source)
at java.awt.Component.dispatchEventImpl(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.LightweightDispatcher.retargetMouseEvent(Unknown Source)
at java.awt.LightweightDispatcher.processMouseEvent(Unknown Source)
at java.awt.LightweightDispatcher.dispatchEvent(Unknown Source)
at java.awt.Container.dispatchEventImpl(Unknown Source)
at java.awt.Window.dispatchEventImpl(Unknown Source)
at java.awt.Component.dispatchEvent(Unknown Source)
at java.awt.EventQueue.dispatchEvent(Unknown Source)
at java.awt.EventDispatchThread.pumpOneEventForHierarchy(Unknown Source)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.pumpEvents(Unknown Source)
at java.awt.EventDispatchThread.run(Unknown Source)
答案 0 :(得分:3)
当您忽略代码存在编译错误并尝试运行它时,会发生以下异常。 (我假设你正在使用Eclipse。在相关的源文件中查找红色错误标记,并检查Problems视图。)
Exception in thread "AWT-EventQueue-0" java.lang.Error:
Unresolved compilation problem:
The method getText() is undefined for the type ActionEvent
奇怪的是嵌入式编译错误消息似乎与您发布的源代码不对应。要么你改变了代码,要么你的构建过程中的一些缺陷会导致你运行陈旧的类文件。
其他几点:
您违反了嵌套类中的Java命名约定。类名应始终以大写字母开头。将“loginButton”更改为“LoginButton”。
您的PasswordForm
课程过多地使用了static
。 static
内部类是正常的,但将password
声明为静态,并将所有逻辑放入静态main
方法将导致长期问题。 (好吧,这段代码显然是实验性的......目前的形式。)
答案 1 :(得分:2)
我希望这会对你有所帮助。
public class PasswordForm {
private static String password = "mypass";
private JTextField usernameInput;
public PasswordForm() {
}
private void init(){
// Basic form create
JFrame frame = new JFrame("Form 1");
frame.setSize(300,300);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
// Creating the grid
JPanel panel = new JPanel(new GridBagLayout());
frame.getContentPane().add(panel, BorderLayout.NORTH);
GridBagConstraints c = new GridBagConstraints();
// Create some elements
usernameInput = new JTextField(10);
c.gridx = 0;
c.gridy = 1;
panel.add(usernameInput,c);
JPasswordField passwordInput = new JPasswordField(10);
c.gridx = 0;
c.gridy = 2;
panel.add(passwordInput,c);
JButton loginInput = new JButton("Login");
c.gridx = 0;
c.gridy = 3;
loginInput.addActionListener(new LoginButton());
panel.add(loginInput,c);
frame.setVisible(true);
}
public static void main(String[] args){
PasswordForm form = new PasswordForm();
form.init();
}
class LoginButton implements ActionListener{
public void actionPerformed(ActionEvent e){
//JTextField usernameInput = (JTextField)e.getSource();
String username = (usernameInput.getText().length()>0?usernameInput.getText():" U have not entered!");
JOptionPane.showMessageDialog(null,"Text is : "+username);
}
}
}
答案 2 :(得分:1)
不确定,但是:
JButton loginInput = new JButton("Login");
JTextField usernameInput = (JTextField)e.getSource();
事件的来源怎么可能是TextField
?这将是事件起源的JButton
。你需要重新设计这个,你的错误就会消失或变得更加清晰。
此外,Stephen C所说的:您提供的错误与您的代码不符。
PS:关于你的问题,作为一个懒惰的Java开发人员,我只需在类级别声明我的userid字段和密码字段,然后直接从事件中访问这些字段。它不是完美的方式,但对于Java初学者来说是可以接受的。答案 3 :(得分:1)
这是@ tomdemuyt答案的附录,因为我只是很快就运行了你的代码。我得到了一个完全不同的错误:
Exception in thread "AWT-EventQueue-0" java.lang.ClassCastException: javax.swing.JButton cannot be cast to javax.swing.JTextField
at StupidCode$loginButton.actionPerformed(PasswordForm.java:54)
at javax.swing.AbstractButton.fireActionPerformed(Unknown Source)
at javax.swing.AbstractButton$Handler.actionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.fireActionPerformed(Unknown Source)
at javax.swing.DefaultButtonModel.setPressed(Unknown Source)
...more stack trace information that I'm not sure will help at this stage
我得到的例外只是将代码复制粘贴到Eclipse中,并添加了必要的导入。所以,正如@Stephen先前指出的那样,你的构建过程中一定有问题,或者如果你没有改变任何东西,那么你的项目设置可能不正确吗?
修改
好吧,我可以让代码做你想做的事。只是代码看起来并不是很好,并且已经过度依赖斯蒂芬所指出的静态类/方法。希望这只是您用来学习Java的实验代码......
如果您将JPasswordField
设为私有和静态,即
private static JPasswordField passwordInput;
//main method below
//...
//main method finished, action listener follows...
在主要方法之外,您的ActionListener
,loginButton
可以“看到”它。这样,您可以执行类似
JOptionPane.showMessageDialog(null,"Text is: "+ new String(passwordInput.getPassword()));
但是这不是一个非常好的构建课程的方法。它有斯蒂芬指出的所有缺陷,可能更多因为我自己不是一个伟大的编码器。只要应该做你想要的东西,如果你想要快速解决问题。
答案 4 :(得分:1)
当我按下按钮显示时,我只想要一个简单的对话框 密码,在C#中会很简单,但不是java ..
这在Java中可能和在C#中一样简单,但您只熟悉一种语言。我对于初学者不知道如何在C#中完成这项工作。无论如何,我稍微修改了你的代码以使其工作(我删除了修复的不相关部分以保持我的答案简短)
public class PasswordForm {
private static String password = "mypass";
public static void main(String[] args){
//Swing operations should happen on the EDT
EventQueue.invokeAndWait( new Runnable(){
public void run(){
//whole UI creation
final JTextField usernameInput = new JTextField(10);
final JPasswordField passwordInput = new JPasswordField(10);
//more UI creation
JButton loginInput = new JButton("Login");
loginInput.addActionListener(new ActionListener(){
public void actionPerformed(ActionEvent e){
JOptionPane.showMessageDialog(null,"Username is:" + usernameInput.getText() + " Password is:" + passwordInput.getText());
}
});
}
} //todo catch the exceptions from the invokeAndWait call
}
}