Question

我已经浏览了整个互联网，无法找到我为什么会发生这种情况。基本上我正在尝试创建一个TextField，它创建一个数字字符串，并将该字符串转换为整数，并将标签更改为数字+1 每当我尝试使用Integer.parseInt它崩溃

这是我的代码

public class dfadsfa {

    private JFrame frame;
    private JTextField textField;

    /**
     * Launch the application.
     */
    public static void main(String[] args) {
        EventQueue.invokeLater(new Runnable() {
            public void run() {
                try {
                    dfadsfa window = new dfadsfa();
                    window.frame.setVisible(true);
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        });
    }

    /**
     * Create the application.
     */
    public dfadsfa() {
        initialize();
    }

    /**
     * Initialize the contents of the frame.
     */

    String Text;
    int Number = Integer.parseInt(Text); //This line Screws it up

    private void initialize() {
        frame = new JFrame();
        frame.setBounds(100, 100, 450, 300);
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.getContentPane().setLayout(null);

        JLabel lblNumber = new JLabel("");
        lblNumber.setBounds(198, 181, 61, 16);
        frame.getContentPane().add(lblNumber);

        textField = new JTextField();
        textField.addActionListener(new ActionListener() {
            public void actionPerformed(ActionEvent e) {
                Text = textField.getText();
                lblNumber.setText(Number+1+"");
            }
        });

        textField.setBounds(163, 140, 130, 26);
        frame.getContentPane().add(textField);
        textField.setColumns(10);


    }

}

这是显示的消息 -

java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:592)
at java.lang.Integer.parseInt(Integer.java:615)
at dfadsfa.<init>(dfadsfa.java:42)
at dfadsfa$1.run(dfadsfa.java:21)
at java.awt.event.InvocationEvent.dispatch(InvocationEvent.java:311)
at java.awt.EventQueue.dispatchEventImpl(EventQueue.java:756)
at java.awt.EventQueue.access$500(EventQueue.java:97)
at java.awt.EventQueue$3.run(EventQueue.java:709)
at java.awt.EventQueue$3.run(EventQueue.java:703)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.ProtectionDomain$JavaSecurityAccessImpl.doIntersectionPrivilege(ProtectionDomain.java:76)
at java.awt.EventQueue.dispatchEvent(EventQueue.java:726)
at java.awt.EventDispatchThread.pumpOneEventForFilters(EventDispatchThread.java:201)
at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:116)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThread.java:105)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:101)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:93)
at java.awt.EventDispatchThread.run(EventDispatchThread.java:82)

Answer 1

您有一个未初始化的String文本。在尝试解析int时，这可以保证失败。

String Text;
int Number = Integer.parseInt(Text);

在尝试解析int之前，您应该初始化此text。

Answer 2

文本未初始化，因此它给出了NumberFormatException。如果传递的字符串不包含可解析的整数，Integer.parseInt（）方法将给出NumberFormatException。

使用

String Text = "0"解决问题

使用Integer.parseInt会崩溃我的程序

2 个答案: