数字的JTextfield验证？

Question

我正在尝试从JTextField 验证我的卷号（整数值的输入）。好吧，我的代码正在编译，但是在运行时它给了我一个错误NumberFormatException。 这是我的验证码

public int rno_vd() {
    int a=0,b=0,c=0,x=0,y=0;
    int vrno =Integer.parseInt(txtRno.getText());
    String r = String.valueOf(vrno);
    if (r.isEmpty()) {
        JOptionPane.showMessageDialog(null,"rno should not be empty");
        a=1;
    }
    else if (Pattern.matches("[a-zA-Z]+",r)) {
        JOptionPane.showMessageDialog(null,"rno should be in digits");
        b=1;
    }
    else if (vrno < 0) {
        JOptionPane.showMessageDialog(null,"rno cannot be negative");
        c=1;
    }
    System.out.println(a + b + c);
    if (a==1 || b==1 || c==1) {
        x=1;
        return x;
    }
    else {
        y=0;
        return y;
    }
}

错误

C:\Users\Hp\Desktop\jproject>javac -cp hibernatejar\*  *.java
Note: DBHandler.java uses unchecked or unsafe operations.
Note: Recompile with -Xlint:unchecked for details.

C:\Users\Hp\Desktop\jproject>java -cp hibernatejar\*;.  Sms
Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: ""
        at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:68)
        at java.base/java.lang.Integer.parseInt(Integer.java:662)
        at java.base/java.lang.Integer.parseInt(Integer.java:770)
        at AddFrame.lambda$new$1(AddFrame.java:80)
        at java.desktop/javax.swing.AbstractButton.fireActionPerformed(AbstractButton.java:1967)
        at java.desktop/javax.swing.AbstractButton$Handler.actionPerformed(AbstractButton.java:2308)
        at java.desktop/javax.swing.DefaultButtonModel.fireActionPerformed(DefaultButtonModel.java:405)
        at java.desktop/javax.swing.DefaultButtonModel.setPressed(DefaultButtonModel.java:262)
        at java.desktop/javax.swing.plaf.basic.BasicButtonListener.mouseReleased(BasicButtonListener.java:279)
        

Answer 1

Komal 在这里，我可以看到您想要验证 JTextField 的滚动编号，即应该只包含整数，不应接受任何其他字符而不是数字... 好吧，我建议您尝试使用 KeyEvent 的 KeyListener 对每个 keyReleased 进行验证。 您按下的每个键都经过验证，如果它的数字是好的，如果不是，它会显示对话框说“只接受数字”。

我正在分享我的代码，希望对您有所帮助

//********function to check if value is numric or not*********
public static boolean isNumeric(String str) { 
        try {  
            Integer.parseInt(str);  
            return true;
        } 
        catch(NumberFormatException e){  
            return false;  
        }  
    }
//************************function ends******************************


//txtRno is my JTextField 

txtRno.addKeyListener(new KeyListener(){
        public void keyPressed(KeyEvent e)
        {
            //code      
        }
        public void keyReleased(KeyEvent e)
        {
            String value = txtRno.getText();
            int l = value.length();
            if(!isNumeric(value) && l>0){//if not numric it will not allow you to edit JTextField and show error message
                txtRno.setEditable(true);
                JOptionPane.showMessageDialog(c,"You need to enter number","ERROR",JOptionPane.ERROR_MESSAGE);
                txtRno.requestFocus();
                txtRno.setText("");
                txtRno.setEditable(true);
                lblError.setText("");
            } 
            else { //else it will take the input as it already number or integer 
                txtRno.setEditable(true);
                lblError.setText("");
            }
        }
        public void keyTyped(KeyEvent e)
        {
            //code
        }
    });

Answer 2

当文本字段为空值时，它正在获取NumberFormatException。

Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: ""

能否先验证文本，然后解析字符串？

    // Validating text input first
    String r = txtRno.getText();

    if (r.isEmpty())
        {JOptionPane.showMessageDialog(null,"rno should not be empty");
            a=1;}
    else if (Pattern.matches("[a-zA-Z]+",r))
        {JOptionPane.showMessageDialog(null,"rno should be in digits");
            b=1;}
    else if (vrno < 0)
        {JOptionPane.showMessageDialog(null,"rno cannot be negative");
            c=1;}

    // Converting to int, if validation is successful
    int vrno =Integer.parseInt(txtRno.getText());