Question

这是代码，当我输入一个字母时，它不显示行.matches而是我得到了一个错误，但是如果我删除了parseFloat rec1并删除了else if（rec1＆lt; total）那么.matches行将请告诉我如何做到这一点，谢谢你提前

CashType c = new CashType（）; c.setVisible（真）;

    c.jButton1.addActionListener(new java.awt.event.ActionListener() {

        public void actionPerformed(java.awt.event.ActionEvent evt) {
         String receive = c.jTextField1.getText();


         float total = total("sellno"+SellNoCount.getText());
         float rec1 = Float.parseFloat(receive); //this is line 1525

            if(!receive.matches("[0-9]+")){
                JOptionPane.showMessageDialog(null,"Enter a Valid Amount");
                c.jTextField1.setText("");

            }

            else if(receive.equalsIgnoreCase("")){
                JOptionPane.showMessageDialog(null,"Enter Amount");
            }
            else if(rec1 < total){

              JOptionPane.showMessageDialog(null,"Insufficient Amount");

            }

//错误

Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "asdasd"
at sun.misc.FloatingDecimal.readJavaFormatString(FloatingDecimal.java:2043)
at sun.misc.FloatingDecimal.parseFloat(FloatingDecimal.java:122)
at java.lang.Float.parseFloat(Float.java:451)
at projectfinal.SellPage$32.actionPerformed(SellPage.java:1525)

Answer 1

从错误消息中，看起来您正在向jTextField1输入“asdasd”。您尝试解析为浮动的值。如果String不是数字，Float.parseFloat(string)将抛出NumberFormatException。在parseFloat()方法中，字符串参数将转换为原始浮点值。

您可以检查输入的值是否为数字，然后将其解析为浮动。

float rec1 = 0;

   if(isNumeric(receive)){
       rec1 = Float.parseFloat(receive);
       if(rec1 < total){

         JOptionPane.showMessageDialog(null,"Insufficient Amount");

       }
   }else {
          JOptionPane.showMessageDialog(null,"Enter a Valid Amount");
            c.jTextField1.setText("");
         }

方法是

 public static boolean isNumeric(String s) {  
    return s.matches("[-+]?\\d*\\.?\\d+");  
}

java.lang.NumberFormatException对于Input String，value是字母

1 个答案: