Java中的异常错误
我创建了一个java程序,可以将十进制转换为二进制,反之亦然。我的十进制到二进制之间没有任何问题。但当我将二进制编码为十进制时,我得到以下错误:Java中的异常错误,java,exception,awt,Java,Exception,Awt,我创建了一个java程序,可以将十进制转换为二进制,反之亦然。我的十进制到二进制之间没有任何问题。但当我将二进制编码为十进制时,我得到以下错误: Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48) at
Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:470)
at java.lang.Integer.parseInt(Integer.java:499)
at converter.actionPerformed(converter.java:42)
at javax.swing.AbstractButton.fireActionPerformed(AbstractButton.java:2028)
at javax.swing.AbstractButton$Handler.actionPerformed(AbstractButton.java:2351)
at javax.swing.DefaultButtonModel.fireActionPerformed(DefaultButtonModel.java:387)
at javax.swing.DefaultButtonModel.setPressed(DefaultButtonModel.java:242)
at javax.swing.plaf.basic.BasicButtonListener.mouseReleased(BasicButtonListener.java:236)
at java.awt.Component.processMouseEvent(Component.java:6382)
at javax.swing.JComponent.processMouseEvent(JComponent.java:3275)
at java.awt.Component.processEvent(Component.java:6147)
at java.awt.Container.processEvent(Container.java:2083)
at java.awt.Component.dispatchEventImpl(Component.java:4744)
at java.awt.Container.dispatchEventImpl(Container.java:2141)
at java.awt.Component.dispatchEvent(Component.java:4572)
at java.awt.LightweightDispatcher.retargetMouseEvent(Container.java:4619)
at java.awt.LightweightDispatcher.processMouseEvent(Container.java:4280)
at java.awt.LightweightDispatcher.dispatchEvent(Container.java:4210)
at java.awt.Container.dispatchEventImpl(Container.java:2127)
at java.awt.Window.dispatchEventImpl(Window.java:2489)
at java.awt.Component.dispatchEvent(Component.java:4572)
at java.awt.EventQueue.dispatchEventImpl(EventQueue.java:704)
at java.awt.EventQueue.access$400(EventQueue.java:82)
at java.awt.EventQueue$2.run(EventQueue.java:663)
at java.awt.EventQueue$2.run(EventQueue.java:661)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:87)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:98)
at java.awt.EventQueue$3.run(EventQueue.java:677)
at java.awt.EventQueue$3.run(EventQueue.java:675)
at java.security.AccessController.doPrivileged(Native Method)
at java.security.AccessControlContext$1.doIntersectionPrivilege(AccessControlContext.java:87)
at java.awt.EventQueue.dispatchEvent(EventQueue.java:674)
at java.awt.EventDispatchThread.pumpOneEventForFilters(EventDispatchThread.java:296)
at java.awt.EventDispatchThread.pumpEventsForFilter(EventDispatchThread.java:211)
at java.awt.EventDispatchThread.pumpEventsForHierarchy(EventDispatchThread.java:201)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:196)
at java.awt.EventDispatchThread.pumpEvents(EventDispatchThread.java:188)
at java.awt.EventDispatchThread.run(EventDispatchThread.java:122)
这是我的密码:
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public class converter extends JFrame implements ActionListener {
JTextField txt1;
JTextField txt2;
JLabel lbl1;
JLabel lbl2;
JButton b1;
JButton b2;
public converter(){
Container c = getContentPane();
JPanel jp = new JPanel();
c.add(jp);
jp.add(lbl1=new JLabel("Decimal: "));
jp.add(txt1=new JTextField(10));
jp.add(lbl2=new JLabel("Binary: "));
jp.add(txt2=new JTextField(10));
jp.add(b1=new JButton("Convert"));
jp.add(b2=new JButton("Clear"));
b1.addActionListener(this);
b2.addActionListener(this);
}
public static void main(String[] args) {
converter cvt = new converter();
cvt.setResizable(false);
cvt.setVisible(true);
cvt.setSize(250,150);
cvt.setTitle("Decimal - Binary Converter");
}
@Override
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
String num = txt1.getText();
int i = Integer.parseInt(num);
if(txt1 != null && e.getSource() == b1){
String z = Integer.toBinaryString(i);
txt2.setText(z);
}
else if(e.getSource() == b2){
txt1.setText("");
txt2.setText("");
}
else if(txt2 != null && e.getSource() == b1){
int x = Integer.parseInt(txt2.getText().trim(), 2);
txt1.setText(""+x);
}
}
}
你能指出哪里不对吗?它的解决方案是什么。检查
txt1
和txt2
值是否为数字。跟踪的第一行显示,当您的程序试图将空字符串(“
)转换为int
)时,出现了一个错误。如果进一步查看跟踪(第5行),则错误发生在actionPerformed
方法中。特别是,这些行:
String num = txt1.getText();
int i = Integer.parseInt(num);
您可以通过以下方法首先检查字符串是否为空来解决此问题:
if (num.length() < 1)
// tell user they must enter a number
if(num.length()<1)
//告诉用户他们必须输入一个数字
您的代码中没有任何边界检查。Aka,您有两个文本输入和一个“转换”功能,但该功能适用于以下所有组合:
- 十进制输入和二进制输入都是给定的
- 十进制输入和二进制输入都被省略
- 给出了十进制输入,省略了二进制输入
- 省略十进制输入,给出二进制输入
Integer.parseInt("")
它会抛出一个NumberFormatException
,正如预期的那样
我将处理以下四种可能的情况:
public static boolean isEmpty(final String str) {
return (str == null || str.trim().equals(""));
}
final String decimalInput = text1.getText();
final String binaryInput = text2.getText();
if(! isEmpty(decimalInput)) {
if( ! isEmpty(binaryInput)) {
// Decimal input and Binary input are both given, show error
} else {
// Decimal input is given, Binary input is omitted, convert to binary
}
} else {
if( isEmpty(binaryInput)) {
// Decimal input and Binary input are both omitted, show error
} else {
// Decimal input is omitted, Binary input is given, convert to decimal
}
}
我突然想到了几件事 您可以捕获异常并显示一条消息,告诉用户他们输入的值无效。您还应该
trim
字段的结果以确保正确
public void actionPerformed(ActionEvent e) {
// TODO Auto-generated method stub
try {
String num = txt1.getText().trim(); // <-- Trim the incoming value
int i = Integer.parseInt(num);
if(txt1 != null && e.getSource() == b1){
String z = Integer.toBinaryString(i);
txt2.setText(z);
}
else if(e.getSource() == b2){
txt1.setText("");
txt2.setText("");
}
// I'm not sure if this a logic error or not, but txt2 is text field...
// Did you think it was the text from the field??
else if(txt2 != null && e.getSource() == b1){
int x = Integer.parseInt(txt2.getText().trim(), 2);
txt1.setText(""+x);
}
} catch (NumberFormatException exp) {
// Display message...
}
}
我尝试了这段代码,但没有出现错误。但我的问题是,当我在txt2字段中输入一个二进制数时,txt1没有显示十进制值。您还存在一些逻辑问题(我的
try catch
有点宽)-检查更新
try {
if (e.getSource() == b1) {
String dec = txt1.getText();
String bin = txt2.getText();
if (dec != null && dec.trim().length() > 0 &&
bin != null && bin.trim().length() > 0) {
// Both fields are filled out?!
} else if (dec != null && dec.trim().length() > 0) {
String value = txt1.getText();
int i = Integer.parseInt(dec);
String z = Integer.toBinaryString(i);
txt2.setText(z);
} else if (bin != null && bin.trim().length() > 0) {
int x = Integer.parseInt(bin, 2);
txt1.setText("" + x);
}
} else if (e.getSource() == b2) {
txt1.setText("");
txt2.setText("");
}
} catch (NumberFormatException exp) {
exp.printStackTrace();
}