线程“main”中的异常 java.lang.NumberFormatException: null
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Exception in thread "main" java.lang.NumberFormatException: null
提问by RequiemDream
I have tried everything I possibly could with this, but I still could not find any solution to this, and here is the error, I tried to change the null in line 16, etc but to no avail, and I'm a new Java programmer so I do not have much experience using this language, all help appreciated!
我已经尝试了所有可能的方法，但我仍然找不到任何解决方案，这是错误，我尝试更改第 16 行中的 null 等但无济于事，我是一个新的 Java程序员所以我没有太多使用这种语言的经验，感谢所有帮助！
Exception in thread "main" java.lang.NumberFormatException: null at java.lang.Integer.parseInt(Integer.java:454) at java.lang.Integer.parseInt(Integer.java:527) at DebugEight4.main(DebugEight4.java:11)
线程“main”中的异常 java.lang.NumberFormatException: null at java.lang.Integer.parseInt(Integer.java:454) at java.lang.Integer.parseInt(Integer.java:527) at DebugEight4.main(DebugEight4.爪哇：11）
import javax.swing.*;
public class DebugEight4
{
public static void main(String[] args)
{
int x = 0, y;
String array[] = new String[100];
String entry;
int i = Integer.parseInt(array[1]);
final String STOP = "XXX";
StringBuffer message = new
StringBuffer("The words in reverse order are\n");
entry = JOptionPane.showInputDialog(null,
"Enter any word\n" +
"Enter " + STOP + " when you want to stop");
while(!(entry.equals(STOP)))
{
array[x] = entry;
x++;
entry = JOptionPane.showInputDialog(null,
"Enter another word\n" +
"Enter " + x + " when you want to stop");
}
for(y = 0; y > 0; ++y)
{
message.append(array[y]);
message.append("\n");
}
JOptionPane.showMessageDialog(null, message);
}
}
回答by MouseLearnJava
The error is caused by the following code:
该错误是由以下代码引起的：
int i = Integer.parseInt(array[1]);
In your code, you just declare the String array as below. But all of the element is null in array.
在您的代码中，您只需如下声明 String 数组。但是array 中的所有元素都是空的。
String array[] = new String[100];
Please initizlize them first.
请先初始化它们。
Actually, variable i is not used in the context, so just remove the code "int i = Integer.parseInt(array[1]);".
实际上，变量 i 没有在上下文中使用，因此只需删除代码“int i = Integer.parseInt(array[1]);”。
And I do not think the following code for for-loop is correct. The only way to stop it is that the index exceed the range of array.
而且我认为 for-loop 的以下代码不正确。阻止它的唯一方法是索引超出数组的范围。
for(y = 0; y > 0; ++y)
{
message.append(array[y]);
message.append("\n");
}
回答by Preet Sangha
There is nothing in the array except empty strings.
除了空字符串外，数组中没有任何内容。
String array[] = new String[100];
//...
int i = Integer.parseInt(array[1]);
an empty string cannot be parsed into an Integer.
空字符串不能解析为整数。
For example:
例如：
String array[] = new String[100]; <<-- Create an array of empty string
array[1] = "1"; <<-- set second item in array to a parseable value
int i = Integer.parseInt(array[1]); <<-- Parse the value out
