Java程序有2个错误

时间:2013-11-07 22:26:37

标签: java arrays exception expression

所以,我正在为我的Java类编写这个代码,我遇到了这个问题,这让我感到烦恼,抱歉,如果它有点长,我是Java编码的新手。

import javax.swing.*;
public class DebugEight4
{
   public static void main(String[] args)
   {
      int x = 0, y;
      String array[] = new String[100];
      String entry;
      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[STOP] = entry;
         entry = JOptionPane.showinputDialog(null,
            "Enter another word\n" +
            "Enter " + STOP + " when you want to stop"); 
      }
      for(y = 0; y > 0; ++y)
      {
         message.append(array[y]);
         message.append("\n");
      }
      JOptionPane.showMessageDialog(null, message);
   }
}

我得到了一个 DebugEight4.java:17:错误:')'当你要停止时,“输入”+“停止”“); (箭头指向STOP和“)^

之间的空格

DebugEight4.java:17:错误:非法启动表达式“Enter”+ STOP“当你想要停止”); (箭头指向')')                                     ^

DebugEight4.java:23:错误:')'预期(当你想要停止时,“输入”+停止+“); (箭头指向;)              ^

这是一个问题,我得到的另一个问题是:

DebugEight4.java:20:错误:不兼容的类型数组[STOP] =条目; required:int found:String(aarow指向STOP

DebugEight4.java:21:错误:找不到符号条目= JOptionPane.showinputDialog(null,symbol:method

对于这篇长篇文章感到抱歉,但正如我所说,我是新手,希望对此有所帮助,谢谢大家!

1 个答案:

答案 0 :(得分:2)

在这一行中,您忘记了+

"Enter " + STOP " when you want to stop"); 

将其更改为:

//              v
"Enter " + STOP + " when you want to stop"); 

此外,只有int可以是数组的索引,但STOPString