Question

此代码应在输入对话框中接收全名字符串示例“Billy Bob Smith”，并在消息对话框中将首字母输出为字母组合示例“BBS”。但由于某种原因，main方法不允许我访问fullName变量。

import javax.swing.*;

public class HardMonogram {
     //---------- ATTRIBUTES ----------//
     private String fullName;
     private String monogram;
     private String first;
     private String middle;
     private String last;


     //---------- METHODS ----------//
     public String getInitial(String seperateName) {
           return seperateName.substring(0, 1);
     }

     public void getSeperateName(String fullName) {
           first  = fullName.substring(0, fullName.indexOf(" "));
           middle = fullName.substring(fullName.indexOf(" ") + 1, fullName.length());
           last   = middle.substring(middle.indexOf(" ") + 1, middle.length());
           middle = middle.substring(0, middle.indexOf(" "));
     }

    public void setMonogram() {
          monogram = getInitial(first)  +
                     getInitial(middle) +
                     getInitial(last);

    JOptionPane.showMessageDialog(null, monogram);
    }

    public static void main(String[] args) {
           myMono.fullName = JOptionPane.showInputDialog(null, "Type in you full name");

           HardMonogram myMono = new HardMonogram();
           myMono.getSeperateName(myMono.fullName);
           myMono.setMonogram();

    }


}

给了我这个构建错误

/Users/aaron/School/Fall 2012/CSCI-C 201/Labs/LB08/HardMonogram.java:33: error: cannot find symbol
    myMono.fullName = JOptionPane.showInputDialog(null, "Type in you full name");
    ^
symbol:   variable myMono
location: class HardMonogram
1 error
[Finished in 1.2s with exit code 1]

这是我的java类介绍，但我不知道为什么我无法访问变量。我显然忽略了一些东西。任何想法？

Answer 1

更新

在再次阅读问题后，您只需要在创建实例后在main方法中移动第一行。

       HardMonogram myMono = new HardMonogram();
       myMono.fullName = JOptionPane.showInputDialog(null, "Type in you full name");
       myMono.getSeperateName(myMono.fullName);
       myMono.setMonogram();

Answer 2

只需在对象声明（myMono.fullName = JOptionPane.showInputDialog(null, "Type in you full name");）之后放置HardMonogram myMono = new HardMonogram();。

Answer 3

MyMono尚未在main方法的第一行声明。将它添加到开头。

public static void main(String[] args) {
      HardMonogram myMono = new HardMonogram();
      myMono.fullName = JOptionPane.showInputDialog(null, "Type in you full name");
      myMono.getSeperateName(myMono.fullName);
      myMono.setMonogram();

}

为什么我不能在main方法中调用私有变量？

3 个答案: