Question

好的 - 我知道我非常接近解决方案，但需要在正确的方向上轻推。我需要用户点击YES并且程序再次开始询问问题。

执行后，我收到以下错误

线程“main”中的异常java.lang.StringIndexOutOfBoundsException：字符串索引超出范围：java.lang.String.charAt为1（未知来源）在Vowel3.main（Vowel3.java:49）

// java class for Panel I/O
import javax.swing.JOptionPane;

// declaration of the class
public class Vowel33
{

    // declaration of main program
    public static void main(String[] args)
    {

    // objects used to store data
    String input_string = null;
    int a_count = 0;
    int e_count = 0;
    int i_count = 0;
    int o_count = 0;
    int u_count = 0;
    int i = 0;
    int yes = 0;

    // 1. display a descriptive message
    String display_message = "This program asks the user for a sentence,\n" 
        + "searches the sentence for all vowels,\n"
        + "and displays the number of times each"
        + "vowel appears in the sentence";
    JOptionPane.showMessageDialog(null, display_message, "Lab 3 Description", JOptionPane.INFORMATION_MESSAGE);



    // 4. visit each String posotion

     do{

        // 3. input the character string

        input_string = JOptionPane.showInputDialog("Enter the sentence to search");

        // 5.  if position i of String is a vowel
        // 6.  increase the appropriate vowel counter
        if (input_string.charAt(i) == 'a' || input_string.charAt(i) == 'A')
        a_count++;

        else if (input_string.charAt(i) == 'e' || input_string.charAt(i) == 'E')
        e_count++;

        else if (input_string.charAt(i) == 'i' || input_string.charAt(i) == 'I')
        i_count++;

        else if (input_string.charAt(i) == 'o' || input_string.charAt(i) == 'O')
        o_count++;

        else if (input_string.charAt(i) == 'u' || input_string.charAt(i) == 'U')
        u_count++;

        i++;

        String display_message1 = input_string                                      // 7. display the String
        + "\n\n" + "has " + input_string.length() + " characters.\n\n"          // 8. display the number of characters
        + "There are \n" 
        + a_count + " a's,\n"                                                   // 9. disaply the number of each vowel
        + e_count + " e's,\n"
        + i_count + " i's,\n"
        + o_count + " o's, and\n"
        + u_count + " u's.\n\n";

        JOptionPane.showMessageDialog(null, display_message1, "Lab 3 Description", JOptionPane.INFORMATION_MESSAGE);

        yes = JOptionPane.showConfirmDialog(null, "Would you like to enter another string?\n\n", "Extra Credit", JOptionPane.YES_NO_OPTION);

    }   while (i < input_string.length());

            if (i == input_string.length())
            {
                yes = JOptionPane.showConfirmDialog(null, "Would you like to enter another string?\n\n", "Extra Credit", JOptionPane.YES_NO_OPTION);
                if (yes == 1)
                {
                    input_string = JOptionPane.showInputDialog("Enter the sentence to search");
                }
            }






    } // end of main
} // end of the class

Answer 1

在你的代码中你有一个条件，如果是是0，那么循环仍然可以执行。因为 yes 永远不会从我能看到的东西重新定义，你基本上有一个无限循环，当我退出字符串长度的边界时，你的代码会出错。

也不确定为什么你在循环的底部有两个JOptionPanes（它会执行x次x = input_string.length（））

考虑类似的事情：

if(i == input_string.length()) {
    //ask the user if they want to enter another string
    if(the user selected yes){
        yes = 1;
        //instantiate again with new input_string
    }
}

另一个注意事项：假设您想要显示一条消息，询问用户是否想要在完成第一个条目的迭代后输入另一个字符串，那么是变量和条件似乎没有实际意义

而循环JOptionPane问题

1 个答案: