地址簿中的逻辑问题，不应有空白条目或相同条目...请帮助

Question

希望你能帮助我编码。

我需要做什么： 1.）我不应该允许用户有一个空白条目。我应该提示他们“没有输入名字！” 2.）我不应该允许用户输入相同的条目。

我不知道如何编码，请帮助。

这是我的完整代码：

public class AddressBook {

private AddressBookEntry entry[];
private int counter;
private String SName;
private int notfound=0;

public static void main(String[] args) {
    AddressBook a = new AddressBook();
    a.entry = new AddressBookEntry[100];
    int option = 0;
    while (option != 5) {
        String content = "Choose an Option\n\n"
                + "[1] Add an Entry\n"
                + "[2] Delete an Entry\n"
                + "[3] Update an Entry\n"
                + "[4] View all Entries\n"
                + "[5] View Specific Entry\n"
                + "[6] Exit";
        option = Integer.parseInt(JOptionPane.showInputDialog(content));
        switch (option) {
            case 1:
                a.addEntry();
                break;
            case 2:
                a.deleteEntry();
                break;
            case 3:
                a.editEntry();
                break;
            case 4:
                a.viewAll();
                break;
            case 5:
                a.searchEntry();
                break;
            case 6:
                System.exit(1);
                break;
            default:
                JOptionPane.showMessageDialog(null, "Invalid Choice!");
        }
    }
}

public void addEntry() {
    entry[counter] = new AddressBookEntry();
    entry[counter].setName(JOptionPane.showInputDialog("Enter name: "));
    entry[counter].setAdd(JOptionPane.showInputDialog("Enter add: "));
    entry[counter].setPhoneNo(JOptionPane.showInputDialog("Enter Phone No.: "));
    entry[counter].setEmail(JOptionPane.showInputDialog("Enter E-mail: "));
    counter++;
}

public void viewAll() {
    String addText = "  NAME\tADDRESS\tPHONE NO.\tE-MAIL ADD\n\n";
    int nonNull = 0;
    for (int i = 0; i < entry.length; i++) {
        if (entry[i] != null) {
            addText = addText + entry[i].getInfo() + "\n";
            nonNull++;
        }
        if (nonNull == counter) {
            break;
        }
    }
    JOptionPane.showMessageDialog(null, new JTextArea(addText));
}

public void searchEntry() {
    SName = JOptionPane.showInputDialog("Enter Name to find: ");
    searchMethod();
}

public void searchMethod() {
    for (int i = 0; i < counter; i++) {
        if (entry[i].getName().equals(SName)) {
            JOptionPane.showMessageDialog(null, entry[i].getInfo2());
            notfound = 0;
            break;
        } else {
            notfound++;
        }
    }
    if (notfound != 0) {
        JOptionPane.showMessageDialog(null, "Name Not Found!");
    }
}
public void editEntry() {
    int notfound = 0;
    SName = JOptionPane.showInputDialog("Enter Name to edit: ");
    for (int i = 0; i < counter; i++) {
        if (entry[i].getName().equals(SName)) {
            entry[i] = new AddressBookEntry();
            entry[i].setName(JOptionPane.showInputDialog("Enter new name: "));
            entry[i].setAdd(JOptionPane.showInputDialog("Enter new add: "));
            entry[i].setPhoneNo(JOptionPane.showInputDialog("Enter new Phone No.: "));
            entry[i].setEmail(JOptionPane.showInputDialog("Enter new E-mail: "));
            notfound = 0;
            break;
        } else {
            notfound++;
        }
    }
    if (notfound != 0) {
        JOptionPane.showMessageDialog(null, "Name Not Found!");
    }
}
public void deleteEntry() {
    SName = JOptionPane.showInputDialog("Enter Name to delete: ");
    for (int i = 0; i < counter; i++) {
        if (entry[i].getName().equals(SName)) {
            JOptionPane.showMessageDialog(null, "Found!");
            entry[i] = null;
            break;
        }
    }
}

}

我的代码发生了什么，它允许用户单击“确定”，即使他还没有输入任何字符......并且它还允许相同的输入......

希望你能帮助我。 ..我很快就需要它，但我仍然不知道要在我的代码中添加什么...请帮助，非常感谢..

Answer 1

我的代码发生了什么，它允许用户单击“确定”，即使他还没有输入任何字符......并且它还允许相同的输入......

showInputDialog 方法允许用户输入任何字符串他想要，或者也什么都没有。如果您不想这样做，您可以使用自制的对话框，修补 JOptionPane （使用子类等），或者在它不是给定字符串之一时再次显示对话框，如下所示：

String input = JOptionPane.showInputDialog(content);
try {
   option = Integer.parseInt(input);
}
catch(NumberFormatException ex) {
   // the user typed nothing, or something that is not an integer
   // TODO: show error message
   // => go to next iteration of while loop.
   continue; 
}
// and now our switch ...