Java JOptionPane.带构造函数的showInputDialog

Question

第一个帖子，我希望我没有显得很无聊。

我在一个Java类，我卡住了一个问题。

要求创建一个类(Contact)，该类具有name、email和phoneNumber的getter和构造函数。然后是一个测试类(TestContact)，它有一个while loop，它不断提示用户，因为点击了OK而没有输入任何内容，点击Enter或名称超过21个字符。

同样，我需要的三个变量(姓名、电子邮件和电话号码)将被输入到同一个输入框中(通过空格进行解析)。

我似乎不知道如何使它工作。我有很多虫子。

首先，我不确定如何设置数组，然后用空格分割它，然后使用该数组设置变量&getter(希望这能理解?)。

此外，由于数组中的NullPointerException和数组索引越界异常，程序不断崩溃。

联系类:

public class Contact 
{
    //Initiating variables
    private String name;
    private String phoneNumber;
    private String eMail;
    //Constructor
    public Contact()
    {
        this.name = getName();
        this.phoneNumber = getPhoneNumber();
        this.eMail = getEMail();
    }
    //Getter for name variable
    public String getName()
    {
        return name;
    }
    //Getter for phoneNumber variable
    public String getPhoneNumber()
    {
        return phoneNumber;
    }
    //Getter for eMail variable
    public String getEMail()
    {
        return eMail;
    }
}

TestContact类:

public class testContact 
{
    public static void main(String[] args) 
    {
        Contact myContact = new Contact();
        String userInput;
        String noUserInput;
        userInput = JOptionPane.showInputDialog("Please enter your First and Last Name, Phone Number, & E-mail: ");
        do  
        {   
            String[] phrases = userInput.split(" ");
            String name = phrases[0] + " " + phrases[1];
            String phoneNumber = phrases[2];
            String eMail = phrases[3];
            if (!userInput.equals(""))
                {   
                    if (name.length() > 21)
                    {
                        String userInput = JOptionPane.showInputDialog("I'm sorry but your name is too long.nPlease enter your First and Last Name, Phone Number, & E-mail: ");
                        String[] phrases = userInput.split(" ");
                        String name = phrases[0] + " " + phrases[1];
                        String phoneNumber = phrases[2];
                        String eMail = phrases[3];
                        JOptionPane.showMessageDialog(null, "Name: "+myContact.getName()+"nPhone Number: "+myContact.getPhoneNumber()+"nE-Mail: "+myContact.getEMail());
                    }
                    else
                    {
                        JOptionPane.showMessageDialog(null, "Name: "+name+"nPhone Number: "+phoneNumber+"nE-Mail: "+eMail);
                    }
                }
            while ((userInput = JOptionPane.showInputDialog("I'm sorry but you didn't enter anything.nPlease enter your First and Last Name, Phone Number, & E-mail: ")) == null)
            {   
                String[] phrases = userInput.split(" ");
                String name = phrases[0] + " " + phrases[1];
                String phone = phrases[2];
                String eMail = phrases[3];
                JOptionPane.showMessageDialog(null, "Name: "+myContact.getName()+"nPhone Number: "+myContact.getPhoneNumber()+"nE-Mail: "+myContact.getEMail());
            }
        }while(userInput != null);
    }
}

注意我改变了TestContact类的周围，使它看起来更漂亮一些，如下图所示。我唯一的问题是如何设置我从字符串数组解析的方法，并放入字符串变量。我如何为构造函数设置这些??

public class testContact 
{
    static String userInput;
    static Contact myContact = new Contact();
    public static void main(String[] args) 
    {
        do  
        {   
            parsing(initialInput());
            if (!userInput.equals(""))
                {   
                    if (myContact.getName().length() > 21)
                    {
                        parsing(nameLengthErrorInput());
                        output();
                    }
                    else
                    {
                        output();
                    }
                }
            else
            {
                parsing(nullErrorInput());
                output();
            }
        }while(userInput != null);
    }
    public static String initialInput()
    {
        userInput = JOptionPane.showInputDialog("Please enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static String nameLengthErrorInput()
    {
        userInput = JOptionPane.showInputDialog("I'm sorry but your name is too long.nPlease enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static String nullErrorInput()
    {
        userInput = JOptionPane.showInputDialog("I'm sorry but you didn't enter anything.nPlease enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static void output()
    {
        JOptionPane.showMessageDialog(null, "Name: "+myContact.getName()+"nPhone Number: "+myContact.getPhoneNumber()+"nE-Mail: "+myContact.getEMail());
    }
    public static void parsing(String userInput)
    {
        String[] phrases = userInput.split(" ");
        String name = phrases[0] + " " + phrases[1];
        String phoneNumber = phrases[2];
        String eMail = phrases[3];
    }
}

我的问题现在是唯一的解析()方法。

Answer 1

有两个主要的错误…

第一…

在Contact构造函数中，您将变量赋值给它们自己…

public Contact() {
    this.name = getName();
    this.phoneNumber = getPhoneNumber();
    this.eMail = getEMail();
}

这几乎等同于说…

public Contact() {
    this.name = this.name;
    this.phoneNumber = this.phoneNumber;
    this.eMail = this.eMail;
}

结果是一样的…

第二…

你在这里得到一个NullPointerException…

if (myContact.getName().length() > 21) {

因为Contact#getName的值从未被赋值(一个有效的)

建议…

第一…

我建议改变Contact构造函数，要求值传递给它…

public Contact(String name, String phoneNumber, String eMail) {
    this.name = name;
    this.phoneNumber = phoneNumber;
    this.eMail = eMail;
}

这将意味着…

static Contact myContact = new Contact();

将不再编译，但您可以将其更改为

static Contact myContact;

相反……

第二…

我建议更改您的parsing方法以返回Contact，例如…

public static Contact parsing(String userInput) {
    Contact contact = null;
    if (userInput != null && userInput.trim().length() > 0) {
        String[] phrases = userInput.split(" ");
        if (phrases.length == 4) {
            String name = phrases[0] + " " + phrases[1];
            String phoneNumber = phrases[2];
            String eMail = phrases[3];
            contact = new Contact(name, phoneNumber, eMail);
        }
    }
    return contact;
}

您还应该防止无效输入。

这意味着每次使用解析方法时都需要分配结果…

myContact = parsing(initialInput());

第三…

每次用户输入失败时，你应该简单地显示一个错误信息，让他们输入新的信息，你在尝试这样做，但是你没有利用现有的错误检查来重新验证输入…

public static void main(String[] args) {
    String errorMsg = "";
    do {
        myContact = parsing(getInput(errorMsg));
        if (myContact != null) {
            if (myContact.getName().length() > 21) {
                myContact = null;
                errorMsg = "<html>I'm sorry but your name is too long.<br>";
            }
        } else {
            errorMsg = "<html>I'm sorry but you didn't enter anything.<br>";
        }
    } while (myContact == null);
    output();
}
public static String getInput(String errorMsg) {
    userInput = JOptionPane.showInputDialog(errorMsg + "Please enter your First and Last Name, Phone Number, & E-mail: ");
    return userInput;
}

Answer 2

我刚刚提交了我的实验结果这是我的最终解决方案

联系类:

public class Contact 
{
    private String name;
    private String phoneNumber;
    private String eMail;
    public Contact(String name, String phoneNumber, String eMail) 
    {
        this.name = name;
        this.phoneNumber = phoneNumber;
        this.eMail = eMail;
    }
    public String getName()
    {
        return name;
    }
    public String getPhoneNumber()
    {
        return phoneNumber;
    }
    public String getEMail()
    {
        return eMail;
    }
}

TestContact类:

import javax.swing.JOptionPane;
public class testContact 
{
    static Contact myContact;
    static Boolean exitBool = true;
    public static void main(String[] args) 
    {       
        myContact = parsing(initialInput());
        do  
        {               
            if (myContact != null)
                {   
                    if (myContact.getName().length() > 21)
                    {
                        myContact = parsing(nameLengthErrorInput());
                        exitBool = false;
                    }
                    else
                    {
                        exitBool = true;
                    }
                }
            else if (myContact == null)
            {
                myContact = parsing(nullErrorInput());
                exitBool = false;
                //output();
            }
        }while(exitBool == false);
        output();
    }    
    public static String initialInput()
    {
        userInput = JOptionPane.showInputDialog("Please enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static String nameLengthErrorInput()
    {
        userInput = JOptionPane.showInputDialog("I'm sorry but your name is too long.nPlease enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static String nullErrorInput()
    {
        userInput = JOptionPane.showInputDialog("I'm sorry but you didn't enter anything.nPlease enter your First and Last Name, Phone Number, & E-mail: ");
        return userInput;
    }
    public static void output()
    {
        JOptionPane.showMessageDialog(null, "Name: "+myContact.getName()+"nPhone Number: "+myContact.getPhoneNumber()+"nE-Mail: "+myContact.getEMail());
    }
    public static Contact parsing(String userInput) {
        Contact contact = null;
        if (userInput != null && userInput.trim().length() > 0) {
            String[] phrases = userInput.split(" ");
                String name = phrases[0] + " " + phrases[1];
                String phoneNumber = phrases[2];
                String eMail = phrases[3];
                contact = new Contact(name, phoneNumber, eMail);
        }
        return contact;
    }
}