这是我的 ID 程序，其中存储名字、姓氏、出生日期和地点、电子邮件和电话号码。如何制作和存储仅包含有效出生日期、电子邮件和电话号码(而不是具有所有属性)的人员对象？
这是我的主要 ID 程序:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class ID {
static List<Oseba> id = new ArrayList<Oseba>();
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int max = 0;
int choice = 0;
boolean isDate = false;
String regEx_Email = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
String regEx_Date = "(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[012])/((19|20)\\d\\d)";
System.out.println("How many IDs would you like to enter? ");
max = sc.nextInt();
System.out.println(" 0. Exit. ");
System.out.println(" 1. Add contact. ");
System.out.println(" 2. Outprint all contacts. ");
choice = sc.nextInt();
while (choice != 0) {
switch (choice) {
case 0:
System.out.println("Goodbye!");
System.exit(0);
case 1:
while (choice != 2) {
System.out.println("Enter First Name: ");
String firstName = sc.next();
System.out.println("Enter Last Name: ");
String lastName = sc.next();
System.out.println("Enter date of birth (dd-mm-yyyy): ");
String date = sc.next();
isDate = date.matches(regEx_Date);
System.out.println("Enter place of birth: ");
String place = sc.next();
System.out.println("Enter email: ");
String email = sc.next();
Pattern p = Pattern.compile(regEx_Email);
Matcher m = p.matcher(email);
if (m.find()) {
System.out.println(email + " is a valid email address.");
} else {
System.out.println(email + " is a invalid email address");
}
System.out.println("Enter phone number:");
String phone = sc.next();
addID(firstName, lastName, date, place, email, phone);
}
break;
case 2:
System.out.println("\n" + ID.id);
break;
default:
System.out.println("Try again.");
break;
}
System.out.println(" 0. Exit. ");
System.out.println(" 1. Add contact. ");
System.out.println(" 2. Outprint all contacts. ");
choice = sc.nextInt();
}
}
private static void addID(String firstName, String lastName, String date, String place, String email, String phone) {
Person p = new Person(firstName, lastName, date, place, email, phone);
id.add(p);
}
}
还有我的 Person 类:
class Person {
String firstName;
String lastName;
String date;
String place;
String email;
String phone;
public Person(String firstName, String lastName, String date, String place, String email, String phone) {
this.firstName = firstName;
this.lastName = lastName;
this.date = date;
this.place = place;
this.email = email;
this.phone = phone;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public String getPlace() {
return place;
}
public void setPlace(String place) {
this.place = place;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getDate() {
return date;
}
public void setDate(String date) {
this.date = date;
}
public String getPhone() {
return phone;
}
public void setPhone(String phone) {
this.phone = phone;
}
public String toString() {
return "First Name: " + firstName + "\n"
+ "Last Name: " + lastName + "\n"
+ "Date of birth: " + date + "\n"
+ "Place of birth: " + place + "\n"
+ "Email: " + email + "\n"
+ "Phone number: " + phone + "\n\n";
}
}
感谢您的帮助。
最佳答案
最好的方法是将您的问题分解为更小的问题。例如，为了验证输入，您必须创建一个方法来负责验证每个案例的输入，而不是直接调用 setter。尝试在任何地方使用这种方法，因为低耦合和高内聚始终是一个要求!我将提供一个关于您的实现的示例，但是，有多种方法可以做到这一点。另外，我不会使用异常，因为我注意到您仍处于起步阶段。
除此之外，为了使其工作，您应该在 Person 类中添加一个默认构造函数(值是预先定义的)。
最后，尽管不建议使用多个 while 循环的方法，因为它会使代码变得更加复杂，但我使用它是为了向您演示如何确保获得正确的输入，例如，如果用户输入如果没有得到验证，那么程序将继续询问用户，直到获得正确的输入。此外，当用户犯错误时，应提供指示以指导他/她。 (这通常是在有异常(exception)的情况下完成的，在我们的例子中，尽管我们提供了简单的控制台打印)。
让我们看看:
public class ID {
static List<Oseba> id = new ArrayList<Oseba>();
\*we create the menu like that in order to avoid multiple lines repetitions *\
private static String menuOptions = "Menu:" + "\nExit - Insert 0"
+ "\nAdd contact - Insert 1" + "\nExit Outprint all contacts - Insert 2";
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int max = 0;
int choice = 0;
boolean isDate = false;
System.out.println("How many IDs would you like to enter? ");
max = sc.nextInt();
Person p = new Person();
while (true) {
print(menuOptions);
choice = sc.nextInt();
switch (choice) {
case 0:
System.out.println("Goodbye!");
System.exit(0);
case 1:
while(true){
String fname = getString("Enter First Name: ");
if(verifyName(fname)){
p.setFirstName(fname);
break;
}
}
while(true){
String lname = getString("Enter Last Name: ");
if(verifyName(lname)){
p.setLastName(lname);
break;
}
}
while(true){
String date = getString("Enter date of birth (dd-mm-yyyy): ");
if(verifyBirthDate(date)){
p.setDate(date);
break;
}
}
while(true){
String birthPlace = getString("Enter place of birth: ");
if(verifyBirthPlace(birthPlace)){
p.setPlace(birthPlace);
break;
}
}
while(true){
String email = getString("Enter email address: ");
if(verifyEmail(email)){
p.setEmail(email);
break;
}
}
while(true){
String phoneNumber = getString("Enter phone number: ");
if(verifyPhoneNumber(phoneNumber)){
p.setPhone(phoneNumber);
break;
}
}
addID(p);
break;
case 2:
System.out.println("\n" + ID.id);
break;
default:
System.out.println("Try again.");
break;
}
print(menuOptions);
choice = sc.nextInt();
}
}
private static void addID(Person prs) {
id.add(prs);
}
public static Boolean verifyName(String name) {
if(!name.matches("[a-zA-Z]+")){
print("\nERROR_MESSAGE:____________The first/last name should contain only letters, everything else is not valid!");
return false;
}else{
return true;
}
}
public static Boolean verifyBirthDate(String date) {
String regEx_Date = "(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[012])/((19|20)\\d\\d)";
if(!date.matches(regEx_Date)){
print("\nERROR_MESSAGE:____________The birth date is not valid!");
return false;
}else{
return true;
}
}
public static Boolean verifyBirthPlace(String birthPlace) {
if(!birthPlace.matches("[a-zA-Z]+")){
print("\nERROR_MESSAGE:____________The birth place is not valid!");
return false;
}else{
return true;
}
}
public static Boolean verifyEmail(String email) {
String regEx_Email = "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";
Pattern p = Pattern.compile(regEx_Email);
Matcher m = p.matcher(email);
if(!m.find()){
print("\nERROR_MESSAGE:____________"+email+" is an invalid email address");
return false;
}else{
return true;
}
}
public static Boolean verifyPhoneNumber(String phoneNumber) {
if(!phoneNumber.matches("[0-9]+")){
print("\nERROR_MESSAGE:____________The phone No. should contain only numbers, everything else is not valid!");
return false;
}else{
return true;
}
}
public static String getString(String msg) {
Scanner in = new Scanner(System.in);
print(msg);
String s = in.nextLine();
return s;
}
public static void print(String s) {
System.out.println(s);
}
}
关于java - 如何将有效输入添加到数组列表中？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20888601/