我要做的是将@字符放在域名之前。我还试图找出如何将域名作为最后4个字符（例如.com）。我该怎么做呢？任何帮助，将不胜感激。
我已将此工作列于此link。
链接中的代码：
import java.util.Scanner;
public class Username {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Must be between 8 and 20 characters.");
System.out.println("Must contain at least one uppercase and lowercase letter.");
System.out.println("Must contain at least one digit. ");
System.out.println("Must contain a special character ] [ ? / < ~ # ! $ % ^ & * ( ) + = } | : ; , > { ");
System.out.println("Must contain @ before the domain");
System.out.println("The only acceptable domains are .com .edu .org .mil .gov .net");
System.out.println("\\n____Please enter your username to access the page. Follow the rules above.____ ");
String input = keyboard.nextLine();
while ((input.length() < 8) || (input.length() > 20))
{
System.out.println("Error! Your input is not valid.");
System.out.println("Please try again.");
keyboard.nextLine();
}
for (int i = 0; i <= input.length(); i++)
{
if(Character.isUpperCase(input.charAt(i)))
{
break;
}
else
{
if(i == input.length())
{
System.out.println("Error: Try again");
input = keyboard.nextLine();
}
}
}
for (int i = 0; i <= input.length(); i++)
{
if(Character.isLowerCase(input.charAt(i)))
{
break;
}
else
{
if(i == input.length())
{
System.out.println("Try again");
input = keyboard.nextLine();
}
}
}
char [] numbers= {\'0\',\'1\',\'2\',\'3\', \'4\',\'5\',\'6\',\'7\',\'8\',\'9\'};
char[] inputArray = input.toCharArray();
for (int i = 0; i < inputArray.length; i++)
{
for (int j = 0; j < numbers.length; j++)
{
if (inputArray[i]== numbers[j])
{
i=inputArray.length;
j=numbers.length;
}
else
{
if(i == inputArray.length-1 && j== numbers.length-1)
{
System.out.println("Try again");
input = keyboard.nextLine();
}
}
}
char [] SpecialCharacter = {\']\',\'[\',\'?\',\'/\',\'<\',\'~\',\'#\',\'.\',\'!\',\'$\',\'%\',\'^\',\'&\',\'*\',\'(\',\')\',\'+\',\'=\',\'}\',\'|\',\'>\',\'{\' };
char[] inputArray2 = input.toCharArray();
for (int k = 0; k < inputArray2.length; k++)
{
for (int l = 0; l < SpecialCharacter.length; l++)
{
if (inputArray2[k]== SpecialCharacter[l])
{
k=inputArray2.length;
l=SpecialCharacter.length;
}
else
{
if(k == inputArray2.length-1 && l == SpecialCharacter.length-1)
{
System.out.println("No...Try Again");
input = keyboard.nextLine();
}
}
}
String domain1 = ".com";
String domain2 = ".edu";
String domain3 = ".org";
String domain4 = ".mil";
String domain5 = ".gov";
String domain6 = ".net";
}
}
}
}
答案 0 :(得分：0)
我会把它放在这里......
在需要的地方插入代码部分。
import java.util.Scanner;
public class Username {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.println("Must be between 8 and 20 characters.");
System.out.println("Must contain at least one uppercase and lowercase letter.");
System.out.println("Must contain at least one digit. ");
System.out.println("Must contain a special character ] [ ? / < ~ # ! $ % ^ & * ( ) + = } | : ; , > { ");
System.out.println("Must contain @ before the domain");
System.out.println("The only acceptable domains are .com .edu .org .mil .gov .net");
System.out.println("\\n____Please enter your username to access the page. Follow the rules above.____ ");
boolean isOK = false;
String input=new String();
while(!isOK){
input=keyboard.nextLine();
if(input.length() < 20 || input.length() > 8) //length check
{
isOK=true;
}
else
{
isOK=false;
continue;
}
if(input.contains('@')) //check domain
{
isOK=true;
}
else
{
isOK=false;
System.out.println("No @ before domain!");
continue;
}
String[] tokens = input.split("@");
String domain=tokens[1];
String username=tokens[0];
//check if contains digit
...
//check uppercase and lowercase in username
...
//check special character in username
...
//split domain
tokens = domain.split(".");
String domainEnding = tokens[1];
//check if the input domain endings are allowed
...
}
}
}
因此，输入部分错误只需使用isOK=false;
continue;
BTW：你为什么在特殊字符和数字数组中使用\'？
您只能使用'1'，'@'等
答案 1 :(得分：0)
如果您只需要验证用户是否输入了有效的电子邮件，那么您应该使用常规版本，这样可以更加轻松快捷。
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class EmailValidator {
private Pattern pattern;
private Matcher matcher;
private static final String EMAIL_PATTERN =
"^(?=.*[0-9]*[a-z]*[A-Z])(?=.*[a-zA-Z])([a-zA-Z0-9]{8,20}+@[A-Za-z0-9]+.(com|org|edu|mil|net))$";
public EmailValidator() {
pattern = Pattern.compile(EMAIL_PATTERN);
}
/**
* Validate hex with regular expression
*
* @param hex
* hex for validation
* @return true valid hex, false invalid hex
*/
public boolean validate(final String hex) {
matcher = pattern.matcher(hex);
return matcher.matches();
}
}