不使用正则表达式的 Java 电子邮件验证

Question

我了解使用 Regex 验证电子邮件只需 3-4 行代码。但是，我正在寻找验证电子邮件 without using Regex. 在某种程度上，代码成功通过了几乎所有的验证，但是，仍然无法弄清楚 -如何避免特殊字符成为电子邮件地址的第一个和最后一个字符。

特殊字符列表 = {'!', '#', '$', '%', '^', '&', '*', '(', ')', '-', '/', '~', '[', ']'} ;

我看到的是：

如果用户名部分 (abc.xyz@gmail.com) 以 任何特殊字符，它应该触发 “无效的电子邮件 地址”错误。域部分也是如此。

例如...下面的email-IDs列表应该打印一个“无效的email ID” error message

#abc.xyz@gmail.com

abc.xyz&@gmail.com

abc.xyz&@!gmail.com

abc.xyz&@gmail.com!

import java.util.Scanner;

public class Email_Validation {

    public static void main(String[] args) {

        // User-input code
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter your email address");
        String email = scan.next();

        // Code to check if email ends with '.' (period sign) 
        boolean checkEndDot  = false;
        checkEndDot = email.endsWith(".");

        // Code to find out last index of '@' sign
        int indexOfAt = email.indexOf('@');
        int lastIndexOfAt = email.lastIndexOf('.');


        //Code to check occurence of @ in the email address  
        int countOfAt = 0;

        for (int i = 0; i < email.length(); i++) {
            if(email.charAt(i)=='@')
                countOfAt++; }


        // Code to check occurence of [period sign i..e, "."] after @ 
        String buffering = email.substring(email.indexOf('@')+1, email.length());
        int len = buffering.length();

        int countOfDotAfterAt = 0;
        for (int i=0; i < len; i++) {
            if(buffering.charAt(i)=='.')
                countOfDotAfterAt++; }


// Code to print userName & domainName
            String userName = email.substring(0, email.indexOf('@'));
            String domainName = email.substring(email.indexOf('@')+1, email.length());

                System.out.println("\n");   

               if ((countOfAt==1) && (userName.endsWith(".")==false)  && (countOfDotAfterAt ==1) &&   
                  ((indexOfAt+3) <= (lastIndexOfAt) && !checkEndDot)) {

                   System.out.println("\"Valid email address\"");}

               else {       
                        System.out.println("\n\"Invalid email address\""); }


                System.out.println("\n");
                System.out.println("User name: " +userName+ "\n" + "Domain name: " +domainName);


    }
}

我该如何解决这个问题？

Answer 1

这个怎么样：

public class EmailMe {
  private static Set<Character> bad = new HashSet<>();

  public static void main(String[] args) {
    char[] specialChars = new char[] {'!', '#', '$', '%', '^', '&', '*', '(', ')', '-', '/', '~', '[', ']'} ;
    for (char c : specialChars) {
      bad.add(c);
    }
    check("#abc.xyz@gmail.com");
    check("abc.xyz&@gmail.com");
    check("abc.xyz&@!gmail.com");
    check("abc.xyz&@gmail.com!");
  }

  public static void check(String email) {
    String name = email.substring(0, email.indexOf('@'));
    String domain = email.substring(email.indexOf('@')+1, email.length());
//    String[] split = email.split("@");
    checkAgain(name);
    checkAgain(domain);
  }


  public static void checkAgain(String part) {
    if (bad.contains(part.charAt(0))) System.out.println("bad start:"+part);
    if (bad.contains(part.charAt(part.length()-1))) System.out.println("bad end:"+part);
  }
}

Answer 2

所以，看看 String API。 http://docs.oracle.com/javase/7/docs/api/java/lang/String.html

具体来说，你有 String.length() 和 String.charAt()。因此，您可以非常非常轻松地从 String 中获取第一个和最后一个字符。您已经在代码中执行此操作；假设你已经得到它。

你可以在这里运行一个很长的 if 语句；

char first = email.charAt(0);
if (first == '!' || first == '#' || <more here>) { 
    return false;
}

但这可能会让人头疼。另一种方法是使用 Set，如果您需要多次检查，这会更有效。 （查找 HashSet 通常很快。）例如，您可以创建一次集合，然后就可以通过 Set.contains(first) 多次使用它。

Answer 3

我使用 Apache Commons 库中的 EmailValidator。
上面的例子，

String email = "anEmailAddress@domain.com";
EmailValidator validator = EmailValidator.getInstance();
if (!validator.isValid(email)) {
   // The email is not valid.
}

或

if (!EmailValidator.getInstance().isValid("anEmailAddress@domain.com")) {
   // The email is not valid.
}

如果你使用的是maven那么你可以使用this dependency，

<dependency>
    <groupId>commons-validator</groupId>
    <artifactId>commons-validator</artifactId>
    <version>1.4.0</version>
    <type>jar</type>
</dependency>

Answer 4

检查下面的代码，我相信它满足您的电子邮件验证。

Apache common lang maven dependency.

<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.4</version>
</dependency>

Java 电子邮件验证实用程序代码。

public class EmailValidationUtility {


    private static final String[] SPECIAL_CHARACTERS_FOR_USERNAME = new String[] {
                "..", "__", ",", "/", "<", ">", "?", ";", ":", "\"", "\"",
                "{", "}", "[", "]", "|", "\\", "!", "@", "#", "$", "%", "^",
                "&", "*", "(", ")", "+", "=", "~", "`" };

    private static final char[] SPECIAL_CHARACTERS_WITH_NUMBERS = new char[] {
                '.', ',', '/', '<', '>', '?', ';', ':', '\'', '\"', '{',
                '}', '[', ']', '|', '\\', '!', '@', '#', '$', '%', '^', '&',
                '*', '(', ')', '-', '_', '+', '=', '~', '`', '1', '2', '3',
                '4', '5', '6', '7', '8', '9', '0' };

    /**
     * Method to validate the input email is valid or not.
     *
     * @param email the email
     * @return true, if is email valid
     */
    public static boolean isEmailValid(String email) {
        String[] emailChunks = StringUtils
                        .splitByWholeSeparatorPreserveAllTokens(email,
                                        "@");

        if (emailChunks.length != 2 || isEmailUserNameInvalid(emailChunks[0])
                        || StringUtils.isBlank(emailChunks[1])) {
            return false;
        }

        String[] domainNames = StringUtils
                        .splitByWholeSeparatorPreserveAllTokens(emailChunks[1],
                                        ".");
        if (domainNames.length < 2) {
            return false;
        }

        int topLevelDomainNameIndex = domainNames.length - 1;
        if (isTopLevelDomainNameInvalid(domainNames[topLevelDomainNameIndex])) {
            return false;
        }

        domainNames = ArrayUtils.remove(domainNames, topLevelDomainNameIndex);

        return (isDomainNameValid(domainNames));
    }



    private static boolean isEmailUserNameInvalid(String emailUserName) {
        return (StringUtils.isBlank(emailUserName) || StringUtils.containsAny(
                        emailUserName, SPECIAL_CHARACTERS_FOR_USERNAME));
    }


    private static boolean isTopLevelDomainNameInvalid(String topLevelDomain) {
        return (StringUtils.isBlank(topLevelDomain) || StringUtils.containsAny(
                        topLevelDomain, SPECIAL_CHARACTERS_WITH_NUMBERS));
    }
    
    private static boolean isDomainNameValid(String[] domainNames) {
        for (String domainName : domainNames) {
            if ((StringUtils.isBlank(domainName) || StringUtils.containsAny(
                            domainName, SPECIAL_CHARACTERS_WITH_NUMBERS))) {
                return false;
            }
        }
        return true;
    }
}