java电话号码验证 [英] java phone number validation

问题描述

这是我的问题：

为电话号码创建一个构造函数，给出xxx-xxx-xxxx或xxx-xxxx格式的字符串作为本地号码。如果格式无效，则抛出异常。

Create a constructor for a telephone number given a string in the form xxx-xxx-xxxx or xxx-xxxx for a local number. Throw an exception if the format is not valid.

所以我想用正则表达式验证它，但我不知道我是否正确执行。我还要扔什么样的例外？我是否需要创建自己的例外？

So I was thinking to validate it using a regular expression, but I don't know if I'm doing it correctly. Also what kind of exception would I have to throw? Do I need to create my own exception?

    public TelephoneNumber(String aString){
        if(isPhoneNumberValid(aString)==true){
            StringTokenizer tokens = new StringTokenizer("-");
            if(tokens.countTokens()==3){
                areaCode = Integer.parseInt(tokens.nextToken());
                exchangeCode = Integer.parseInt(tokens.nextToken());
                number = Integer.parseInt(tokens.nextToken());
            }
            else if(tokens.countTokens()==2){
                exchangeCode = Integer.parseInt(tokens.nextToken());
                number = Integer.parseInt(tokens.nextToken());
            }
            else{
                //throw an excemption here
            }
        }

    }


 public static boolean isPhoneNumberValid(String phoneNumber){
     boolean isValid = false;

     //Initialize reg ex for phone number.
    String expression = "(\\d{3})(\\[-])(\\d{4})$";
    CharSequence inputStr = phoneNumber;
    Pattern pattern = Pattern.compile(expression);
    Matcher matcher = pattern.matcher(inputStr);
    if(matcher.matches()){
        isValid = true;
     }
        return isValid;
    }

对不起，是的，这是作业。对于此分配，唯一有效的格式是xxx-xxx-xxxx和xxx-xxxx，在这种情况下，所有其他格式（xxx）xxx-xxxx或xxxxxxxxxx均无效。

Hi sorry, yes this is homework. For this assignments the only valid format are xxx-xxx-xxxx and xxx-xxxx, all other formats (xxx)xxx-xxxx or xxxxxxxxxx are invalid in this case.

我想知道我的正则表达式是否正确

I would like to know if my regular expression is correct

推荐答案

所以我在想使用正则表达式验证它，但我不知道我是否正确使用它。

它确实看起来过于复杂。此外，匹配 xxx-xxx-xxxx 或 xxx-xxxx 其中 x 是一个数字，可以用（\\\\ {3} - ）{1,2} \\d {4}更好地完成。要了解有关正则表达式的更多信息，我建议您浏览 http://regular-expressions.info 。

It indeed looks overcomplicated. Also, matching xxx-xxx-xxxx or xxx-xxxx where x is a digit can be done better with "(\\d{3}-){1,2}\\d{4}". To learn more about regex I recommend to go through http://regular-expressions.info.

我还要抛出什么样的异常？我是否需要创建自己的例外？

A ValidatorException 似乎很直接。

public static void isPhoneNumberValid(String phoneNumber) throws ValidatorException {
    if (!phoneNumber.matches(regex)) {
        throws ValidatorException("Invalid phone number");
    }
}

如果您不想自己创建一个一些原因，那么我可能会选择 IllegalArgumentException ，但我仍然不建议这样做。

If you don't want to create one yourself for some reasons, then I'd probably pick IllegalArgumentException, but still, I don't recommend that.

那个说，这个验证当然不包括国际和/或外部电话号码。除非这是真正的功课，否则我建议重新考虑验证。

That said, this validation of course doesn't cover international and/or external telephone numbers. Unless this is really homework, I'd suggest to rethink the validation.

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