Replacing user portion of email address in java

You have

user.nick@domain.com

and 开发者_开发问答result should be:

******@domain.com

Currently I'm doing it this way:

public static String removeUserFromEmail(String email) {
    StringBuffer sbEmail = new StringBuffer(email);
    int start = sbEmail.indexOf("@");
    sbEmail.delete(0, start);
    return "******" + sbEmail.toString();
}

Is there something simpler or more elegant?

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i would be inclined to run indexOf on email string before putting it in the stringbuffer...

int start = email.indexOf( '@' );

if( start == -1 )
{
   // handle invalid e-mail
}
else
{
   return "*****" + email.substring( start );
}

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Nothing wrong with that solution, although I have two suggestions:

1) Use StringBuilder instead of StringBuffer unless you need to synchronize access between multiple threads. There is a performance penalty associated with StringBuffer that for this application is likely unnecessary.

2) One of the benefits of StringBuilder/Buffer is avoiding excessive string concatenations.

Your return line converts the Buffer to a string, and then concatenates. I would probably do this instead:

int start = email.indexOf("@");

if (start < 0) {
    return "";  // pick your poison for the error condition
}

StringBuilder sbEmail = new StringBuilder(email);
sbEmail.replace(0, start, "******");
return sbEmail.toString();

FYI - my solution is really just some thoughts on your current use of StringBuffer (which are hopefully helpful). I would recommend Konstantin's solution for this simple string exercise. Simple, readable, and it gives you the opportunity to handle the error condition.

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"some.user@domain.com".replaceAll("^[^@]+", "******");

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Looks OK. Better check if indexOf returns -1.

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public static String removeUserFromEmail(String email) {
    String[] pieces = email.split("@");
    return (pieces.length > 1 ? "******" + pieces[1] : email);
}

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You could use a regex, but your solution seems fine to me. Probably faster than the regex too.