Using Recursion with Username matching

Question

I am implementing a String matching algorithm for a username database. My method takes an existing Username database and a new username that the person wants and it checks to see if the username is taken. if it is taken the method is supposed to return the username with a number that isn't taken in the database.

Example:

"Justin","Justin1", "Justin2", "Justin3"

Enter "Justin"

return: "Justin4" since Justin and Justin with the numbers 1 thru 3 are already taken.

In my code sample below, newMember returns Justin1 even though it already exists--where is the mistake?

public class UserName {

    static int j = 0;

    static String newMember(String[] existingNames, String newName){
        boolean match = false;

        for(int i = 0; i < existingNames.length; i++){
            if(existingNames[i] == (newName)){
                match = true;
            }

        }
        if(match){
            j++;
            return newMember(existingNames, newName + j);
        }
        else{
            return newName; 
        }
    }

    public static void main(String[] args){

        String[] userNames = new String[9];
        userNames[0] = "Justin1";
        userNames[1] = "Justin2";
        userNames[2] = "Justin3";
        userNames[3] = "Justin";

        System.out.println(newMember(userNames, "Justin"));

        // I don't understand why it returns Justin1 when the name is already taken 
        // in the array.
    }
}

[How do I compare strings in Java?](http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java) — William Morrison, Aug 15 '13 at 19:21
You have done the same old mistake, that has been resolved about a thousands times on SO. — Rohit Jain, Aug 15 '13 at 19:21
@Ruchira its in his code. For future reference, please include question in your post outside of your code, OP. — William Morrison, Aug 15 '13 at 19:22
First thing is first, please indent your code properly so it's easy enough to read. — ThePerson, Aug 15 '13 at 19:24

krsteeve · Accepted Answer · 2013-08-15T19:34:48.370

2

This line:

if(existingNames[i] == (newName)){

Should become

if(existingNames[i].equals(newName)){

In general, always use equals instead of == for Strings in Java.

edited Aug 15 '13 at 19:34

answered Aug 15 '13 at 19:24

krsteeve

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I used Equals but I get a null pointer exception. I don't understand what causes the null pointer exception – user2510809 Aug 15 '13 at 19:33
@user2510809 Your array has length 9 but you're only filling in 4 values. Try changing the length of the array to 4? – krsteeve Aug 15 '13 at 19:36
@user2510809 Or don't use an array at all and use a `List` instead. – arshajii Aug 15 '13 at 19:38
@arshajii yes that would be even better, just trying to fix the immediate problem :) – krsteeve Aug 15 '13 at 19:38
Should probably use .equalsIgnoreCase() – riddle_me_this Apr 06 '15 at 17:22

arshajii · Answer 2 · 2013-08-15T19:41:53.537

1

I would change your approach to this. I think it would be better to use a Map<String, Integer>. For example, the fact that "Justin" - "Justin3" are taken usernames can be represented by the map:

{"Justin": 3}

To check if a username is taken, check if it's a key in the map. To get the "next" username for a specific taken username, get the value corresponding to the name from the map and add 1. Something like this:

static String newMember(Map<String, Integer> existingNames, String newName) {
    if (existingNames.containsKey(newName)) {
        int newNum = existingNames.get(newName) + 1;
        existingNames.put(newName, newNum);
        return newName + newNum;
    }

    existingNames.put(newName, 0);
    return newName;    
}

Oh, and use .equals() instead of == when comparing strings :)

edited Aug 15 '13 at 19:41

answered Aug 15 '13 at 19:24

arshajii

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@boxed__l What do you mean? – arshajii Aug 15 '13 at 19:30
Aren't you splitting `justin4` into `justin(key):4(value)` ? – boxed__l Aug 15 '13 at 19:33
@boxed__l Not really: the map holds the username:max_number pairs, where max_number is the largest number attached to that username so far. – arshajii Aug 15 '13 at 19:35
this is a very interesting approach I will try this. But didn't you mean to put an else statement after the If statement? – user2510809 Aug 15 '13 at 19:42
@user2510809 No, since there is a `return` at the end of the `if`-clause. – arshajii Aug 15 '13 at 20:14
oh so a return ends the function immediately ? – user2510809 Aug 15 '13 at 20:33
@user2510809 Yes, that's right. – arshajii Aug 15 '13 at 20:38

Marco Corona · Answer 3 · 2013-08-15T19:34:52.517

0

You're not comparing the strings correctly, and you're adding the integer 1 to a string. Comparing strings is done with .equals. In order to concatenate the integer onto the end of the string:

static Integer j=0;

return newMember(existingNames, newName + j.toString());

edited Aug 15 '13 at 19:34

answered Aug 15 '13 at 19:21

Marco Corona

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score 0 · Answer 4 · edited Aug 15 '13 at 19:24

0

When comparing strings in Java, you should generally use the equals() methods to do so.

edited Aug 15 '13 at 19:24

tostao

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answered Aug 15 '13 at 19:22

Surveon

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score 0 · Answer 5 · answered Aug 15 '13 at 19:23

0

*Never ever compare equality of Strings with ==. use equals() *

answered Aug 15 '13 at 19:23

Ruchira Gayan Ranaweera

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score 0 · Answer 6 · answered Aug 15 '13 at 19:23

Use .equals to compare strings in Java, not ==.

== will determine if the string references are the same which they almost certainly will not be. (There are rare cases when they may be due to the intern pool.)

.equals will determine if the content of the strings are the same.

score 0 · Answer 7 · answered Aug 15 '13 at 19:26

0

Using == with strings compares the references, not the underlying objects. Use str.equals().

answered Aug 15 '13 at 19:26

blgt

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score 0 · Answer 8 · answered Aug 15 '13 at 19:31

I guess you have a function somewhere in your application answering if a username already exists, I call this usernameExists(String username) returning true or false.

String getOkUsername (String wantedUsername) {
  //if it's free, return it!
  if(!usernameExists(wantedUsername)) {
      return wantedUsername;
  };

  //OK, already taken, check for next available username 
  int i=1;
  while(usernameExists(wantedUsername+Integer.toString(i))) {
      i++;}
  return wantedUsername + Integer.toString(i);
}

score -1 · Answer 9 · answered Aug 15 '13 at 19:23

-1

It is much easier to do it in SQL:

select ... where ... LIKE username%

answered Aug 15 '13 at 19:23

Germann Arlington

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Using Recursion with Username matching

9 Answers9