How to assign multiple values to a single string in Java with an array?

Question

Currently creating a MUD game as my first serious project in Java, I already have the lore and background set up I just need to start the actual code. I started by first trying to create sets of enemies like this:

//                       Name,     HP,  MP, ATK
enemies[] = {"Weak Skeleton:",    100,   0, 100;
             "Blue Slime:",        50,   0,  25;
             "Novice Rogue Mage:", 25, 100,   0;}

but I obviously can't do that in Java just like that. Is there a way to assign strings a value (kind of like a key-value pair in python, but multiple values assigned to one string) in Java and if so is there a way for me to access and manipulate the integer values assigned to the string?

You probably need to use Map<String, String[]> or Map<String, List<String>> to store that. — ernest_k
– ernest_k, Commented Nov 13, 2018 at 6:00
Create an Enemy class with hp, mp and atk fields. Then create a collection of Enemy objects. No point in doing OO if you're not gonna use objects. — Robby Cornelissen
– Robby Cornelissen, Commented Nov 13, 2018 at 6:01
I would create a class that contains these fields, and then have an Array of this new Class — Scary Wombat
– Scary Wombat, Commented Nov 13, 2018 at 6:01
The "easy" way to do this is with Object[] but you get no type safety that way. — markspace
– markspace, Commented Nov 13, 2018 at 6:01

bvrakvs · Accepted Answer · 2018-11-13 06:18:44Z

2

The most proper to do this is to have an object representing the combination of values you have;

public class Enemy {

    private String name;
    private Integer healthPoint;
    private Integer manaPoint;
    private Integer attackPower;

    // getter, setter, constructors
    // example constructor
    public Enemy(String name, Integer hp, Integer mp, Integer ap) {
        this.name = name;
        this.healthPoint = hp;
        this.manaPoint = mp;
        this.attackPower = ap;
    }
}

Then represent this with a map if you'd like, by doing;

List<Enemy> enemies = new ArrayList<>();
enemies.add(new Enemy("Weak Skeleton", 100, 0, 100));
enemies.add(new Enemy("Blue Slime", 50, 0, 25));
enemies.add(new Enemy("Novice Rogue Mage", 25, 100, 0));
Map<String, Enemy> enemyMap = enemies.stream().collect(Collectors.groupingBy(Enemy::getName));

So if you'd need to access and edit any value, it is easy via;

enemyMap.get("Weak Skeleton").setAttackPower(75);

or to add a new enemy;

Enemy dragonBoss = new Enemy("Dragon Boss", 1000, 500, 250);
enemyMap.put(dragonBoss.getName(), dragonBoss);

Though, I'd suggest using some enumeration to identify each enemy, trying to access with name String is not the best practice.

public enum EnemyType { 

    WEAK_SKELETON("Weak Skeleton"), BLUE_SLIME("Blue Slime"), NVC_ROGUE_MAGE("Novice Rogue Mage"); 

    private String name; 

    public String getName() { 
        return this.name; 
    }

    private EnemyType (String name) {
        this.name = name;
    } 
}

And using it as name/key in map.

You'd be better off with Lombok, to chain setters if you'd want, plus it reduces a lot of boilerplate code (getters, setters, constructors). Check @Data with @Accessors(chain = true)

edited Nov 13, 2018 at 6:18

answered Nov 13, 2018 at 6:07

bvrakvs

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1 Comment

Samuel Quansah Over a year ago

Thanks for the answer. It was a lot more comprehensive than I was expecting it to be. You even linked some resources to make this endeavor easier and take up less code. I'll look into implementing maps and classes(for enemies) in my project.

talex · Accepted Answer · 2018-11-13 06:03:32Z

2

You can use Map in java.

Map<String,List<Integer>> enemies1 = new HashMap<>();
enemies1.put("str1",list1);
enemies1.put("str2",list2);
enemies1.put("str3",list3);

edited Nov 13, 2018 at 6:03

talex

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answered Nov 13, 2018 at 6:01

Tran Ho

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Comments

markspace · Accepted Answer · 2019-07-19 04:28:25Z

2

Two ways: one is just to use an array. To get "loose" semantics in Java just use Object[]. This works but isn't great.

Object[][] enemies1  = {{"Weak Skeleton:", 100, 0, 100},
                        {"Blue Slime:", 50, 0, 25},
                        {"Novice Rogue Mage:", 25, 100, 0}};

A rather better idea as mentioned is to create a more semantic type like Enemy and create new objects of that type.

Enemy[] enemies1 = { new Enemy( "Weak Skeleton:", 100, 0, 100 ),
                     new Enemy( "Blue Slime:", 50, 0, 25 ),
                     new Enemy( "Novice Rogue Mage:", 25, 100, 0 ) };

edited Jul 19, 2019 at 4:28

answered Nov 13, 2018 at 6:05

markspace

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Comments

talex · Accepted Answer · 2018-11-13 06:08:19Z

1

You can create class that represents enemy

public class Enemy {
    private int hp;
    private int mp;
    private int atk;

    public Enemy(int hp, int mp, int atk) {
        this.hp = hp;
        this.mp = mp;
        this.atk = atk;
    }

    public int getHp() {
        return hp;
    }

    public void setHp(int hp) {
        this.hp = hp;
    }

    public int getMp() {
        return mp;
    }

    public void setMp(int mp) {
        this.mp = mp;
    }

    public int getAtk() {
        return atk;
    }

    public void setAtk(int atk) {
        this.atk = atk;
    }
}

and then use it like this:

    HashMap<String, Enemy> map = new HashMap<>();

    map.put("Weak Skeleton", new Enemy(100, 0, 100));
    map.put("Blue Slime", new Enemy(50, 0, 25));

answered Nov 13, 2018 at 6:08

talex

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2 Comments

markspace Over a year ago

I wouldn't split up the name and the stats though. That seems the wrong direction.

talex Over a year ago

@markspace from design point of view I agree with you, but I tried to provide minimal example.

cse · Accepted Answer · 2018-11-13 06:18:27Z

1

You need to change the approach to think about data as you are changing the language.
What you want to achieve, can be achieve in following way:

 class Data{
    //Choose your access-specified as per your need
    public int hp, mp, atk;
    public Data(int h, int m, int a)
    { hp = h; mp=m; atk=a;}
 }
 ...
 HashMap<String,Data> dataSet = new HashMap<>();
 dataSet.put("val1", new Data(50,40,30));
 ...
 Data d = dataSet.get("val1");
 System.out.println("Value of HP is:" + d.hp);
...

answered Nov 13, 2018 at 6:18

cse

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