Randomly select from array then assign selection to a variable

Question

I have a function that randomly selects a monster from an object. Here's the function:

var travel = function(direction) {
        var newRoom = rooms[currentRoom.paths[direction]];
        if (!newRoom) {
            $("<p>You can't go that way.</p>").properDisplay();
        }
        else {
            currentRoom = newRoom;
            $("<p>You are now in the " + currentRoom.name + " Room.</p>").properDisplay();
                if (currentRoom.hasMonsters) {
                    function pickRand() {
                        var monsterArray = Object.keys(monsters);   
                        var randomKey = Math.floor(Math.random() * monsterArray.length);
                        return $("<p>Holy Crap!  There's a " + monsterArray[randomKey] + " in here!</p>").properDisplay();  
                    }
                    pickRand();

                }
        }
    };

Here's the object:

var monsters = {
    zombie:  {
        hitPoints: 10,
        loot: "magic knife"
    },
    skeleton: {
        hitPoints: 15,
        loot: "magic shield"
    },
    ghoul:  {
        hitPoints: 12,
        loot: "magic helm"
    }
};

It's set up to randomly select "Zombie", "Skeleton", or "Ghoul." Everything works fine. How do I take whatever was randomly selected and save it to a variable?

I've tried a couple things like:

var beast = pickRand();

and

var beast = monsterArray;

But no luck. What am I missing?

it adds some CSS and prepends some text in the game. It's more cosmetic than anything. — megler
– megler, Commented Nov 28, 2015 at 20:23
Your pickRand function will be returning the return value of properDisplay. Why not just return monsterArray[randomKey] — louisbros
– louisbros, Commented Nov 28, 2015 at 20:26
If I do a return, how do I call it later? Sorry, still trying to learn JS. What I'm trying to do is assign the random value to a variable so I can use it later when you decide to fight the monster. I want to be able to call whichever monster was randomly generated by name. (zombie, ghoul, etc). — megler
– megler, Commented Nov 28, 2015 at 20:33

louisbros · Accepted Answer · 2015-11-28 21:02:46Z

1

It looks like the problem is that you're returning the return value of the properDisplay function.

If you used an array of monsters rather than a map you could store all of the monster information when you select a random one:

var monsters = [
    {
        name: 'zombie',
        hitPoints: 10,
        loot: "magic knife"
    },
    {
        name: 'skeleton',
        hitPoints: 15,
        loot: "magic shield"
    },
    {
        name: 'ghoul',
        hitPoints: 12,
        loot: "magic helm"
    }
];

function pickRand(arr) {
    var index = Math.floor(Math.random() * arr.length);
    return arr[index];
}

var monster = pickRand(monsters);

Now that you have your monster you could display it:

$("<p>Holy Crap!  There's a " + monster.name + " in here!</p>").properDisplay();

Demo: https://jsfiddle.net/louisbros/3cu9spfd/

edited Nov 28, 2015 at 21:02

answered Nov 28, 2015 at 20:53

louisbros

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6 Comments

megler Over a year ago

Here is the fiddle I have @louisbros. jsfiddle.net/eut1h5m7 When I call the variable, it says [object object]. The only thing I can think is I placed the variable incorrectly? The monster is called when you "go east" at the beginning of the game.

megler Over a year ago

my bad. I missed the display of monster.name. Thanks @louisbros. All set!

megler Over a year ago

Is there any way to have the variable remember the same randomly selected monster for a period of time? ie. it randomly selects Zombie - how do I have it remember Zombie until I want it to randomly generate another monster? Does that make sense? I get that everytime the function is called, it'll randomly generate again. I need it to temporarily remember it's selection.

louisbros Over a year ago

Yes you can keep the value each time by giving the monster variable a greater scope. For example, declare the variable outside the function and set it inside the function.

megler Over a year ago

OK, I understand declaring outside the function. You do that in your example. I'm vague on "set it inside the function" I did this:

function pickRand(arr) { 			var index = Math.floor(Math.random() * monsters.length); 			return arr[index]; 			var monster = pickRand(monsters); 		}; 		 		var monster = pickRand(monsters);

But no luck. I'm sure it's the inside the function part I'm messing up.

|

Tatiana Tylosky · Accepted Answer · 2015-11-28 20:14:29Z

0

var beast = pickRand(); should totally work. You just need to assign the variable right when you call it?

var travel = function(direction) {
        var newRoom = rooms[currentRoom.paths[direction]];
        if (!newRoom) {
            $("<p>You can't go that way.</p>").properDisplay();
        }
        else {
            currentRoom = newRoom;
            $("<p>You are now in the " + currentRoom.name + " Room.</p>").properDisplay();
                if (currentRoom.hasMonsters) {
                    function pickRand() {
                        var monsterArray = Object.keys(monsters);   
                        var randomKey = Math.floor(Math.random() * monsterArray.length);
                        return $("<p>Holy Crap!  There's a " + monsterArray[randomKey] + " in here!</p>").properDisplay();  
                    }
                    var beast = pickRand();
                }
        }
    }

answered Nov 28, 2015 at 20:14

Tatiana Tylosky

112 bronze badges

2 Comments

megler Over a year ago

Here's why I thought it didn't work. I used that and just to see if it would call correctly, I added alert(beast); and it outputs objectObject. I'm guessing that means it's calling the object incorrectly...

Tatiana Tylosky Over a year ago

maybe try alert(beast[0])? You may need to look at the first item in that object

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