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我学习了一些javascript并且遇到了一个绊脚石。我想从我的数组中选择2个随机值并将它们保存到一个名为Deal的变量（我猜这将是另一个数组）中，我可以获得一个随机值但不是2 ..我的代码如下

var deckOfCards = []; //declaring an empty array to put Cards into later

function Card(name, value, altValue) { //function that defines a card object ( key part is "this")
this.cardName = name;
this.cardValue = value;
this.cardAltValue = altValue || false;
}

var suits = ['Hearts', 'Diamonds', 'Spades', 'Clubs'];
var values = ['Ace', 'Two', 'Three', 'Four', 'Five', 'Six', 'Seven', 'Eight', 'Nine', 'Ten', 'Jack', 'Queen', 'King'];

for(var i in suits) //loops through each of the suits
{
 for(var j in values) { // loops through each of the cards for the suit

if(values[j] != 'Ace') {//j is the index within the array,
    var cardValue = parseInt(j, 10) + 1; 
    if(cardValue > 10) cardValue = 10;
    deckOfCards.push(new Card(values[j] +' of ' +suits[i], cardValue));//adding   a new card object to deckOfCards
} else {
    deckOfCards.push(new Card(values[j] +' of ' +suits[i], 1, 11));// adding alternate value to deckOfCards
}

   }

  }


var Deal = deckOfCards[Math.floor(Math.random()*deckOfCards.length)];;
console.log(Deal);

任何人都可以给我一个指针吗？

谢谢

4

5 回答 5

2 
function randomCard(deck, numToExtract) {
    var extracted = [];
    for (var i = 0; i < numToExtract; i++){
        extracted.push(deck.splice(Math.floor(Math.random()*deckOfCards.length, 1));
    }
    return extracted;
}

返回值将是从牌组中提取的牌数组，牌组数组将取出这些牌。您需要跟踪卡片的位置并确保没有卡片“消失”。

于 2013-10-29T15:40:03.797 回答
2

您可能会考虑洗牌整副牌（将所有 52 张牌随机化），然后处理变得简单地从牌堆中拉出最上面的牌。

  • 声明所有卡片的数组，deckOfCards[]，定义为卡片列表，
  • 宣布洗牌，ShuffleDeck[]
  • 从deckOfCards[]中随机挑选所有卡片，复制到ShuffleDeck[]
  • 这笔交易变成了从 ShuffleDeck 中拉出顶牌

一个可能有帮助的观察是，洗牌后的牌组只需要在定义的牌组中有一个索引，因此它只是一组随机的值 0..51（或您喜欢的 1..52）。

//define your cards
deckOfCards[] = ... //left up to you

//define Dealt[], to make checking whether already shuffled, dealt
card = 0 //or 1, however you prefer
while (card<52) {
    Dealt[card] =0; card++;
}


//Shuffle
card = 0 //or 1, however you prefer
while (card<52) {
    //try to pick a card not yet shuffled
    pick = Math.floor(Math.random()*deckOfCards.length);
    while (Dealt[pick]) { //used, pick again
        pick = Math.floor(Math.random()*deckOfCards.length);
    }
    Dealt[card] =1; //used
    ShuffleDeck[card] =pick; card++;
}

//now dealing becomes simply
top=0
//define a function to deal a card
var deal = function() {
    if( !(top<52) ) {
        return -1; //or, handle error here
    }
    result = top;
    top++;
    return result; //work with indices
    //or
    //return deckOfCards[ShuffleDeck[result]];
}
carddealt = deal();

//deal two cards to a Hand,
Hand[player].push( deal() );
Hand[player].push( deal() );

//suppose you want to deal four hands, five cards each
round = 0;
while (round < 5 ) {
    hand = 0;
    while (hand < 4 ) {
        Hand[hand].push( deal() );
    }
}

或者，您可以在发牌时从 shuffled[] 中丢弃每张牌。

于 2013-10-29T15:59:44.573 回答
1

尝试这个：

var Deal = deckOfCards[Math.floor(Math.random()*deckOfCards.length)];
var Deal2 = deckOfCards[Math.floor(Math.random()*deckOfCards.length)];
while (Deal2.equals(Deal)) { // if you want Deal and Deal2 have to different
    Deal2 = deckOfCards[Math.floor(Math.random()*deckOfCards.length)];
}

您还需要实现该equals功能。

function Card(name, value, altValue) {
    this.cardName = name;
    this.cardValue = value;
    this.cardAltValue = altValue || false;

    equals = function(otherObject) {
        return (otherObject.cardName == this.cardName && otherObject.cardValue == this.cardValue && otherObject.cardAltValue == this.cardAltValue)
    }
}
于 2013-10-29T15:29:30.793 回答
1 
var cardsHeld = {};
for( var x = 0; x < 2; x++ ) {
    var newCard = deckOfCards[Math.floor(Math.random()*deckOfCards.length)];
    //If Not found within current CardsHeld Obj, Append.
    if ( !cardsHeld.hasOwnProperty( newCard.cardName )) {
        cardsHeld[newCard.cardName] = newCard;
    }
}

console.log( cardsHeld );
/**
 * Prints:
 *    Object
 *       Five of Spades: Card
 *       Two of Hearts: Card
 */

小提琴：http: //jsfiddle.net/MguBJ/1/

于 2013-10-29T15:53:07.583 回答
1

你有几个选择。您可以第二次对整个牌组调用相同的随机函数，然后确保第二个值与第一个值不同（使用while循环），或者您实际上可以第一次从牌组中取出卡片，使用某些东西像splice，那么你可以一直调用它，直到牌堆空了。

该splice选项可能是更好的选项。

于 2013-10-29T15:30:32.383 回答