Assigning A Point value to images to calculate A score












0















If I have the images of a standard 52 card deck and wanted to create a game in which 4 players randomly select a card from the deck and whoever has the highest value card wins. How would I assign values to the card images? I currently have a random number being assigned to pick the card for each player...



Button p1 = new Button("Player 1");

p1.setOnAction(new EventHandler<ActionEvent>() {

@Override
public void handle(ActionEvent event) {
int n = rand.nextInt(52)+0;
deckView.setImage(cards.get((n)%cards.size()));
event.consume();
p1.setDisable(true);
Collections.shuffle(cards);

}

});


deck view is an object that displays the card above player one. As shown below:
OverView of game



Do I need create 52 card objects and assign them values there? If I do that, do I need to find another way to find a random value using the objects of the cards instead of N? Here are the instructions if I did not explain clearly.



(The winner of the game is the player who won the most rounds. Each round, a player plays one card. The player with the highest value card wins the round. 2 is the lowest value and Ace is the highest value card. If multiples of the same card value is drawn, the suits determine the winner. Spades > Clubs > Diamonds > Hearts")










share|improve this question























  • Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

    – fabian
    Nov 25 '18 at 22:52











  • Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

    – raywrigh
    Nov 25 '18 at 23:06


















0















If I have the images of a standard 52 card deck and wanted to create a game in which 4 players randomly select a card from the deck and whoever has the highest value card wins. How would I assign values to the card images? I currently have a random number being assigned to pick the card for each player...



Button p1 = new Button("Player 1");

p1.setOnAction(new EventHandler<ActionEvent>() {

@Override
public void handle(ActionEvent event) {
int n = rand.nextInt(52)+0;
deckView.setImage(cards.get((n)%cards.size()));
event.consume();
p1.setDisable(true);
Collections.shuffle(cards);

}

});


deck view is an object that displays the card above player one. As shown below:
OverView of game



Do I need create 52 card objects and assign them values there? If I do that, do I need to find another way to find a random value using the objects of the cards instead of N? Here are the instructions if I did not explain clearly.



(The winner of the game is the player who won the most rounds. Each round, a player plays one card. The player with the highest value card wins the round. 2 is the lowest value and Ace is the highest value card. If multiples of the same card value is drawn, the suits determine the winner. Spades > Clubs > Diamonds > Hearts")










share|improve this question























  • Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

    – fabian
    Nov 25 '18 at 22:52











  • Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

    – raywrigh
    Nov 25 '18 at 23:06
















0












0








0


1






If I have the images of a standard 52 card deck and wanted to create a game in which 4 players randomly select a card from the deck and whoever has the highest value card wins. How would I assign values to the card images? I currently have a random number being assigned to pick the card for each player...



Button p1 = new Button("Player 1");

p1.setOnAction(new EventHandler<ActionEvent>() {

@Override
public void handle(ActionEvent event) {
int n = rand.nextInt(52)+0;
deckView.setImage(cards.get((n)%cards.size()));
event.consume();
p1.setDisable(true);
Collections.shuffle(cards);

}

});


deck view is an object that displays the card above player one. As shown below:
OverView of game



Do I need create 52 card objects and assign them values there? If I do that, do I need to find another way to find a random value using the objects of the cards instead of N? Here are the instructions if I did not explain clearly.



(The winner of the game is the player who won the most rounds. Each round, a player plays one card. The player with the highest value card wins the round. 2 is the lowest value and Ace is the highest value card. If multiples of the same card value is drawn, the suits determine the winner. Spades > Clubs > Diamonds > Hearts")










share|improve this question














If I have the images of a standard 52 card deck and wanted to create a game in which 4 players randomly select a card from the deck and whoever has the highest value card wins. How would I assign values to the card images? I currently have a random number being assigned to pick the card for each player...



Button p1 = new Button("Player 1");

p1.setOnAction(new EventHandler<ActionEvent>() {

@Override
public void handle(ActionEvent event) {
int n = rand.nextInt(52)+0;
deckView.setImage(cards.get((n)%cards.size()));
event.consume();
p1.setDisable(true);
Collections.shuffle(cards);

}

});


deck view is an object that displays the card above player one. As shown below:
OverView of game



Do I need create 52 card objects and assign them values there? If I do that, do I need to find another way to find a random value using the objects of the cards instead of N? Here are the instructions if I did not explain clearly.



(The winner of the game is the player who won the most rounds. Each round, a player plays one card. The player with the highest value card wins the round. 2 is the lowest value and Ace is the highest value card. If multiples of the same card value is drawn, the suits determine the winner. Spades > Clubs > Diamonds > Hearts")







javafx






share|improve this question













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share|improve this question










asked Nov 25 '18 at 22:26









raywrighraywrigh

62




62













  • Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

    – fabian
    Nov 25 '18 at 22:52











  • Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

    – raywrigh
    Nov 25 '18 at 23:06





















  • Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

    – fabian
    Nov 25 '18 at 22:52











  • Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

    – raywrigh
    Nov 25 '18 at 23:06



















Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

– fabian
Nov 25 '18 at 22:52





Do the card image names follow a certain pattern, e.g. S10.png for ten of spades and HQ.png for queen of hearts?

– fabian
Nov 25 '18 at 22:52













Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

– raywrigh
Nov 25 '18 at 23:06







Yes each are labeled accordingly 10_of_spades.png or ace_of_hearts.png

– raywrigh
Nov 25 '18 at 23:06














1 Answer
1






active

oldest

votes


















0














Just keep a list of the values / suits somewhere. If you put them into ascending order, you can quite easily create the cards in order of ascending value by using a nested loop for iterating over values and suits respectively, the outer loop being the one iterating the values.



As for drawing cards: You probably don't want repetition, so you should remove a card from the deck after drawing it.



public enum Suit {
HEART("hearts"), SPADE("spades"), CLUB("clubs"), DIAMOND("diamonds");

private final String imageString;

Suit(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}

}




public enum CardValue {
TWO("2"),
THREE("3"),
FOUR("4"),
FIVE("5"),
SIX("6"),
SEVEN("7"),
EIGHT("8"),
NINE("9"),
TEN("10"),
JACK("jack"),
QUEEN("queen"),
KING("king"),
ACE("ace");

private final String imageString;

CardValue(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}
}




public final class Card implements Comparable<Card> {
private final CardValue value;
private final Suit suit;

private final int cardValue;

public Card(Suit suit, CardValue value) {
if (suit == null || value == null) {
throw new IllegalArgumentException();
}
this.suit = suit;
this.value = value;
this.cardValue = value.ordinal() * 4 + suit.ordinal();
}

public CardValue getValue() {
return value;
}

public Suit getSuit() {
return suit;
}

@Override
public int compareTo(Card o) {
return cardValue - o.cardValue;
}

public Image createImage() {
return new Image(value.getImageString() + "_of_" + suit.getImageString() + ".png");
}

@Override
public int hashCode() {
return cardValue;
}

@Override
public boolean equals(Object obj) {
return obj != null
&& (obj instanceof Card)
&& (((Card)obj).cardValue == cardValue);
}

}




private final Random random = new Random();


Create deck:



private final List<Card> cards;

{
cards = new ArrayList<>(suits.length * values.length);
Suit suits = Suit.values();
CardValue values = CardValue.values();
List<Card> cards = new ArrayList<>(suits.length * values.length);

for (CardValue val : values) {
for (Suit suit : suits) {
cards.add(new Card(suit, val));
}
}
}


put all cards on a deck and shuffle



// create copy to avoid having to recreate all the cards when putting all cards back in the deck
List<Card> deck = new ArrayList<>(cards);
Collections.shuffle(deck, random);


Remove "topmost" card from the deck (in this case the last one in the list for efficiency):



Card card = deck.remove(deck.size() - 1);


Instead you could also not use Collections.shuffle on deck and instead remove a random card every time you draw a card:



Card card = deck.remove(random.nextInt(deck.size()));


The Card class provides a method for creating an image for it and comparing the card values.






share|improve this answer
























  • So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

    – raywrigh
    Nov 26 '18 at 0:05











  • You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

    – fabian
    Nov 26 '18 at 0:13











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Just keep a list of the values / suits somewhere. If you put them into ascending order, you can quite easily create the cards in order of ascending value by using a nested loop for iterating over values and suits respectively, the outer loop being the one iterating the values.



As for drawing cards: You probably don't want repetition, so you should remove a card from the deck after drawing it.



public enum Suit {
HEART("hearts"), SPADE("spades"), CLUB("clubs"), DIAMOND("diamonds");

private final String imageString;

Suit(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}

}




public enum CardValue {
TWO("2"),
THREE("3"),
FOUR("4"),
FIVE("5"),
SIX("6"),
SEVEN("7"),
EIGHT("8"),
NINE("9"),
TEN("10"),
JACK("jack"),
QUEEN("queen"),
KING("king"),
ACE("ace");

private final String imageString;

CardValue(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}
}




public final class Card implements Comparable<Card> {
private final CardValue value;
private final Suit suit;

private final int cardValue;

public Card(Suit suit, CardValue value) {
if (suit == null || value == null) {
throw new IllegalArgumentException();
}
this.suit = suit;
this.value = value;
this.cardValue = value.ordinal() * 4 + suit.ordinal();
}

public CardValue getValue() {
return value;
}

public Suit getSuit() {
return suit;
}

@Override
public int compareTo(Card o) {
return cardValue - o.cardValue;
}

public Image createImage() {
return new Image(value.getImageString() + "_of_" + suit.getImageString() + ".png");
}

@Override
public int hashCode() {
return cardValue;
}

@Override
public boolean equals(Object obj) {
return obj != null
&& (obj instanceof Card)
&& (((Card)obj).cardValue == cardValue);
}

}




private final Random random = new Random();


Create deck:



private final List<Card> cards;

{
cards = new ArrayList<>(suits.length * values.length);
Suit suits = Suit.values();
CardValue values = CardValue.values();
List<Card> cards = new ArrayList<>(suits.length * values.length);

for (CardValue val : values) {
for (Suit suit : suits) {
cards.add(new Card(suit, val));
}
}
}


put all cards on a deck and shuffle



// create copy to avoid having to recreate all the cards when putting all cards back in the deck
List<Card> deck = new ArrayList<>(cards);
Collections.shuffle(deck, random);


Remove "topmost" card from the deck (in this case the last one in the list for efficiency):



Card card = deck.remove(deck.size() - 1);


Instead you could also not use Collections.shuffle on deck and instead remove a random card every time you draw a card:



Card card = deck.remove(random.nextInt(deck.size()));


The Card class provides a method for creating an image for it and comparing the card values.






share|improve this answer
























  • So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

    – raywrigh
    Nov 26 '18 at 0:05











  • You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

    – fabian
    Nov 26 '18 at 0:13
















0














Just keep a list of the values / suits somewhere. If you put them into ascending order, you can quite easily create the cards in order of ascending value by using a nested loop for iterating over values and suits respectively, the outer loop being the one iterating the values.



As for drawing cards: You probably don't want repetition, so you should remove a card from the deck after drawing it.



public enum Suit {
HEART("hearts"), SPADE("spades"), CLUB("clubs"), DIAMOND("diamonds");

private final String imageString;

Suit(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}

}




public enum CardValue {
TWO("2"),
THREE("3"),
FOUR("4"),
FIVE("5"),
SIX("6"),
SEVEN("7"),
EIGHT("8"),
NINE("9"),
TEN("10"),
JACK("jack"),
QUEEN("queen"),
KING("king"),
ACE("ace");

private final String imageString;

CardValue(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}
}




public final class Card implements Comparable<Card> {
private final CardValue value;
private final Suit suit;

private final int cardValue;

public Card(Suit suit, CardValue value) {
if (suit == null || value == null) {
throw new IllegalArgumentException();
}
this.suit = suit;
this.value = value;
this.cardValue = value.ordinal() * 4 + suit.ordinal();
}

public CardValue getValue() {
return value;
}

public Suit getSuit() {
return suit;
}

@Override
public int compareTo(Card o) {
return cardValue - o.cardValue;
}

public Image createImage() {
return new Image(value.getImageString() + "_of_" + suit.getImageString() + ".png");
}

@Override
public int hashCode() {
return cardValue;
}

@Override
public boolean equals(Object obj) {
return obj != null
&& (obj instanceof Card)
&& (((Card)obj).cardValue == cardValue);
}

}




private final Random random = new Random();


Create deck:



private final List<Card> cards;

{
cards = new ArrayList<>(suits.length * values.length);
Suit suits = Suit.values();
CardValue values = CardValue.values();
List<Card> cards = new ArrayList<>(suits.length * values.length);

for (CardValue val : values) {
for (Suit suit : suits) {
cards.add(new Card(suit, val));
}
}
}


put all cards on a deck and shuffle



// create copy to avoid having to recreate all the cards when putting all cards back in the deck
List<Card> deck = new ArrayList<>(cards);
Collections.shuffle(deck, random);


Remove "topmost" card from the deck (in this case the last one in the list for efficiency):



Card card = deck.remove(deck.size() - 1);


Instead you could also not use Collections.shuffle on deck and instead remove a random card every time you draw a card:



Card card = deck.remove(random.nextInt(deck.size()));


The Card class provides a method for creating an image for it and comparing the card values.






share|improve this answer
























  • So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

    – raywrigh
    Nov 26 '18 at 0:05











  • You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

    – fabian
    Nov 26 '18 at 0:13














0












0








0







Just keep a list of the values / suits somewhere. If you put them into ascending order, you can quite easily create the cards in order of ascending value by using a nested loop for iterating over values and suits respectively, the outer loop being the one iterating the values.



As for drawing cards: You probably don't want repetition, so you should remove a card from the deck after drawing it.



public enum Suit {
HEART("hearts"), SPADE("spades"), CLUB("clubs"), DIAMOND("diamonds");

private final String imageString;

Suit(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}

}




public enum CardValue {
TWO("2"),
THREE("3"),
FOUR("4"),
FIVE("5"),
SIX("6"),
SEVEN("7"),
EIGHT("8"),
NINE("9"),
TEN("10"),
JACK("jack"),
QUEEN("queen"),
KING("king"),
ACE("ace");

private final String imageString;

CardValue(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}
}




public final class Card implements Comparable<Card> {
private final CardValue value;
private final Suit suit;

private final int cardValue;

public Card(Suit suit, CardValue value) {
if (suit == null || value == null) {
throw new IllegalArgumentException();
}
this.suit = suit;
this.value = value;
this.cardValue = value.ordinal() * 4 + suit.ordinal();
}

public CardValue getValue() {
return value;
}

public Suit getSuit() {
return suit;
}

@Override
public int compareTo(Card o) {
return cardValue - o.cardValue;
}

public Image createImage() {
return new Image(value.getImageString() + "_of_" + suit.getImageString() + ".png");
}

@Override
public int hashCode() {
return cardValue;
}

@Override
public boolean equals(Object obj) {
return obj != null
&& (obj instanceof Card)
&& (((Card)obj).cardValue == cardValue);
}

}




private final Random random = new Random();


Create deck:



private final List<Card> cards;

{
cards = new ArrayList<>(suits.length * values.length);
Suit suits = Suit.values();
CardValue values = CardValue.values();
List<Card> cards = new ArrayList<>(suits.length * values.length);

for (CardValue val : values) {
for (Suit suit : suits) {
cards.add(new Card(suit, val));
}
}
}


put all cards on a deck and shuffle



// create copy to avoid having to recreate all the cards when putting all cards back in the deck
List<Card> deck = new ArrayList<>(cards);
Collections.shuffle(deck, random);


Remove "topmost" card from the deck (in this case the last one in the list for efficiency):



Card card = deck.remove(deck.size() - 1);


Instead you could also not use Collections.shuffle on deck and instead remove a random card every time you draw a card:



Card card = deck.remove(random.nextInt(deck.size()));


The Card class provides a method for creating an image for it and comparing the card values.






share|improve this answer













Just keep a list of the values / suits somewhere. If you put them into ascending order, you can quite easily create the cards in order of ascending value by using a nested loop for iterating over values and suits respectively, the outer loop being the one iterating the values.



As for drawing cards: You probably don't want repetition, so you should remove a card from the deck after drawing it.



public enum Suit {
HEART("hearts"), SPADE("spades"), CLUB("clubs"), DIAMOND("diamonds");

private final String imageString;

Suit(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}

}




public enum CardValue {
TWO("2"),
THREE("3"),
FOUR("4"),
FIVE("5"),
SIX("6"),
SEVEN("7"),
EIGHT("8"),
NINE("9"),
TEN("10"),
JACK("jack"),
QUEEN("queen"),
KING("king"),
ACE("ace");

private final String imageString;

CardValue(String imageString) {
this.imageString = imageString;
}

public String getImageString() {
return imageString;
}
}




public final class Card implements Comparable<Card> {
private final CardValue value;
private final Suit suit;

private final int cardValue;

public Card(Suit suit, CardValue value) {
if (suit == null || value == null) {
throw new IllegalArgumentException();
}
this.suit = suit;
this.value = value;
this.cardValue = value.ordinal() * 4 + suit.ordinal();
}

public CardValue getValue() {
return value;
}

public Suit getSuit() {
return suit;
}

@Override
public int compareTo(Card o) {
return cardValue - o.cardValue;
}

public Image createImage() {
return new Image(value.getImageString() + "_of_" + suit.getImageString() + ".png");
}

@Override
public int hashCode() {
return cardValue;
}

@Override
public boolean equals(Object obj) {
return obj != null
&& (obj instanceof Card)
&& (((Card)obj).cardValue == cardValue);
}

}




private final Random random = new Random();


Create deck:



private final List<Card> cards;

{
cards = new ArrayList<>(suits.length * values.length);
Suit suits = Suit.values();
CardValue values = CardValue.values();
List<Card> cards = new ArrayList<>(suits.length * values.length);

for (CardValue val : values) {
for (Suit suit : suits) {
cards.add(new Card(suit, val));
}
}
}


put all cards on a deck and shuffle



// create copy to avoid having to recreate all the cards when putting all cards back in the deck
List<Card> deck = new ArrayList<>(cards);
Collections.shuffle(deck, random);


Remove "topmost" card from the deck (in this case the last one in the list for efficiency):



Card card = deck.remove(deck.size() - 1);


Instead you could also not use Collections.shuffle on deck and instead remove a random card every time you draw a card:



Card card = deck.remove(random.nextInt(deck.size()));


The Card class provides a method for creating an image for it and comparing the card values.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 25 '18 at 23:55









fabianfabian

53k115373




53k115373













  • So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

    – raywrigh
    Nov 26 '18 at 0:05











  • You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

    – fabian
    Nov 26 '18 at 0:13



















  • So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

    – raywrigh
    Nov 26 '18 at 0:05











  • You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

    – fabian
    Nov 26 '18 at 0:13

















So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

– raywrigh
Nov 26 '18 at 0:05





So if the cards are already in a folder, each having its own file name, instead of creating the cards how would I assign values?

– raywrigh
Nov 26 '18 at 0:05













You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

– fabian
Nov 26 '18 at 0:13





You could of course also take the file names, split them and compare the parts, but this is harder to implement. Furthermore if you include the images as resources in a jar, it's much more problematic to list them. For this reason I recommend creating the image urls based on the card values as shown in the answer.

– fabian
Nov 26 '18 at 0:13




















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