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No se pueden obtener valores de la declaración de cambio para insertar en una matriz

Pasé horas tratando de resolver este ejercicio y me quedé estancado en este punto. No puedo obtener valores de mi declaración de cambio para enviar a una matriz para las variables "displayVal" y "val".

 const getDeck = () => { const suits = ['Spades','Hearts','Diamonds','Clubs']; let cards = []; for(let s=0; s < suits.length; s++){ for(let d = 0; d < 13; d++){ let displayVal; let val; switch (d) { case (d === 1): displayVal = "Ace"; val = 11; break; case (d >=2 && d <= 10): displayVal = d; val = d; break; case (d === 11): displayVal = "Jack" val = 10; break; case (d === 12): displayVal = "Queen" val = 10; break; case (d === 13): displayVal = "King" val = 10; } cards.push({ val, displayVal, suit: suits[s] } ); } } return cards; } const deck = getDeck(); console.log(`Deck length equals 52? ${deck.length === 52}`);
about 4 years ago · Juan Pablo Isaza
2 answers
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0

Uso de la rama if-else

En realidad, está utilizando condiciones, así que continúe con if-else ramificado

 const getDeck = () => { const suits = ["Spades", "Hearts", "Diamonds", "Clubs"]; let cards = []; for (let s = 0; s < suits.length; s++) { for (let d = 0; d < 13; d++) { let displayVal; let val; if (d === 1) { displayVal = "Ace"; val = 11; } else if (d >= 2 && d <= 10) { displayVal = d; val = d; } else if (d === 11) { displayVal = "Jack"; val = 10; } else if (d === 12) { displayVal = "Queen"; val = 10; } else if (d === 13) { displayVal = "King"; val = 10; } else { displayVal = ""; val = 0; } cards.push({ val, displayVal, suit: suits[s], }); } } return cards; }; const deck = getDeck(); console.log(deck); console.log(`Deck length equals 52? ${deck.length === 52}`);

Uso de la caja del interruptor

 const getDeck = () => { const suits = ['Spades','Hearts','Diamonds','Clubs']; let cards = []; for(let s=0; s < suits.length; s++){ for(let d = 0; d < 13; d++){ let displayVal; let val; switch (true) { case (d === 1): displayVal = "Ace"; val = 11; break; case (d >=2 && d <= 10): displayVal = d; val = d; break; case (d === 11): displayVal = "Jack" val = 10; break; case (d === 12): displayVal = "Queen" val = 10; break; case (d === 13): displayVal = "King" val = 10; default: displayVal ="" val =0 } cards.push({ val, displayVal, suit: suits[s] } ); } } return cards; } const deck = getDeck(); console.log(deck); console.log(`Deck length equals 52? ${deck.length === 52}`);
about 4 years ago · Juan Pablo Isaza Report

0

Hay una forma / enfoque alternativo que se puede usar aquí.

Fragmento de código

 const getDeck = () => { const suits = ['Spades','Hearts','Diamonds','Clubs']; const getValueOf = x => x === 1 ? 11 : x > 10 ? 10 : x; const displayValMap = {1: 'Ace', 11: 'Jack', 12: 'Queen', 13: 'King'}; return suits.map( suit => ( [ ...Array(13) .keys() ] .map(x => x + 1) .map(d => ({ val: getValueOf(d), displayVal: d in displayValMap ? displayValMap[d] : d, suit })) ) ).flat(); } const deck = getDeck(); console.log(`Deck length equals 52? ${deck.length === 52}`); console.log(deck);

Explicación

  • Configurar formas de obtener val y displayVal (en el fragmento anterior, se usa getValueOf - un método de una línea; y displayValMap un objeto)
  • Iterar a través de los suits
  • Use ... spread-operator para generar una matriz de longitud 13, luego obtenga las claves de la matriz.
  • Use .map para hacer que la matriz generada comience desde 1, en lugar de 0
  • Para cada elemento de la matriz, cree un objeto con val , displayVal y accesorios de suit
  • Y, finalmente, use .flat() para devolver todos los objetos a una sola matriz en lugar de matrices anidadas.

Otra forma puede ser:

 const getDeck = () => ['Spades','Hearts','Diamonds','Clubs'].map( suit => ( [ ...Array(13) .keys() ] .map(d => ({ val: (d === 0 ? 10 : d > 9 ? 9 : d) + 1, displayVal: {0: 'Ace', 10: 'Jack', 11: 'Queen', 12: 'King'}[d] || d + 1, suit })) ) ).flat(); console.log(getDeck().length); console.log(getDeck());

about 4 years ago · Juan Pablo Isaza Report
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