如何從數組 1 中刪除項目並將項目推送到數組 2

Question

我嘗試的是檢查 state.newCardArray 並根據索引刪除該項目，然后將其推送到 state.arrayFoldedCards。 這是一個紙牌游戲，所以如果 newCardArray 為空並且沒有贏家或輸家，那么游戲就會停止......

const initialState = {
  newCardArray: ['a', 'b', 'c']; // after removing the array looks like ['a', 'c']
  arrayFoldedCards: [] // pushing to the array resulted in an updated array that looks like ['b']
}

export const game = (state = initialState, action) => {
  switch (action.type) {
    case types.GET_CARD:
      {
        const getRandomNumber = Math.floor(state.newCardArray.length * Math.random());
        console.log(state.newCardArray);
        console.log(state.arrayFoldedCards);
        return {
          ...state,
          randomNumber: getRandomNumber,
          arrayFoldedCards: state.newCardArray[getRandomNumber].push(state.arrayFoldedCards.pop())
        }
      }
  }
}

Answer 1

看起來你想要像 splice/concat 這樣的東西

var idxToRemove = 1; //Removing 'b'
var arr0 = ['a', 'b', 'c']; 
var arr1 = []; //Array to move to

return arr1.concat(arr0.splice(idxToRemove, 1));

Splice 返回一個已刪除元素的數組，以便您可以將它們連接在一起。 Concat 合並兩個數組而不發生變異。

Answer 2

 case types.GET_CARD: { const indexToRemove = Math.floor(state.currentCardArray.length * Math.random()); const updatedArray = state.arrayFoldedCards.concat(state.currentCardArray.splice(indexToRemove, 1)); return { ...state, arrayFoldedCards: updatedArray } }