Question

我的主要父组件中具有以下状态：

this.state = {
    playableCards: [],
    openedCard: null,
    offeredChips: 0,
    activePlayer: 0, // first player is 0, second player is 1
    players: [
        {
            name: "player1",
            remainingChips: 11,
            cards: [],
            score: null
        },
        {
            name: "player2",
            remainingChips: 11,
            cards: [],
            score: null
        }
    ]
};

现在，我有一些方法可以更改状态的不同属性。例如：

takeCard = () => {
    const {
        activePlayer,
        players,
        playableCards,
        offeredChips,
        openedCard
    } = this.state;

    if(openedCard) {

        // Add card to active player
        let playersClone = [...players];
        playersClone[activePlayer].cards = [
            ...playersClone[activePlayer].cards,
            openedCard
        ];

        // Add any offered chips to active player
        playersClone[activePlayer].remainingChips += offeredChips;

        this.setState({ players: playersClone }, () =>
            this.calculateScore(activePlayer)
        );

        // Remove card from deck
        this.setState({
            playableCards: playableCards.filter(function(card) {
                return card !== openedCard;
            })
        });

        // Change active player
        const nextPlayer = activePlayer === 0 ? 1 : 0;
        this.setState({ activePlayer: nextPlayer });

        // Reset offered chips to 0
        this.setState({ offeredChips: 0 });

        // Reset opened card
        this.setState({ openedCard: null });

    } else {
        console.log("Open a card first!");
    }

};

如您所见，许多属性仅通过单击事件即可更改（此方法附加到单击事件中）。我想知道这是否是正确的做法，还是应该合并所有setState()？

Answer 1

可以调用多个setState，因为React在setState之前内部进行批处理，因此只会调用render一次。就是说，您在写setState时出错，从而使批处理忽略更改或设置不正确的值的几率很高（例如，您可能基于先前的值针对相同的键调用setState两次，并且可能期望得到与结果不同的结果你得到）。因此，建议您在处理所有值后调用一次setState

    // Add card to active player
    let playersClone = [...players];
    playersClone[activePlayer].cards = [
        ...playersClone[activePlayer].cards,
        openedCard
    ];

    // Add any offered chips to active player
    playersClone[activePlayer].remainingChips += offeredChips;

    const playableCards = playableCards.filter(function(card) {
            return card !== openedCard;
    })


    // Change active player
    const nextPlayer = activePlayer === 0 ? 1 : 0;

    // Reset offered chips to 0
    // Reset opened card
    // Remove card from deck
    this.setState({ 
          openedCard: null,
          offeredChips: 0, 
          playableCards, 
          players: playersClone
    }, () =>
        this.calculateScore(activePlayer)
    );

Answer 2

您可以更改状态的多个属性。

this.setState({ openedCard: null, offeredChips: 0, activePlayer: nextPlayer });

Answer 3

您可以更改状态的多个属性。this.setState({value1: 0, value2: 0})

如何更改状态中的多个属性（同时）？

3 个答案: