如何处理多个按钮的状态?

Chr*_*s O 2 javascript state button reactjs

我有一个引导网格,其中每个网格项都是从一个对象数组中填充的,但是在每个网格项之后我想要一个投票按钮.如何通过单独维护每个按钮上的状态来实现这一点,即当单击按钮1时,文本应从"投票"变为"投票",而其他文本仍为"投票".

点击按钮时,所有这些按钮都变为'Voted'

class Items extends Component {
    constructor(props) {
        super(props);
        this.state = { hasVoted: false };

        this.OnClick = this.OnClick.bind(this);
    }

    OnClick() {
        this.setState(prevState => ({
            hasVoted: !prevState.hasVoted
        }));
    }

    render() {
        const Item = teasers.items.map(item =>
            <Col key={item.nid}>
                <span>
                    {itemType}
                </span>

                <a href={item.path}>
                    <Image src={item.image.src} title={item.productType} />
                    <span>
                        {item.Title}
                    </span>
                    <div className={teasersStyle.copy}>
                        {" "}{item.Copy}>
                    </div>
                </a>

                <div
                    className={this.state.hasVoted ? "active" : "notactive"}
                    onClick={this.OnClick}
                >
                    {this.state.hasVoted ? "Voted" : "Vote"}
                </div>
            </Col>
        );
        return (
            <div>
                <Grid>
                    <Row>
                        {Item}
                    </Row>
                </Grid>
            </div>
        );
    }
}

export default Items;
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And*_*rew 7

我为你创建了一个简单的例子:

class App extends React.Component {
    constructor() {
        super();
        this.onClick = this.onClick.bind(this);
        this.state = {
            arr: [
                { name: "first", isActive: true },
                { name: "second", isActive: true },
                { name: "third", isActive: true },
                { name: "fourth", isActive: true }
            ]
        };
    }
    onClick(index) {
        let tmp = this.state.arr;
        tmp[index].isActive = !tmp[index].isActive;
        this.setState({ arr: tmp });
    }
    render() {
        return (
            <div>
                {this.state.arr.map((el, index) =>
                    <div key={index} onClick={() => this.onClick(index)}>
                        name: {el.name} / isActive: {el.isActive ? "true" : "false"}
                    </div>
                )}
            </div>
        );
    }
}
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检查小提琴并在您的情况下实施它.

  • 谢谢安德鲁,效果很好,但有没有一种简单的方法可以在单击时使所有其他人成为假? (3认同)
  • 非常感谢安德鲁 (2认同)