如何通过react处理多个按钮的状态？

IT技术 javascript reactjs button state
2021-05-13 06:31:10

我有一个引导网格，其中每个网格项都由一组对象填充，但在每个网格项之后我想要一个投票按钮。我如何通过分别维护每个按钮的状态来实现这一点，即当单击按钮 1 时，文本应从“投票”更改为“投票”，而其他文本保持为“投票”。

在单击按钮的那一刻，所有这些都更改为“已投票”

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;
1个回答

我为您创建了一个简单的示例：

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>
        );
    }
}

检查小提琴并在您的情况下实施它。

处理此问题的另一种方法是将活动按钮的索引保持在状态：

class App extends React.Component {

state = {
    users: [
    { name: "John" },
    { name: "Sarah" },
    { name: "Siri" },
    { name: "Jim" },
    { name: "Simon" },
  ],
  activeIndex: 0,
}

render() {
    const { users, activeIndex } = this.state;

    return (
      <div>
        {users.map((u, i) => (
          <div
            className={i === activeIndex ? 'active' : ''}
            onClick={() => this.setState({ activeIndex: i })}
            key={u.name}
          >
            {u.name}
          </div>
        ))}
      </div>
    )
  }
}

https://jsfiddle.net/846tfe3u/

其它你可能感兴趣的问题
上一篇登录时重定向 - React.js 下一篇超高精度和高精度印刷