回调防止setState

Question

我有一个页面，有一个子组件用于浏览按钮，在父组件上，通过回调我用浏览文件设置状态。

export default class Attachment extends React.Component {
  state = {
    attachment: {},
  };

  handleAddAttachment = file => {
    this.setState({ attachment: file });
    this.props.handleChange('attachment', file); // this causes the previous line to not working.
  };

  render() {
    const { attachment } = this.state;

    return (
      <Fragment>
        <div>
          <div>
            Do you have
            <br />
            something to <LineBreak />
            show me?
          </div>
          <div css={attach}>Upload attachments here</div>
          <AttachmentButton  handleAddAttachment={this.handleAddAttachment} />
          <AttachedFile attachment={attachment} />
        </div>
      </Fragment>
    );
  }
}

  makeHandleChange = (pageName, change) => {
    this.setState({
      ticket: { ...this.state.ticket, [pageName]: change },
    });
  };

Answer 1

我不知道这为什么不起作用，但您可能想尝试一个解决方案：将函数作为第二个参数传递给SetState。

请参见这里的示例：react set state callback正确传递参数的方法