防止onmouseout触发子元素

Question

鉴于以下内容:

render() {

  const onMouseOver = (event) => {
    this.setState({ isHovering: true });
  };

  const onMouseOut = (event) => {
    this.setState({ isHovering: false });
  };

  const open = this.state.isHovering ? true : false;

  return (
    <Nav className={styles.TopNav} bsStyle="pills" activeKey={1}>
      <NavDropdown
        className={dropDownClasses}
        eventKey={1}
        open={open}
        title="Cards"
        id="nav-dropdown"
        onMouseEnter={onMouseOver}
        onMouseOut={onMouseOut}>
        <MenuItem eventKey="1.1">Action</MenuItem>
        <MenuItem eventKey="1.2">Another action</MenuItem>
      </NavDropdown>
      <NavItem className={styles.navLink} eventKey={2}>Blog</NavItem>
    </Nav>
  );

当鼠标悬停在NavDropdown的子节点上时,如何防止onMouseOut触发.

我可以在onMouseOut中检测我是否超过了孩子吗？

Answer 1

这不是React特有的.mouseover和mouseout事件冒泡,因此处理程序也会触发元素的子元素.mouseenter并mouseleave没有泡沫.

所以你应该听mouseleave.