在 React 中单击组件外部时更改状态

问题描述

我有一个下拉菜单，如下图所示:

I have a dropdown as is shown in the following image:

当我单击文件夹图标时，它会打开和关闭，因为 showingProjectSelector 属性处于设置为 false 的状态.

When I click the folder icon it opens and closes because showingProjectSelector property in the state that is set to false.

  constructor (props) {
    super(props)
    const { organization, owner, ownerAvatar } = props
    this.state = {
      owner,
      ownerAvatar,
      showingProjectSelector: false
    }
  }

当我单击该图标时，它会正确打开和关闭.

When I click the icon, it opens and closes properly.

<i
  onClick={() => this.setState({ showingProjectSelector: !this.state.showingProjectSelector })}
  className='fa fa-folder-open'>
</i>

但是我想要做的是在我点击下拉菜单时关闭它.我如何在不使用任何库的情况下做到这一点?

But what I'm trying to do is to close the dropdown when I click outside it. How can I do this without using any library?

这是整个组件:https://jsbin.com/cunakejufa/edit?js,输出

推荐答案

你可以尝试利用 onBlur:

<i onClick={...} onBlur={() => this.setState({showingProjectSelector: false})}/>

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