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。
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。
<i
onClick={() => this.setState({ showingProjectSelector: !this.state.showingProjectSelector })}
className='fa fa-folder-open'>
</i>
但我想做的是当我点击下拉菜单外部时关闭下拉菜单。我怎样才能在不使用任何库的情况下做到这一点?
这是整个组件:https://jsbin.com/cunakejufa/edit?js,输出 https://jsbin.com/cunakejufa/edit?js,output
你可以尝试利用onBlur
:
<i onClick={...} onBlur={() => this.setState({showingProjectSelector: false})}/>
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