Question: I'm using your react-native-popup-menu for logout button.
when button clicked the authentication deleted and screen will go to login .
but the menu still left.

How to close this menu when screen switched?

```js
<Menu>
  <MenuTrigger>
    <Icon
      name='more-vert'
      color='#fff'
    />
  </MenuTrigger>
  <MenuOptions>
    <MenuOption value={1}>
      <Text onPress={() => {
        this.props.onLogout()
      }}>logout</Text>
    </MenuOption>
  </MenuOptions>
</Menu>

```

*Originally asked by bexoss on react-native-popup-menu GitHub.*

Answer: 
All other answers here will work. Here is another (simpler) solution to the problem depending on what are your requirements.

You are handling logout event in Text component and therefore handlers to close menu are not triggered. Try to pass it to the `onSelect` property of MenuOption:

```js
<MenuOption onSelect={() => this.props.onLogout()}>
  <Text>logout</Text>
</MenuOption>

```

Note: If you returned `false` from your handler, menu would not close.
As mentioned in their documentation, you should use the `visible` prop in your `Menu` component.
You will need to adapt your code like this: 

```
<Menu opened={this.state.opened}>
  <MenuTrigger onPress={() => this.setState({ opened: true })}>
    <Icon
      name='more-vert'
      color='#fff'
    />
  </MenuTrigger>
  <MenuOptions>
    <MenuOption value={1}>
      <Text onPress={() => {
        this.props.onLogout()
        this.setState({ opened: false })
      }}>logout</Text>
    </MenuOption>
  </MenuOptions>
</Menu>

```

And I think you will need to define the default state of your component like this:

```
 constructor(props) {
   super(props)
   this.state = { opened: false }
 }

```

You can also find a full example [here](https://github.com/instea/react-native-popup-menu/blob/master/examples/ControlledExample.js)