I have a menu in React with an option to Log out. I keep the session state in the Redux store. When I click on the Log out option I want to dispatch an action to destroy the session in the Store and the App will render the content hiding any non-public UI.
This is my render() method in the Menu component:
render() {
const { customer } = this.props
const options = customer.isLoggedIn ? [
{ key: 'p1', name: 'Private Page 1', ref: '/private-page-1' },
{ key: 'p2', name: 'Private Page 2', ref: '/private-page-2' },
{ key: 'l', name: 'Logout', ref: '/' },
] : [
{ key: 'l', name: 'Login', ref: '/' },
]
return (
<div className="App-menu">
<ul>
{
options.map( ( opt ) => {
return (
<li key={ opt.key }>
<Link to={ opt.ref }>{ opt.name }</Link>
</li>
)
} )
}
</ul>
</div>
)
}
Is it a better way to modify the Store state than adding and onClick handler in the Link component for the log out?
What is the best approach to follow uni-directional data flow best practice with React+Redux?
Thanks! :)