Call a function after state changes - react-native

Question

I'm building a React Native app and when one button is pressed I want to call two functions. The first one will make a get call and set the state loading: true, the second one will show a popup with the result of that get call.
I am calling the second function only if loading === false but it is executed immediately after the first one before the state can change, because loading is false by default. I can resolve this with setTimeout but I was wondering if there was a cleaner way to do this.
onPress() {
this.props.getUsers();
setTimeout(() => {
if (this.props.loading === false) {
this.props.popUpVisible();
}
}, 1000);
}

Answer 1

You can create callback function for that
getUsers = (callback) => {
//do whatever you want
//when it's done
callback();
}
In onPress function
onPress = () => {
this.props.getUsers(() => {
if (this.props.loading === false) {
this.props.popUpVisible();
}
});
}

Answer 2

setState Function can take two param:
setState(updater, callback)
setState({loading:true},() => {
//this fires once state.loading === true
})

Answer 3

Use getDerivedStateFromProps. It always fire when component's props change.
Below is the example.
class EmailInput extends Component {
state = {
email: this.props.defaultEmail,
prevPropsUserID: this.props.userID
};
static getDerivedStateFromProps(props, state) {
// Any time the current user changes,
// Reset any parts of state that are tied to that user.
// In this simple example, that's just the email.
if (props.userID !== state.prevPropsUserID) {
return {
prevPropsUserID: props.userID,
email: props.defaultEmail
};
}
return null;
}
// ...
}