How to avoid navigating to other screen multiple times - react-native

Question

When press on any button on my React Native App to navigate to a different screen multiple times, then it will redirected to the next screen multiple times.
My sample code is:
// This is my button click event
myMethod()
{
this.props.navigation.navigate("ScreenName")
}
I am using react-navigation to navigate through my app.
How can I fix this behaviour?

Answer 1

I think there are a few ways this could be done. Perhaps recording when the navigation has occurred and preventing it from navigating multiple times.
You may also want to consider resetting hasNavigated after an amount of time etc as well.
// Somewhere outside of the myMethod scope
let hasNavigated = false
// This is my button click event
myMethod()
{
if (!hasNavigated) {
this.props.navigation.navigate("ScreenName")
hasNavigated = true
}
}

Answer 2

This react-navigation issue contains a discussion about this very topic, where two solutions were proposed.
The first, is to use a debouncing function such as Lodash's debounce that would prevent the navigation from happening more than once in a given time.
The second approach, which is the one I used, is to check on a navigation action, whether it is trying to navigate to the same route with the same params, and if so to drop it.
However, the second approach can only be done if you're handling the state of the navigation yourself, for example by using something like Redux.
Also see: Redux integration.

Answer 3

One of solution is custom custom components with adds debounce to onPress:
class DebounceTouchableOpacity extends Component {
constructor(props) {
super(props);
this.debounce = false;
}
_onPress = () => {
if (typeof this.props.onPress !== "function" || this.debounce)
return;
this.debounce = true;
this.props.onPress();
this.timeoutId = setTimeout(() => {
this.debounce = false;
}, 2000);
};
componentWillUnmount() {
this.timeoutId && clearTimeout(this.timeoutId)
}
render() {
const {children, onPress, ...rest} = this.props;
return (
<TouchableOpacity {...rest} onPress={this._onPress}>
{children}
</TouchableOpacity>
);
}
}
another: wrap onPress function into wrapper with similar behavior
const debounceOnPress = (onPress, time) => {
let skipCall = false;
return (...args) => {
if (skipCall) {
return
} else {
skipCall = true;
setTimeout(() => {
skipCall = false;
}, time)
onPress(...args)
}
}
}