I have implemented react native tab navigator and added 4 screens to it.
I post some record to api in the second screen and i want to have the updated record in the 4th screens where i am getting updated records..
Useeffect only gets targeted only once, and when i put something in it's argument it gives me strange behavior.
I want useeffect to reload and call the api to get latest items in the 4th screen without putting anything in it's arguement(empty argument)
Any help would be highly appreciated.
Try doing this ;
useEffect(() => {
const unsubscribe = navigation.addListener("focus", () => {
makeApiCall();
});
return unsubscribe;
}, [navigation]);
Get navigation in component's arguments(destructuring)
like below;
const My4thTab = ({ navigation }) => {
}
This way useEffect will trigger only once, every time you come on this screen
but make sure to clear the previous state where you store your data, otherwise, there could be a record duplication.
Hope it helps :)
First-of-all I apologise because I'm new to this concept of redux haha so bare with me here :)
I have implemented a very simple redux in my app
I have one screen here ListAnItem
Here there is a button.
Add size when pressing this they get navigated to another screen called SizeSelector
SizeSelector
here there is a button called dispatch size when they press this button it runs the following.
dispatch({type: "ADD_SIZE", payload: size })
this works perfectly with useSelector.
once this dispatch has been run I use navigation.goBack() to send the user back to the ListAnItem screen.
but once they come to this screen I want to add this useSelector redux state value to a useState hook something like this;
const [sizeState, setSizeState] = React.useState(null)
how can I change this state the minute something has been added to useSelector value of ADD_SIZE.
if that makes sense 😂.
Thanks,
Arnav.
In my react-native project, I have three checkboxes, I need to track the state of those checkboxes so I use an object with key-value (value is boolean) to represent the states of all three checkboxes and use useState hook to manage them. Here is my code:
import React, { useState, useEffect } from 'react';
...
const MyScreen = ({ navigation }) => {
// initially, all checkboxes are checked
const initialCheckBoxState = {
0: true,
1: true,
2: true,
};
const [checkBoxesState, setCheckBoxesState] = useState(initialCheckBoxState);
useEffect(() => {
return () => {
console.log('Screen did unmount');
};
}, [checkBoxesState]);
return (
<View>
...
<SectionList
sections={options}
renderItem={({ index, item }) => (
<CheckBox
onPress={() => {
const checkBoxesStateCopy = { ...checkBoxesState };
checkBoxesStateCopy[index] = !checkBoxesStateCopy[index];
setCheckBoxesState(checkBoxesStateCopy);
}}
/>
)}
/>
...
</View>
);
};
I omitted code that is not the concern of my problem. As you can see, for each item I draw one CheckBox component.
In practice, there are always three items (i.e. three check boxes to show). At the beginning I declared initialCheckBoxState, each key-pair represents the state of the checkbox of each. In the onPress callback of Checkbox I toggle each check box state & update the checkBoxesState by hook method setCheckBoxesState as a whole.
Everything works fine at runtime, my screen is re-rendered when toggling checkbox state, UI shows the status of checkboxes correctly. But issue comes when I navigate back to the previous screen and navigate back to this screen, all checkboxes states are back to the initial states.
So, why the checkboxes states are not reserved?
P.S. previous screen and MyScreen are under the same stack navigator. User press a button of previous screen to navigate to MyScreen. From MyScreen user can go to previous screen by pressing the "headerLeft" button
First lets answer the question:
why the checkboxes states are not reserved?
This component is handling its state completely independent, the state is created & handled inside and no values are passed-in from outside. what does it mean? this component has its initial state value inside of itself, it doesn't use any prop or anything else to initialize the state. everytime this component gets created, state is again initialized with that value. so that's the reason you lose all changes done to checkboxes, because when you leave this screen(component) , it gets unmounted(we'll talk about this in next question) and because all values are just handled inside, every data (containing checkboxes state) will be lost.
So now lets talk about this:
is react-native supposed to reserve the state when come back to the screen?
short answer is No. Every component is destroyed when unmounted including their state and data.
Now lets answer why
screens are still on the stack in memory, not destroyed?
Usually developers use a package like react-navigation or RNRF(which is built on top of react-navigation) for react navigation, most of times we don't care about how they handle this navigation logic, we just use the interface the provided us. each of these packages may have their own way to handle navigation. providing full answer to determine why exactly the screen in still in memory needs full code review and sure lots of debugging but i guess there are 2 possibilities. first as i said maybe the package you are using keeps the unmounted screens in memory at least for a while for some reason. the 2nd is a common react community issue which is Unmounted component still in memory which you can check at: https://github.com/facebook/react/issues/16138
And at last lets answer the question:
how do i keep checkboxes state even with navigating back and losing component containing their state?
This doesn't have just one way to that but simple and short answer is move your state out of the that component, e.g move it out to the parent component or a global variable.
to make it more clear lets explain like this: imagine screen A is always mounted, then you go in B and there you can see some checkboxes and you can modify the states. if the state is handled completely inside B, if you navigate back from screen B to A you lose all changes because B is now unmounted. so what you should do it to put checkboxes states in A screen then pass the values down to B. and when modifying the values, you modify A state. so when B gets unmounted all changes are persistant because you have them in A.
other approached exists as well, you can create a global singleton object named globalState. then put values needed to share between multiple screens there. if you prefer redux or mobx you can use them. one of their usages is when you have some data that you need to share between mutiple screens, these data are independent from where you are at and will persist.
This explanation is from official react-navigation documentation:
Consider a stack navigator with screens A and B. After navigating to
A, its componentDidMount is called. When pushing B, its
componentDidMount is also called, but A remains mounted on the stack
and its componentWillUnmount is therefore not called.
When going back from B to A, componentWillUnmount of B is called, but
componentDidMount of A is not because A remained mounted the whole
time.
https://reactnavigation.org/docs/navigation-lifecycle/#example-scenario
Your MyScreen screen is equivalent to screen B from the example, which means you can expect your screen to stay mounted if you navigate forward, but not backwards.
Its simple, just add a keyExtractor to your SectionList component, which would uniquely identify each checkbox, so that react knows which one to re-render on update.
You'll want to use AsyncStorage to persist data to the device. State variables will be cleared any time the component unmounts.
AsyncStorage docs:
https://react-native-community.github.io/asaync-storage/
import AsyncStorage from '#react-native-community/async-storage';
//You can only store string values so convert objects to strings:
const storeData = async (value) => {
try {
const jsonValue = JSON.stringify(value)
await AsyncStorage.setItem('#storage_Key', jsonValue)
} catch (e) {
// saving error
}
}
const getData = async () => {
try {
const jsonValue = await AsyncStorage.getItem('#storage_Key')
return jsonValue != null ? JSON.parse(jsonValue) : null;
} catch(e) {
// error reading value
}
}
UPDATE -
State is not being persisted due to the nature of React Component lifecycles. Specifically, when you navigate away from a screen the lifecycle method componentWillUnmount is called.
Here's an excerpt from the docs:
componentWillUnmount() is invoked immediately before a component is unmounted and destroyed. Perform any necessary cleanup in this method, such as invalidating timers, canceling network requests, or cleaning up any subscriptions that were created in componentDidMount().
...Once a component instance is unmounted, it will never be mounted again.
This means any values stored in state will be destroyed as well and upon navigating back to the screen ComponentDidMount will be called which is where you may want to assign persisted values back to state.
Two possible approaches aside from AsyncStorage that may work for some use cases to persist data across screens is using Context or a singleton.
I did some changes in my second screen and comeback to my first screen the data is not updated. After reloading app the data is updated. How to update the data from second screen to first screen navigation without refresh or reload
What i believe without any code is that if you do navigation.goBack() or navigation.navigate() it doesnt call the api if its in your componentDidMount, what you can try is adding an eventlistener called onFocus so that whenever screen is focused you call that :
like this in your componentDidMount
componentDidMount(){
this.focusListener = this.props.navigation.addListener('didFocus', () => {
// The screen is focused
// Calling action to reset current day index to 1
this.getItineryData();
});
}
Hope it helps
Example:
Lets say you have a screen with list items loaded into this.state.items.
Now you have a button that navigates you to a new screen.
On this screen you want to append an item to that list.
Once you dismiss that screen, you'll be back at the screen with the original list items.
What is the best way to achieve the goal of updating that list with the new item?
Should this be done by:
1. Passing the new screen the this.state object.
2. By calling an update function on the old screen? (Function probably passed as a prop to the new screen.)
3. Using the StackNavigator's prop in function?
4. Another way?
I appreciate any help.
You should pass a callback that updates the state through the navigate action:
this.props.navigation.navigate('NewScreen', {
addItem: item => this.setState(prevState => ({ items: prevState.items.concat([item]) })),
});
Then call it in the new screen to add an item:
this.props.navigation.state.params.addItem(newItem);
Using a state management library like mobx if your component state is extracted into Stores, the second screen component can update the store and all components that injects this store will be auto updated.