How to call loading function with React useEffect only once

JavascriptReactjsReact Hooks

Javascript Problem Overview


The useEffect React hook will run the passed in function on every change. This can be optimized to let it call only when the desired properties change.

What if I want to call an initialization function from componentDidMount and not call it again on changes? Let's say I want to load an entity, but the loading function doesn't need any data from the component. How can we make this using the useEffect hook?

class MyComponent extends React.PureComponent {
    componentDidMount() {
        loadDataOnlyOnce();
    }
    render() { ... }
}

With hooks this could look like this:

function MyComponent() {
    useEffect(() => {
        loadDataOnlyOnce(); // this will fire on every change :(
    }, [...???]);
    return (...);
}

Javascript Solutions


Solution 1 - Javascript

If you only want to run the function given to useEffect after the initial render, you can give it an empty array as second argument.

function MyComponent() {
  useEffect(() => {
    loadDataOnlyOnce();
  }, []);

  return <div> {/* ... */} </div>;
}

Solution 2 - Javascript

TL;DR

useEffect(yourCallback, []) - will trigger the callback only after the first render.

Detailed explanation

useEffect runs by default after every render of the component (thus causing an effect).

When placing useEffect in your component you tell React you want to run the callback as an effect. React will run the effect after rendering and after performing the DOM updates.

If you pass only a callback - the callback will run after each render.

If passing a second argument (array), React will run the callback after the first render and every time one of the elements in the array is changed. for example when placing useEffect(() => console.log('hello'), [someVar, someOtherVar]) - the callback will run after the first render and after any render that one of someVar or someOtherVar are changed.

By passing the second argument an empty array, React will compare after each render the array and will see nothing was changed, thus calling the callback only after the first render.

Solution 3 - Javascript

useMountEffect hook

Running a function only once after component mounts is such a common pattern that it justifies a hook of its own that hides implementation details.

const useMountEffect = (fun) => useEffect(fun, [])

Use it in any functional component.

function MyComponent() {
    useMountEffect(function) // function will run only once after it has mounted. 
    return <div>...</div>;
}

About the useMountEffect hook

When using useEffect with a second array argument, React will run the callback after mounting (initial render) and after values in the array have changed. Since we pass an empty array, it will run only after mounting.

Solution 4 - Javascript

We have to stop thinking in component-life-cycle-methods (i.e. componentDidMount). We have to start thinking in effects. React effects are different from old-style class-life-cycle-methods.

By default effects run after every render cycle, but there are options to opt out from this behaviour. To opt out, you can define dependencies that mean that an effect is only carried out when a change to one of the dependencies is made.

If you explicitly define, that an effect has no dependecy, the effect runs only once, after the first render-cycle.

1st solution (with ESLint-complaint)

So, the first solution for your example would be the following:

function MyComponent() {

    const loadDataOnlyOnce = () => {
      console.log("loadDataOnlyOnce");
    };

    useEffect(() => {
        loadDataOnlyOnce(); // this will fire only on first render
    }, []);
    return (...);
}

But then the React Hooks ESLint plugin will complain with something like that:

React Hook useEffect has missing dependency: loadDataOnlyOnce. Either include it or remove the dependency array.

At first this warning seems annoying, but please don't ignore it. It helps you code better.

2nd solution (the right way, if dependency is not dependent on component)

If we add loadDataOnlyOnce to the dependency array, our effect will run after every render-cycle, because the reference of loadDataOnlyOnce changes on every render, because the function is destroyed(garbarge-collected) and a new function is created, but that's exactly what we don't want.

We have to keep the same reference of loadDataOnlyOnce during render-cycles.

So just move the function-definition above:

const loadDataOnlyOnce = () => {
  console.log("loadDataOnlyOnce");
};

function MyComponent() {
    useEffect(() => {
        loadDataOnlyOnce(); // this will fire only on first render
    }, []);
    return (...);
}

3rd solution (the right way, if dependency is dependent on component)

If the dependency of the effect (loadDataOnlyOnce), is dependent on the component (need props or state), there's React's builtin useCallback-Hook.

An elementary sense of the useCallback-Hook is to keep the reference of a function identical during render-cycles.

function MyComponent() {
    const [state, setState] = useState("state");

    const loadDataOnlyOnce = useCallback(() => {
      console.log(`I need ${state}!!`);
    }, [state]);

    useEffect(() => {
        loadDataOnlyOnce(); // // this will fire only when loadDataOnlyOnce-reference changes
    }, [loadDataOnlyOnce]);
    return (...);
}

Solution 5 - Javascript

Pass an empty array as the second argument to useEffect. This effectively tells React, quoting the docs:

> This tells React that your effect doesn’t depend on any values from props or state, so it never needs to re-run.

Here's a snippet which you can run to show that it works:

function App() {
  const [user, setUser] = React.useState(null);

  React.useEffect(() => {
    fetch('https://randomuser.me/api/')
      .then(results => results.json())
      .then(data => {
        setUser(data.results[0]);
      });
  }, []); // Pass empty array to only run once on mount.
  
  return <div>
    {user ? user.name.first : 'Loading...'}
  </div>;
}

ReactDOM.render(<App/>, document.getElementById('app'));

<script src="https://unpkg.com/[email protected]/umd/react.development.js"></script>
<script src="https://unpkg.com/[email protected]/umd/react-dom.development.js"></script>

<div id="app"></div>

Solution 6 - Javascript

function useOnceCall(cb, condition = true) {
  const isCalledRef = React.useRef(false);

  React.useEffect(() => {
    if (condition && !isCalledRef.current) {
      isCalledRef.current = true;
      cb();
    }
  }, [cb, condition]);
}

and use it.

useOnceCall(() => {
  console.log('called');
})

or

useOnceCall(()=>{
  console.log('Fetched Data');
}, isFetched);

Solution 7 - Javascript

I like to define a mount function, it tricks EsLint in the same way useMount does and I find it more self-explanatory.

const mount = () => {
  console.log('mounted')
  // ...

  const unmount = () => {
    console.log('unmounted')
    // ...
  }
  return unmount
}
useEffect(mount, [])

Solution 8 - Javascript

leave the dependency array blank . hope this will help you understand better.

   useEffect(() => {
      doSomething()
    }, []) 

empty dependency array runs Only Once, on Mount

useEffect(() => {
  doSomething(value)
}, [value])  

pass value as a dependency. if dependencies has changed since the last time, the effect will run again.

useEffect(() => {
  doSomething(value)
})  

no dependency. This gets called after every render.

Solution 9 - Javascript

Here is my version of Yasin's answer.

import {useEffect, useRef} from 'react';

const useOnceEffect = (effect: () => void) => {
  const initialRef = useRef(true);

  useEffect(() => {
    if (!initialRef.current) {
      return;
    }
    initialRef.current = false;
    effect();
  }, [effect]);
};

export default useOnceEffect;

Usage:

useOnceEffect(
  useCallback(() => {
    nonHookFunc(deps1, deps2);
  }, [deps1, deps2])
);

Solution 10 - Javascript

This does not answer your question exactly but may have the same intended affect of only running a function once and after the first render. Very similar to the componentDidMount function. This uses useState instead of useEffect to avoid dependency lint errors. You simply pass a self-executing anonymous function as the first argument to useState. As an aside, I'm not sure why React doesn't simply provide a hook that does this.

import React, { useState } from "react"

const Component = () => {

  useState((() => {
    console.log('componentDidMountHook...')
  }))

  return (
    <div className='component'>Component</div>
  )
}

export default Component

Solution 11 - Javascript

Incase you just call the function in useeffect after render you add an empty array as the second argument for the useeffect

useEffect=(()=>{
   functionName(firstName,lastName);
},[firstName,lastName])

calling a custom hooks

adding a empty array for rendering

Solution 12 - Javascript

window.onpageshow works even if the user presses the back button to navigate to the page, unlike passing an empty array as second argument of the use-effect hook which does not fire when returning to the page via the back button (thus not on every form of initial page load).

 useEffect(() => {    
     window.onpageshow = async function() {
      setSomeState(false)
      let results = await AsyncFunction() 
       console.log(results, 'Fires on on first load, 
        refresh, or coming to the page via the back button')
    };
 };

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionD&#225;vid Moln&#225;rView Question on Stackoverflow
Solution 1 - JavascriptTholleView Answer on Stackoverflow
Solution 2 - JavascriptEdan ChetritView Answer on Stackoverflow
Solution 3 - JavascriptBen CarpView Answer on Stackoverflow
Solution 4 - JavascriptwuarminView Answer on Stackoverflow
Solution 5 - JavascriptYangshun TayView Answer on Stackoverflow
Solution 6 - JavascriptYasin TazeogluView Answer on Stackoverflow
Solution 7 - JavascriptecoologicView Answer on Stackoverflow
Solution 8 - JavascriptShuhad zamanView Answer on Stackoverflow
Solution 9 - Javascriptglinda93View Answer on Stackoverflow
Solution 10 - Javascriptjb.darkcharmView Answer on Stackoverflow
Solution 11 - JavascriptVigneshView Answer on Stackoverflow
Solution 12 - JavascriptGreggory WileyView Answer on Stackoverflow