Reputation: 137
i got a issue with my code. My function "getNames" rerender may times, but i want it to render once? have you got any clue ?
import grounds from './../../UballersGroundsData.json';
export default function Groundlist() {
function getNames(jsonObj){
for(let item in jsonObj){
console.log("item = " + item);
for(let property in jsonObj[item] ){
console.log(jsonObj[item]);
// if (property === "groundName"){
// console.log(jsonObj[item][property]);
// }
}
}
}
return(
<div>
<h1>Hello world!</h1>
<ul>
{getNames(grounds)}
</ul>
</div>
)
}
Thank you !
Upvotes: 1
Views: 186
Reputation: 943
You should never call a function inside return scope of render. It's normal for a component to re-render without proper treatment.
Taking in mind the other 2 answers
You can use the full power of useEffect, useCallback and React.memo to prevent anything from re-render.
import React from 'react';
import grounds from './../../UballersGroundsData.json';
function Groundlist() {
// initiate state
const [names, setNames] = React.useState([]);
// This will prevent the Function from recalculate - useCallback
const getNames = React.useCallback(function (jsonObj) {
for(let item in jsonObj){
console.log("item = " + item);
for(let property in jsonObj[item] ){
console.log(jsonObj[item]);
// if (property === "groundName"){
// console.log(jsonObj[item][property]);
// }
}
}
}, []);
// Will make function run only once and nevermore - useEffect
React.useEffect(() => {
setNames(getNames());
}, [])
return(
<div>
<h1>Hello world!</h1>
<ul>
{names.map(a => <li>{a}</li>)}
</ul>
</div>
)
}
// Will prevent React from try to re-render without changing in props, so as your component has no props, will never re-render without yourself unmounting first
export default React.memo(Groundlist);
In another cases you can control exact when the component should recalculate your names using the last argument of functions
useCallback(() => {}, []) //<---
For example
useCallback(() => {}, [updateState]);
when updateState change the function will be recreated.
Upvotes: 0
Reputation: 1677
Try using useMemo
and useCallBack
you want to optimize your react app.
React Official docs clearly described how to use it: useMemo
useCallBack
Upvotes: 0
Reputation: 19813
You can use useMemo react hook to memoize the returned value i.e. skip unnecessary / heavy calculations due to change in other state, props or context variables.
Example:
import { useMemo } from "react"
export default function Groundlist(props) {
const grounds = props.data // if grounds is passed as props from Parent component
const groundsMemo = useMemo(() => {
// do all the heavy calculations here
// (e.g. do the work of getNames function)
// and return some JSX or Array (data)
// returned value will be memoized;
// means it will be re-calculated only if "grounds" changes
// Hence, no unnecessary calls to getNames (heavy calculations)
}, [grounds])
return (
<div>
{/* Use this if groundsMemo is JSX */}
<ul>{groundsMemo}</ul>
{/* Use this if groundsMemo is an Array (data) */}
<ul>{groundsMemo.map(item => <li key={some_key}>
{item.property}
</li>)}</ul>
</div>
)
}
Upvotes: 1
Reputation: 612
You should put your function inside of a useEffect hook, then set it to a state hook with useState. Then, map out the list items for your list (assuming you are returning an array from your function). If you want it to only run getNames on the first render, you would set it up the useEffect hook with an empty dependency array. Code should look something like this:
import React, { useEffect, useState } from 'react'
import grounds from './../../UballersGroundsData.json';
export default function Groundlist() {
const [names, setNames] = useState([]) // Initial state with empty array
useEffect(() => {
function getNames(jsonObj){
// your function logic here...
}
const result = getNames(grounds) // Call your function
setNames(result) // set it to names state hook
}, []) // Empty array here means it will only use the useEffect on the first render.
return(
<div>
<h1>Hello world!</h1>
<ul>
{Array.from(names).map(name => <li>{name}</li>)}
</ul>
</div>
)
}
Upvotes: 1