TypeError: Cannot read property ‘children’ of undefined

Typeerror: Cannot Read Property ‘Children’ Of Undefined

We Are Going To Discuss About TypeError: Cannot read property ‘children’ of undefined. So lets Start this Javascript Article.

How to resolve TypeError: Cannot read property ‘children’ of undefined

So on Main.js, when you are using this kinda of syntax (functional component) the way you are getting the props is not correct and not only that, you can only have one children per component.
So first things first, you can correct this part in many different ways, you can either use destructuring or with only one parameter
const Main = (prop1, prop2, prop3) => { ... }
Option 1
const Main = ({ prop1, prop2, prop3, children }) => { ... }
Option 2
const Main = (props) => { console.log(props.props1) console.log(props.props2) console.log(props.children) ... }
But to your case, on the App.js all components/tags inside the Main component, they are all one children, so instead of what you are doing, you should do
// Main.js ... <div className="container" style={{margin: '10px', border: '2px solid black'}}> {props.children} // OR if you are using destructuring {children} </div>
That way you are getting everything inside <Main> ... </Main> you be rendered inside the <div className="container ...>

TypeError: Cannot read property ‘children’ of undefined

So on Main.js, when you are using this kinda of syntax (functional component) the way you are getting the props is not correct and not only that, you can only have one children per component.
So first things first, you can correct this part in many different ways, you can either use destructuring or with only one parameter
const Main = (prop1, prop2, prop3) => { ... }
Option 1
const Main = ({ prop1, prop2, prop3, children }) => { ... }
Option 2
const Main = (props) => { console.log(props.props1) console.log(props.props2) console.log(props.children) ... }
But to your case, on the App.js all components/tags inside the Main component, they are all one children, so instead of what you are doing, you should do
// Main.js ... <div className="container" style={{margin: '10px', border: '2px solid black'}}> {props.children} // OR if you are using destructuring {children} </div>
That way you are getting everything inside <Main> ... </Main> you be rendered inside the <div className="container ...>

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So on Main.js, when you’re utilizing this kinda of syntax (practical part) the best way you might be getting the props will not be appropriate and never solely that, you’ll be able to solely have one kids per part.

So first issues first, you’ll be able to appropriate this half in many alternative methods, you’ll be able to both use destructuring or with just one parameter

const Main = (prop1, prop2, prop3) => { ... }

Option 1

const Main = ({ prop1, prop2, prop3, kids }) => { ... }

Option 2

const Main = (props) => { 
  console.log(props.props1)
  console.log(props.props2)
  console.log(props.kids)
  ...
}

But to your case, on the App.js all elements/tags contained in the Main part, they’re all one kids, so as a substitute of what you might be doing, you need to do

// Main.js
...
<div className="container" model={{margin: '10px', border: '2px stable black'}}>
    {props.kids}
    // OR in case you are utilizing destructuring
    {kids}
</div>

That method you might be getting every thing inside <Main> ... </Main> you be rendered contained in the <div className="container ...>

So This is all About This Tutorial. Hope This Tutorial Helped You. Thank You.

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