Reusable Modal Component React Typescript - reactjs

Question

I have a component which has a button within it, like so -
<Button variant="primary" disabled={checkAccepted} onClick={openModal}>Send</Button>
I would like this button to, when it is active, to open up a modal when clicked. I am unsure how to do this and have been messing around with props but can't seem to figure it out. I also want the modal to be reusable so that any content can be passed in the modal body.I am thinking how do I open up the modal from within my openModal function?
I tried returning it like so -
const openModal = () => {
return (
<Modal>
<ModalBody>*Pass in swappable content here*</ModalBody>
</Modal>
)
}
But that doesn't seem to work. I am sure I am missing something.

Answer 1

You can't return components from an event handler. The way to handle events in react is almost always to alter the state of your application which triggers a re-render. In your case you need to keep track of the open state of your modal.
This can be done either in a controlled way (you keep track of the open state yourself and pass it to your <Modal> component as a prop) or in an uncontrolled way (the <Modal> component manages the open state itself). The second approach requires that you provide e.g. an element to render to your Modal component that acts as a trigger:
const MyModal = ({ children, trigger }) => {
const [modal, setModal] = useState(false);
const toggle = () => setModal(!modal);
return (
<div>
{React.cloneElement(trigger, { onClick: toggle })}
<Modal isOpen={modal} toggle={toggle}>
<ModalBody>{children}</ModalBody>
</Modal>
</div>
);
};
Then you can use it like that:
<MyModal trigger={<Button variant="primary">Send</Button>}>
<p>This is the content.</p>
</MyModal>
Or you can implement it in a controlled way. This is more flexible as it allows you to render the triggering element anywhere:
const MyModal = ({ children, isOpen, toggle }) => (
<div>
<Modal isOpen={isOpen} toggle={toggle}>
<ModalBody>{children}</ModalBody>
</Modal>
</div>
);
Usage Example:
function App() {
const [isOpen, setIsOpen] = useState(false);
const toggle = () => setIsOpen(!isOpen);
return (
<div className="App">
<Button variant="primary" onClick={toggle}>
Send
</Button>
<MyModal isOpen={isOpen} toggle={toggle}>
<p>This is the content.</p>
</MyModal>
</div>
);
}

Answer 2

You should pass the function which triggers the modal to your <Button /> component as prop. Then, in your component, you want to add the onClick event. You can't set an onClick event to the <Button />. It will think of onClick as a prop being passed to <Button />. Within <Button /> you can set the onClick event to an actual <button> element, and use the function which was passed in as a prop on that event.
You can use state to keep track of when the modal button is clicked. Your function can look like: (I am using class based components here, but you can do the same thing with functional components)
buttonClickedHandler = () => {
this.setState({isModalButtonClicked: !this.state.isModalButtonClicked});
}
Then, you can set the Modal component,
<Modal isShow={this.state.isModalButtonClicked} modalButton={this.buttonClickedHandler}>
<div> ...set contents of modal</div>
</Modal>
<button onClick={this.buttonClickedHandler}>Show Modal</button>
So, within the Modal component, you can have something like this:
<React.Fragment>
<Backdrop showModal={this.props.isShow} clicked={this.props.modalButton}/>
{this.props.children}
</React.Fragment>
Backdrop is basically the greyed out background. You can also set an onClick event to listen to when the backdrop is clicked.