How to show a popup modal on page load in React js functional component.basically when the page loads the popup or modal should be visible automatically without any click
CodePudding user response:
Most modals will have a prop to specify whether or not the modal should be open. You can simply set this to true
on initialization.
Consider an example using a Dialog
from the React UI library MUI:
const SimpleDialog = (props) =>
{
const [open, setOpen] = useState(true)
return (
<Dialog open = {open}>
<DialogTitle>Title</DialogTitle>
<DialogContent>...</DialogContent>
</Dialog>
)
}
The open
prop of the Dialog
is based on the value of the state variable, which is initialized as true
. You can change this value as needed through various Dialog
actions. To learn more about MUI Dialogs, see the docs.
For information on useState
, see the React docs.
CodePudding user response:
class App extends React.Component {
constructor(props) {
super(props);
this.state = {
modalState: true
};
this.handleShow = this.handleShow.bind(this);
}
handleShow() {
this.setState({ modalState: !this.state.modalState });
}
render() {
return (
<div>
<div className={"modal fade" (this.state.modalState ? " show d-block" : " d-none")} tabIndex="-1" role="dialog">
<div className="modal-dialog" role="document">
<div className="modal-content">
<div className="modal-header">
<h5 className="modal-title">My Profile</h5>
<button type="button" className="close" onClick={this.handleShow}>
<span>×</span>
</button>
</div>
<div className="modal-body">...</div>
<div className="modal-footer">
<button type="button" className="btn btn-secondary" onClick={this.handleShow}>Close</button>
<button type="button" className="btn btn-primary">Save changes</button>
</div>
</div>
</div>
</div>
</div>
);
}
}
ReactDOM.render(, document.getElementById('root'));
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.4.1/css/bootstrap.min.css">