I was recently looking for a nice solution to this without adding React-Bootstrap to my project (as Bootstrap 4 is about to be released).
This is my solution: https://jsfiddle.net/16j1se1q/1/
let Modal = React.createClass({
componentDidMount(){
$(this.getDOMNode()).modal('show');
$(this.getDOMNode()).on('hidden.bs.modal', this.props.handleHideModal);
},
render(){
return (
<div className="modal fade">
<div className="modal-dialog">
<div className="modal-content">
<div className="modal-header">
<button type="button" className="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 className="modal-title">Modal title</h4>
</div>
<div className="modal-body">
<p>One fine body…</p>
</div>
<div className="modal-footer">
<button type="button" className="btn btn-default" data-dismiss="modal">Close</button>
<button type="button" className="btn btn-primary">Save changes</button>
</div>
</div>
</div>
</div>
)
},
propTypes:{
handleHideModal: React.PropTypes.func.isRequired
}
});
let App = React.createClass({
getInitialState(){
return {view: {showModal: false}}
},
handleHideModal(){
this.setState({view: {showModal: false}})
},
handleShowModal(){
this.setState({view: {showModal: true}})
},
render(){
return(
<div className="row">
<button className="btn btn-default btn-block" onClick={this.handleShowModal}>Open Modal</button>
{this.state.view.showModal ? <Modal handleHideModal={this.handleHideModal}/> : null}
</div>
);
}
});
React.render(
<App />,
document.getElementById('container')
);
The main idea is to only render the Modal component into the React DOM when it is to be shown (in the App components render function). I keep some 'view' state that indicates whether the Modal is currently shown or not.
The 'componentDidMount' and 'componentWillUnmount' callbacks either hide or show the modal (once it is rendered into the React DOM) via Bootstrap javascript functions.
I think this solution nicely follows the React ethos but suggestions are welcome!