到目前为止，这是我所拥有的，但是我正在为手柄而苦苦挣扎。
@FXML
public void link() {
hypTerms.setOnaction(new EventHandler<actionEvent>() {
@Override
public void handle(actionEvent e) {
;
}
});
}
JavaFX中是否有功能允许您单击超链接并在浏览器（例如Chrome或Firefox）中打开它？
•
问答
ether1984 回答：JavaFX中是否有功能允许您单击超链接并在浏览器（例如Chrome或Firefox）中打开它？
这应该有效：
Hyperlink link = new Hyperlink("Click me");
String url = "https://www.google.com/";
link.setOnAction(a->getHostServices().showDocument(url));
它将自动在默认浏览器中打开页面。
,您可以尝试处理所有异常。
export class App extends Component {
state = {
sideDrawerOpen: false
};
drawerToggleClickHandler = () => {
this.setState((prevState) => {
return { sideDrawerOpen: !prevState.sideDrawerOpen };
});
};
sidedrawerToggleClickHandler = () => {
this.setState({ sideDrawerOpen: false });
}
backdropClickHandler = () => {
this.setState({ sideDrawerOpen: false });
};
componentDidMount() {
this.setState({ sideDrawerOpen: false });
};
render() {
let backdrop;
if (this.state.sideDrawerOpen) {
backdrop = <Backdrop click={this.backdropClickHandler} />
}
return (
<div className="App_margin">
<Router>
<div className='App'>
<Nav drawerClickHandler={this.drawerToggleClickHandler} />
<SideDrawer sidedrawerClickHandler={this.sidedrawerToggleClickHandler} show={this.state.sideDrawerOpen} />
{ backdrop }
< Switch >
<Route path='/setup_page' component={setup_page} exact />
<Route path='/main_page' component={main_page} />
<Route path='/settings_page' component={settings_page} />
<Route component={Error} />
</Switch>
</div>
</Router>
</div>
);
}
}