JavaFX手风琴将进一步扩展面板

Question

In Accordion it is possible to set the selected Pane with accordion.setExpandedPane , and it is also possible to get all the children with accordion.getPanes() .

However I'm struggling to find how to implement "select next Pane" functionality without explicitly extending the TitledPane class for the Panes, and maintaining an index manually via accordion.expandedPaneProperty() and some custom implementation.

A Proposed solution would be to add a listened to expandedPaneProperty() :

    accordion.expandedPaneProperty().addListener((observable, oldValue, newValue) -> {
        if(null != newValue){
            idx = 0;
            for(TitledPane whytho: this.getPanes()){
                if(!whytho.equals(newValue))idx++;
                else break;
            }
            selectedIndex = idx;
        }
    });

Is there an easier way for this?

Answer 1

You can query the index of the currently expanded pane using List#indexOf(Object) . Then you just need to set the expanded pane to the pane at index ± 1 .

private void expandPrevious(Accordion acc) {
    int index = acc.getPanes().indexOf(acc.getExpandedPane());
    int newIndex = Math.max(index - 1, 0);
    acc.setExpandedPane(acc.getPanes().get(newIndex));
}

private void expandNext(Accordion acc) {
    int index = acc.getPanes().indexOf(acc.getExpandedPane());
    int newIndex = Math.min(index + 1, acc.getPanes().size() - 1);
    acc.setExpandedPane(acc.getPanes().get(newIndex));
}

The above doesn't perform any wrap around logic; in other words, calling expandNext while the last pane is expanded won't expand the first pane—rather nothing will change. It also doesn't handle the case where getExpandedPane() returns null .