JavaFX排序列表视图

Question

我的应用程序中有一个ListView，我想对条目进行排序。我还希望列表自动排序，如果一个新的条目被添加。

为此，我使用sortedlist。Java API表示“ObservableList中的所有更改都会立即传播到SortedList中”。

public class Test extends Application
{
    public static final ObservableList names = FXCollections.observableArrayList();

    public static void main(String[] args) {
        launch(args);
    }

    @Override
    public void start(Stage primaryStage) {
        final ListView listView = new ListView(names);
        listView.setPrefSize(200, 250);
        listView.setEditable(true);

        names.addAll("Brenda", "Adam", "Williams", "Zach", "Connie", "Donny", "Lynne", "Rose", "Tony", "Derek");

        listView.setItems(names);
        SortedList<String> sortedList = new SortedList(names);
        sortedList.setComparator(new Comparator<String>(){
            @Override
            public int compare(String arg0, String arg1) {
                return arg0.compareToIgnoreCase(arg1);
            }
        });

        for(String s : sortedList)
            System.out.println(s);
        System.out.println();

        names.add("Foo");
        System.out.println("'Foo' added");
        for(String s : sortedList)
            System.out.println(s);

        StackPane root = new StackPane();
        root.getChildren().add(listView);
        primaryStage.setScene(new Scene(root, 200, 250));
        primaryStage.show();
    }
}

Adam
Brenda
Connie
Derek
Donny
Lynne
Rose
Tony
Williams
Zach

'Foo' added
Adam
Brenda
Connie
Derek
Donny
Foo    <--
Lynne
Rose
Tony
Williams
Zach

Answer 1

您需要在ListView中使用SortedList。即。做

    final ListView listView = new ListView();
    listView.setPrefSize(200, 250);
    listView.setEditable(true);

    names.addAll("Brenda", "Adam", "Williams", "Zach", "Connie", "Donny", "Lynne", "Rose", "Tony", "Derek");

    SortedList<String> sortedList = new SortedList(names);
    listView.setItems(sortedList);