JavaFX ListView多选
发布时间:2020-12-15 00:26:58 所属栏目:Java 来源:网络整理
导读:我想从ListView中选择多个项目.它会对鼠标点击做出反应.我试过这个: selectedLogsList.addAll(logsListView.getSelectionModel().getSelectedItems()); 但它给了我重复. 所以我尝试了这个: logsListView.getSelectionModel().selectedItemProperty().addLi
我想从ListView中选择多个项目.它会对鼠标点击做出反应.我试过这个:
selectedLogsList.addAll(logsListView.getSelectionModel().getSelectedItems()); 但它给了我重复. logsListView.getSelectionModel().selectedItemProperty().addListener( new ChangeListener<String>() { @Override public void changed(ObservableValue<? extends String> ov,String old_val,String new_val) { if(selectedLogsList.contains(new_val)) { selectedLogsList.remove(new_val); } else { selectedLogsList.add(new_val); } } }); 我也想,如果选择了单元格,它应该被标记为(不同的背景颜色) 解决方法
如果您只想知道选择了哪些项目,请查看我的示例.看看setOnMouseClicked()方法..
package application; import javafx.application.Application; import javafx.collections.FXCollections; import javafx.collections.ObservableList; import javafx.event.Event; import javafx.event.EventHandler; import javafx.scene.Scene; import javafx.scene.control.ListView; import javafx.scene.control.SelectionMode; import javafx.scene.layout.Pane; import javafx.stage.Stage; public class Main extends Application { @Override public void start(Stage primaryStage) { try { Pane root = new Pane(); Scene scene = new Scene(root,600,600); scene.getStylesheets().add(getClass().getResource("application.css").toExternalForm()); ListView<String> listView = new ListView<String>(); ObservableList<String> list = FXCollections.observableArrayList(); listView.setItems(list); list.add("item1"); list.add("item2"); list.add("item3"); listView.getSelectionModel().setSelectionMode(SelectionMode.MULTIPLE); listView.setOnMouseClicked(new EventHandler<Event>() { @Override public void handle(Event event) { ObservableList<String> selectedItems = listView.getSelectionModel().getSelectedItems(); for(String s : selectedItems){ System.out.println("selected item " + s); } } }); list.add("item4"); root.getChildren().add(listView); primaryStage.setScene(scene); primaryStage.show(); } catch(Exception e) { e.printStackTrace(); } } public static void main(String[] args) { launch(args); } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |