JavaFX中TableView选中预约行后打开对应详情表单的问题

我用JavaFX开发GUI，表格里有两种预约类型，对应不同的visitTypeId。数据初始化和表格列配置代码如下：

现有代码 #

数据初始化 #

ObservableList<IAppointmentModel> initialData() {
    IAppointmentModel apt1 = new IAppointmentModel("10/31/2023", "Test Apt 1", 1, "Dr. Smith");
    IAppointmentModel apt2 = new IAppointmentModel("2/8/2023", "Test Apt 2", 2, "Dr. Smith");
    return FXCollections.<IAppointmentModel>observableArrayList(apt1, apt2);
}

表格列配置 #

aptData.setCellValueFactory(new PropertyValueFactory<IAppointmentModel, String>("aptDate"));
referringMd.setCellValueFactory(new PropertyValueFactory<IAppointmentModel, String>("referringMd"));
visitType.setCellValueFactory(new PropertyValueFactory<IAppointmentModel, String>("visitType"));
tableView.setItems(initialData());

用户右键点击预约行时，会弹出带“View Details”选项的上下文菜单，需求是点击该选项后根据预约类型打开对应表单（比如Test Apt 1打开FormA，Test Apt 2打开FormB）。但我尝试了两种实现，点击后均无反应：

尝试的两种无效实现 #

第一种（基于visitTypeId判断） #

viewButton.setOnAction(new EventHandler<ActionEvent>() {
    @Override
    public void handle(ActionEvent event) {
        IAppointmentModel item = tableView.getSelectionModel().getSelectedItem();
        try {
            if(item.equals(visitTypeId.equals(1))){
                Parent popUp;
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/VisitDetailsGait.fxml")));
                Stage stage1 = new Stage();
                stage1.setTitle("GAIT Visit Details:   ");
                stage1.setScene(new Scene(popUp, 800, 680));
                stage1.show();
            }else if(item.equals(visitTypeId.equals(2))){
                Parent popUp;
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/VisitDetailsUE.fxml")));
                Stage stage1 = new Stage();
                stage1.setTitle("UE Visit Details:   ");
                stage1.setScene(new Scene(popUp, 800, 680));
                stage1.show();
            }
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
});

第二种（基于visitType文本判断） #

viewButton.setOnAction(new EventHandler<ActionEvent>() {
    @Override
    public void handle(ActionEvent event) {
        IAppointmentModel item = tableView.getSelectionModel().getSelectedItem();
        try {
            if(item.equals(visitType.getText().equals("Test Apt 1"))){
                Parent popUp;
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/FormA.fxml")));
                Stage stage1 = new Stage();
                stage1.setTitle("GAIT Visit Details:   ");
                stage1.setScene(new Scene(popUp, 800, 680));
                stage1.show();
            }else if(item.equals(visitType.getText().equals("Test Apt 2"))){
                Parent popUp;
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/FormB.fxml")));
                Stage stage1 = new Stage();
                stage1.setTitle("UE Visit Details:   ");
                stage1.setScene(new Scene(popUp, 800, 680));
                stage1.show();
            }
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
});

问题分析与修正方案 #

两种实现的核心错误都是条件判断逻辑错误：

• 第一种中item.equals(visitTypeId.equals(1))是将IAppointmentModel对象和布尔值（visitTypeId.equals(1)的返回结果）比较，永远不会相等，条件永远不触发。
• 第二种中item.equals(visitType.getText().equals("Test Apt 1"))同样是对象和布尔值比较，且visitType.getText()获取的是表格列本身的文本，而非选中行的预约类型文本。

修正后的代码（基于visitTypeId判断） #

假设IAppointmentModel类有getVisitTypeId()方法获取预约类型ID：

viewButton.setOnAction(new EventHandler<ActionEvent>() {
    @Override
    public void handle(ActionEvent event) {
        IAppointmentModel item = tableView.getSelectionModel().getSelectedItem();
        // 先判断选中项是否为空
        if (item == null) {
            return;
        }
        try {
            Parent popUp;
            Stage stage1 = new Stage();
            if(item.getVisitTypeId() == 1){
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/VisitDetailsGait.fxml")));
                stage1.setTitle("GAIT Visit Details:   ");
            } else if(item.getVisitTypeId() == 2){
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/VisitDetailsUE.fxml")));
                stage1.setTitle("UE Visit Details:   ");
            } else {
                // 处理未知类型，可提示或直接返回
                return;
            }
            stage1.setScene(new Scene(popUp, 800, 680));
            stage1.show();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
});

修正后的代码（基于visitType文本判断） #

假设IAppointmentModel类有getVisitType()方法获取预约类型名称：

viewButton.setOnAction(new EventHandler<ActionEvent>() {
    @Override
    public void handle(ActionEvent event) {
        IAppointmentModel item = tableView.getSelectionModel().getSelectedItem();
        if (item == null) {
            return;
        }
        try {
            Parent popUp;
            Stage stage1 = new Stage();
            String visitTypeName = item.getVisitType();
            if("Test Apt 1".equals(visitTypeName)){
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/FormA.fxml")));
                stage1.setTitle("GAIT Visit Details:   ");
            } else if("Test Apt 2".equals(visitTypeName)){
                popUp = FXMLLoader.load(Objects.requireNonNull(GaitApplication.class.getResource("Visits/FormB.fxml")));
                stage1.setTitle("UE Visit Details:   ");
            } else {
                return;
            }
            stage1.setScene(new Scene(popUp, 800, 680));
            stage1.show();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }
});

额外注意事项：

1. 必须确保IAppointmentModel类提供了对应的getter方法（如getVisitTypeId()、getVisitType()），否则PropertyValueFactory无法正常工作，同时修正后的代码也无法获取对应属性。
2. 添加了选中项为空的判断，避免空指针异常。
3. 可根据实际需求补充未知预约类型的处理逻辑。

内容的提问来源于stack exchange，提问作者Kameron