滑动另一个容器时不需要的向左滑动可滑动容器

Question

我在页面中添加了一些可滑动的容器。可滑动的容器有左右按钮，顶部组件是一个多按钮，可将用户带到另一个页面。当用户返回带有可滑动容器的页面并在不同的可滑动容器上向左（仅向左）滑动时，第一个容器也会向左打开。图片和示例代码如下：-

//Swipeable containers bugs
public void SwipeableContainerPage(Form mainForm){
    
    Form formSC = new Form("Swipeable Container", new BorderLayout());
    formSC.getToolbar().setBackCommand("", e -> mainForm.showBack());
    
    Container northcnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    Container centercnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    Container southcnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    
    MultiButton mb_1 = new MultiButton("MB1");
    Button b_l1 = new Button("LB1");
    Button b_r1 = new Button("RB1");
    
    MultiButton mb_2 = new MultiButton("MB2");
    Button b_l2 = new Button("LB2");
    Button b_r2 = new Button("RB2");
    
    MultiButton mb_3 = new MultiButton("MB3");
    Button b_l3 = new Button("LB3");
    Button b_r3 = new Button("RB3");
            
    SwipeableContainer sc_1 = new SwipeableContainer(b_l1, b_r1, mb_1);
    SwipeableContainer sc_2 = new SwipeableContainer(b_l2, b_r2, mb_2);
    SwipeableContainer sc_3 = new SwipeableContainer(b_l3, b_r3, mb_3);
    
    mb_1.addActionListener(new ActionListener() {
    @Override
        public void actionPerformed(ActionEvent e) {
            GoToAndBackPage(formSC);
        }        
    });
    
    mb_2.addActionListener(new ActionListener() {
    @Override
        public void actionPerformed(ActionEvent e) {
            GoToAndBackPage(formSC);
        }        
    });
    
    mb_3.addActionListener(new ActionListener() {
    @Override
        public void actionPerformed(ActionEvent e) {
            GoToAndBackPage(formSC);
        }        
    });
    
    centercnt.add(sc_1);
    centercnt.add(sc_2);
    centercnt.add(sc_3);        
    
    formSC.add(BorderLayout.NORTH, northcnt).add(BorderLayout.CENTER, centercnt).add(BorderLayout.SOUTH, southcnt);  
    formSC.show();
    
}
//page for SC to go to
public void GoToAndBackPage(Form mainForm){
    Form formGoTo = new Form("Go To Page", new BorderLayout());
    formGoTo.getToolbar().setBackCommand("", e -> mainForm.showBack());
    
    Container northcnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    Container centercnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    Container southcnt = new Container(new BoxLayout(BoxLayout.Y_AXIS));
    
    Label l_welcome = new Label("Welcome to the Go To Page");
    
    centercnt.add(l_welcome);
    
    formGoTo.add(BorderLayout.NORTH, northcnt).add(BorderLayout.CENTER, centercnt).add(BorderLayout.SOUTH, southcnt);  
    formGoTo.show();        
}

Answer 1

这听起来像是一个错误，理想情况下，您应该 file an issue 使用测试用例。但试试这个解决方法。

将处理导航的代码包装到串行调用中的不同表单中。例如。如果代码如下所示：

myMultiButton.addActionListener(e -> showOtherForm(...));

然后这样做：

myMultiButton.addActionListener(e -> callSerially(showOtherForm(...)));

我的理论是按压将可滑动组件带到“可能”被拖动的状态。但是因为导航事件恰好很快按钮没有收到释放事件（我们现在处于不同的形式）并且一切都丢失了。

因此，当您返回表单时，您现在有两个“收费”按钮。通过使用 callSerially，您可以有效地刷新事件队列，并确保在移动到下一个 Form.

之前，当前 Form 上的每个人都处理了释放事件

滑动另一个容器时不需要的向左滑动可滑动容器

Unwanted swipe left on swipeable container when swiping another

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