检索当前的活动屏幕名称Blackberry

Question

My application is a tab based one. For example, app has got 3 tabs - A, B & C. If the currently opened tab is "A", when the user again tries to click on the same tab, app shouldn't do anything.

I tried to do this using the following code snippet

final Screen currentActiveScreen = UiApplication.getUiApplication().getActiveScreen();

        if (newScreen == currentActiveScreen)
        {
            return;
        }

New screen is the screen that the user is trying to navigate to.

But this code is not firing and when I tried to debug, I found that some random number is also coming with the screen and hence the code is not entering the if loop.

The values in the condition comes like this and is returning false. AScreen@497293b7 == AScreen@2f25c01e

So, Is there anything wrong in what I'm trying to do ? How can we get the screen name only.

Thanks in advance.

Answer 1

There's not much to go on here, you may need to add more detail.

Those numbers you're seeing are the hashcodes of the objects. If newScreen refers to a new instance of a screen, == wont evaluate to true. There are 2 choices that come to mind, either override the equals method of your custom screen; or make A,B, and C different classes (they most likely are already), then use instanceof

eg.

ScreenA newScreen = new ScreenA();
if (currentActiveScreen instanceof ScreenA) return;

Answer 2

There is no such thing as a screen name in BlackBerry Java. And Kevin is right - you cannot compare the screens directly (for the reasons explained in his reply). You need to define your own discerning mechanism to check the type of the screen you would navigate to before even creating it.

If you have 3 tabs, you can pre-create 3 screens (one per tab), store them in three members of your Object, then simply remove and add the needed screen when the user presses something. Before doing that, you can indeed compare the screen you are trying to bring up with the current one.

Something like this:

Screen tabA = new MyScreenForTabA();
Screen tabB = new MyScreenForTabB();
Screen tabC = new MyScreenForTabC();
//...
// Somewhere else in the code
//...
    Screen nextScreen;
    switch (chosenTabLetter) { // Just throwing an idea
        case 'A':
            nextScreen = tabA;
            break;
        case 'B':
            nextScreen = tabB;
            break;
        case 'C':
            nextScreen = tabC;
            break;
    }
    if (nextScreen == currentScreen) {
        return;
    }

This way you can juggle the three screens without constantly creating new ones.