android - 如何在viewflipper的android中检查以前的值

Question

我有八个 View flipper，其中有 4 个问题和 4 个答案。每当单击特定的 viewflipper 时，就会显示这些问题和答案。我的问题是，当第一次单击任何 viewflipper 时，当时无事可做，在单击 viewflipper 后，下一个单击的 viewflipper 值想要检查哪个相同(相等)或不同。(如果问题是 5+3 那么答案应该是 8)。

条件:

1. if I am first time Q is clicked then next time another viewflipper which have Q is clicked nothing check operation is necessary and nothing hide .

2. if I am first time Q is clicked then next time A is clicked retrive value from textview of Q and check is it right or not, if it is right hide both Q and A.

3. if I am first time A is clicked then next timeQ is clicked retrive value from textview of A and check is it right or not, if it is right hide both Q and A.

4. if I am first time A is clicked then next time another viewflipper which haveA is clicked nothing check operation is necessary nothing hide.

5. if I am first time Q is clicked then next time same viewflipper which have Q is clicked nothing check operation is necessary nothing hide.

6. if I am first time A is clicked then next time same viewflipper which have A is clicked nothing check operation is necessary nothing hide.

那么，如何检查当前可点击的 ViewFlipper 和上一个点击的 ViewFlipper。我不知道如何检查问题。

看看这个，我想，我的问题更清楚了。

编辑:12 月 16 日

@Override
public void onClick(View v) {

    switch (v.getId()) {
    case R.id.vf1:
        try {
            if (click) {
                Message msg = new Message();
                msg.what = 1;
                delayHandler.sendMessageDelayed(msg, DELAYTIME);

                vFilpper1.showNext();


        } catch (Exception e) {
            e.printStackTrace();
        }
        counter++;
        break;
    case R.id.vf2:
        try {
            Message msg = new Message();
            msg.what = 2;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 2");

            vFilpper2.showNext();


        } catch (Exception e) {
            e.printStackTrace();
        }
        break;
    case R.id.vf3:
        try {
            Message msg = new Message();
            msg.what = 3;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 3");
            vFilpper3.showNext();


        } catch (Exception e) {
            e.printStackTrace();
        }
        break;

    case R.id.vf4:
        try {
            Message msg = new Message();
            msg.what = 4;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 4");
            vFilpper4.showNext();

        } catch (Exception e) {
            e.printStackTrace();
        }
        break;
    case R.id.vf5:
        try {
            Message msg = new Message();
            msg.what = 5;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 5");
            vFilpper5.showNext();

        } catch (Exception e) {
            e.printStackTrace();
        }
        break;
    case R.id.vf7:
        try {
            Message msg = new Message();
            msg.what = 7;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 7");
            vFilpper7.showNext();


        } catch (Exception e) {
            e.printStackTrace();
        }
        break;
    case R.id.vf8:
        try {
            Message msg = new Message();
            msg.what = 8;
            delayHandler.sendMessageDelayed(msg, DELAYTIME);
            Log.d("viewfilpper", "VFlipper 8");
            vFilpper8.showNext();


        } catch (Exception e) {
            e.printStackTrace();
        }
        break;

Answer 1

问题是您的代码缺少先前选择的 ViewFlipper 的状态。我只能提供部分代码，因为我不知道您的其余代码。

在此处有您的代码 fragment 的类中，“public void onClick(View v) {”方法中，您想要这样的东西:

public class Puzzle extends Activity{ //or whatever your class is called

    private int previousFlipperID = -1; //this stores the previous state of the flipper that was selected

    @Override
    public void onClick(View v) {
        //.... your code, but modified
            // changes
        if(previousFlipperID == -1){
            //this is the VERY FIRST CLICK, using -1 to denotate that nothing was previously selected
            previousFlipperID = v.getId();
            return;  
        }
            // end changes

        switch (v.getId()) {
        case R.id.vf1:
        try {
            if (click) {

            // changes
                switch(previousFlipperID){
                case 0: 
                    //do your logic here that you stated. note that at this point in the code
                    // you are at the point where your PREVIOUS flipper was the first flipper
                    // and your CURRENT flipper is also the first one, since v.getId() is R.id.vf1
                    break;
                case 1:  //some logic, but the CURRENT flipper is the first flipper, while PREVIOUS flipper was the second flipper
                    ...
                case 7: //the previous flipper selected was the last flipper
                    break;
                }
            // end changes

                Message msg = new Message();
                msg.what = 1;
                delayHandler.sendMessageDelayed(msg, DELAYTIME);

                vFilpper1.showNext();

        } catch (Exception e) {
            e.printStackTrace();
        }
        counter++;
        break;

        // changes
        previousFliperID = v.getId(); //update the previous flipper
        // end changes

    }
}

在我的代码中，特别要查找“//changes”。你会看到我使用了一个整数来存储前一个脚蹼。然后我检查它是否是第一次的方法是检查前一个脚蹼是否为 -1。然后在最后，确保将当前的鳍状肢 ID 设置为以前的鳍状肢 ID，以便下次更新“以前的”鳍状肢 ID。

另外，请注意，我在其中有一个嵌套的 switch 语句，因为您需要一堆额外的检查来对发生的事情执行您的逻辑，具体取决于当前和之前的脚蹼是什么。

答案是根据我所获信息的最佳能力(也是我的最佳理解)，所以我希望这会有所帮助。

附言这么说我感到内疚，但如果我的回答是正确的，请检查我的答案，因为我实际上需要一些赏金点数来提出赏金问题，但我认为我没有足够的点数。谢谢!和节日快乐