Android ViewFlipper 如何添加重复的子项和子项更改每个中的特定编辑文本

发布于 2024-10-08 09:38:25 字数 1479 浏览 6 评论 0原文

我在 Android 应用程序中设置了一个 ViewFlipper,它将托管一系列窗口来显示消息。每个窗口应该对应一组不同的消息,类似于多个打开的聊天。对于每个窗口,我使用相同的 window.xml 视图将视图带到屏幕上,它还有需要编辑的 EditText 变量。

作为参考,我正在创建和添加子项,如下所示:

 public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);  
        detector = new GestureDetector(this,this);
        setContentView(R.layout.viewflip);
        flipper = (ViewFlipper) findViewById(R.id.viewflip);

        addChild(flipper);
        addChild(flipper);
}

private void addChild(ViewFlipper flip){
     int index=0;
     View view = getView();

     if(flip.getChildCount()==0){
         flip.addView(view,index);
     }
     else{
     flip.addView(view,flip.getChildCount());
 }

 }

 private View getView(){
     LayoutInflater inflater = this.getLayoutInflater();
     View view = inflater.inflate(R.layout.window, null);        
     return view;
 }

如您所见,我基本上是在复制视图(我不确定这是否是我的设计的正确方法)。因此,如果我在添加子项后在 onCreate 函数中执行某些操作,例如

EditText messageHistoryText = (EditText) findViewById(R.id.messageHistory);         
        messageHistoryText.append("Testing :\n");

我在两个窗口上看到文本。

我认为这样的东西会更好:

View v1 = flipper.getChildAt(1);
        EditText messageHistoryText2 = (EditText) v1.findViewById(R.id.messageHistory);
        messageHistoryText2.append("Testing2 :\n");

但是当我使用那个时,我根本看不到任何东西。也许添加孩子时出现错误。也许我无法使用相同的视图,或者我可能以错误的方式有选择地更改 EditText。

尖端?

I have set up a ViewFlipper in my Android application that will host a series of windows to display messages. Each window should correspond to a different set of messages, similar to multiple open chats. For each window I am using the same window.xml view to bring the view onto the screen, it also has the variable for the EditText need to edit.

For reference I am creating and adding children as follows:

 public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);  
        detector = new GestureDetector(this,this);
        setContentView(R.layout.viewflip);
        flipper = (ViewFlipper) findViewById(R.id.viewflip);

        addChild(flipper);
        addChild(flipper);
}

private void addChild(ViewFlipper flip){
     int index=0;
     View view = getView();

     if(flip.getChildCount()==0){
         flip.addView(view,index);
     }
     else{
     flip.addView(view,flip.getChildCount());
 }

 }

 private View getView(){
     LayoutInflater inflater = this.getLayoutInflater();
     View view = inflater.inflate(R.layout.window, null);        
     return view;
 }

As you can see, I am basically duplicating the view (I'm not sure this is the proper approach to my design). So if I were to do something in the onCreate function after adding the children, such as

EditText messageHistoryText = (EditText) findViewById(R.id.messageHistory);         
        messageHistoryText.append("Testing :\n");

I see the text on both windows.

I thought that something like this would be better:

View v1 = flipper.getChildAt(1);
        EditText messageHistoryText2 = (EditText) v1.findViewById(R.id.messageHistory);
        messageHistoryText2.append("Testing2 :\n");

but when I use that one, I don't see anything at all. Perhaps there is a mistake on adding the children. Perhaps I can't use the same view, or perhaps I am selectively changing an EditText in an incorrect way.

Tips?

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星星的轨迹 2024-10-15 09:38:25

尝试以下操作,

EditText messageHistoryText = flip.getChildAt(j).findViewById(R.id.messageHistory);

其中 j 表示 ViewFlipper 中的子位置。

Try the following,

EditText messageHistoryText = flip.getChildAt(j).findViewById(R.id.messageHistory);

where j represents the child position in ViewFlipper.

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