我有一个gridView,我在弹出窗口中显示(gridview是透明布局,继承自linearlayout,只有一个部分透明的背景).我永远无法让这个GridView的OnItemClick执行.当我触摸gridview中的图像时,它似乎被点击(图像bachgrond更改),但OnItemClick未被调用.

下面是我的适配器的代码和包含gridView的弹出视图.谢谢！

//Adapter

公共类ImageAdapter扩展BaseAdapter {
    private Context mContext;
    private int itemBackground;

public ImageAdapter(Context c) {
    mContext = c;

  //---setting the style---
    TypedArray a = c.obtainStyledAttributes(R.styleable.Gallery1);
    itemBackground = a.getResourceId(R.styleable.Gallery1_android_galleryItemBackground, 0);
    a.recycle();
}

….

public View getView(int position, View convertView, ViewGroup parent) {
    ImageView imageView;
    if (convertView == null) {
        imageView = new ImageView(mContext);
    } else {
        imageView = (ImageView) convertView;
    }
    imageView.setImageResource(images[position]);
    imageView.setScaleType(ImageView.ScaleType.CENTER_INSIDE);
    imageView.setBackgroundResource(itemBackground);
    return imageView;

}
public Integer[] images = {
        R.drawable.sound1,
        R.drawable.sound2,
        R.drawable.sound3,
        R.drawable.sound4
};

}

//////////In Activity, onCreate////////

...

final LayoutInflater inflater=(LayoutInflater)this.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
    final TransparentLayout musicGrid = (TransparentLayout) inflater.inflate(R.layout.gridviewpopup, null, false);
    final GridView gView = (GridView) musicGrid.findViewById(R.id.music_gridview);
    gView.setAdapter(new ImageAdapter(this));

    final PopupWindow soundSelectorWindow = new PopupWindow(this);
    soundSelectorWindow.setContentView(musicGrid);
    soundSelectorWindow.setTouchable(true);

    gView.setOnItemClickListener(new OnItemClickListener() 
    {
        public void onItemClick(AdapterView<?> parent, View v, int position, long id) 
        {   
            //NEVER GETS HERE
            soundSelectorWindow.dismiss();

        }
    }); 

解决方法:

如果删除soundSelectorWindow.setTouchable(true)会发生什么？