Android Gridview - 需要打开不同的意图

时间:2011-05-16 23:02:07

标签: android gridview android-intent

只是Android开发的初学者。我目前有一个GridView设置作为我的应用程序的主菜单。我设法使用OnItemClickListener()获得了一个意图。但是,单击时我的所有6个图像都转到相同的Activity类。如何为每个图像提供不同活动的意图?

MainActivity:

public void onCreate(Bundle savedInstanceState)
{
   super.onCreate(savedInstanceState);
   setContentView(R.layout.main);

   GridView gridview = (GridView) findViewById(R.id.gridview);
   gridview.setAdapter(new ImageAdapter(this));
   gridview.setOnItemClickListener(new OnItemClickListener() {

       public void onItemClick(AdapterView<?> parent, View v, int position, long id) {
           Toast.makeText(nclApp2.this, "" + position, Toast.LENGTH_SHORT).show();
           Intent i = new Intent(nclApp2.this, Screen2.class);

           startActivity(i);
       }
   });

ImageAdapter:

public class ImageAdapter extends BaseAdapter {
private Context mContext;

public ImageAdapter(Context c) {
    mContext = c;
}

public int getCount() {
    return mThumbIds.length;
}

public Object getItem(int position) {
    return null;
}

public long getItemId(int position) {
    return 0;
}

// create a new ImageView for each item referenced by the Adapter
public View getView(int position, View convertView, ViewGroup parent) {
    ImageView imageView;
    if (convertView == null) {  // if it's not recycled, initialize some attributes
        imageView = new ImageView(mContext);
        imageView.setLayoutParams(new GridView.LayoutParams(85, 85));
        imageView.setScaleType(ImageView.ScaleType.CENTER_CROP);
        imageView.setPadding(8, 8, 8, 8);
    } else {
        imageView = (ImageView) convertView;
    }

    imageView.setImageResource(mThumbIds[position]);
    return imageView;
}

// references to our images
private Integer[] mThumbIds = {
        R.drawable.icon, R.drawable.icon,
        R.drawable.icon, R.drawable.icon,
        R.drawable.icon, R.drawable.icon,

};

class MyOnClickListener implements OnClickListener
{
 private final int position;

 public MyOnClickListener(int position)
 {
  this.position = position;
 }

 public void onClick(View v)
 {

 }
}

编辑:经过几个小时的搜索,我发现这有效!:

public void onItemClick(AdapterView<?> parent, View v, int position, long id) {
           Toast.makeText(nclApp2.this, "" + position, Toast.LENGTH_SHORT).show();
           //Intent i = new Intent(nclApp2.this, Screen2.class);

           Intent myIntent = null;
           if(position == 0){
               myIntent = new Intent(v.getContext(), Screen1.class);
           }
           if(position == 1){
               myIntent = new Intent(v.getContext(), Screen2.class);
           }
           if(position ==2){
               myIntent = new Intent(v.getContext(), Screen3.class);
           }
           startActivity(myIntent);
       }

1 个答案:

答案 0 :(得分:3)

你有

public void onItemClick(AdapterView<?> parent, View v, int position, long id) {

所以请使用int position部分......

switch (position) {
case 0:
// start one activity
break;
case 1:
// start another activity
break;
// etc.