来自互联网的Gridview图像

Question

我想在Gridview中显示来自互联网的图像。当我运行我的程序没有在错误目录下载，但没有图像显示在我的屏幕..

ImageAdapter.java：

public class ImageAdapter extends BaseAdapter { 
private Context mContext; 
Bitmap bmImg; 
ImageView imageView; 

public String[] mThumbIds = { 
     "http://i724.photobucket.com/albums/ww246/bnhenson/squishy.jpg", "http://cdn4.teen.com/wp-content/uploads/2012/07/paranorman-zayn-malik-tweets.jpg", 
     "http://xc0.xanga.com/08fe2b7a30c37281496919/m224298958.jpg" 
}; 

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

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

@Override 
public Object getItem(int position) { 
    return mThumbIds[position]; 
} 

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

@Override 
public View getView(int position, View convertView, ViewGroup parent) {   
    imageView = new ImageView(mContext); 
    downloadFile(mThumbIds[position]); 
    imageView.setScaleType(ImageView.ScaleType.CENTER_CROP); 
    imageView.setLayoutParams(new GridView.LayoutParams(135, 135)); 
    imageView.setPadding(0, 0, 1, 0); 
    return imageView; 
} 
void downloadFile(String fileUrl) { 

    AsyncTask<String, Object, String> task = new AsyncTask<String, Object, String>() { 

     @Override 
     protected String doInBackground(String... params) { 
      URL myFileUrl = null; 
      try { 
       myFileUrl = new URL(params[0]); 
      } catch (MalformedURLException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 
      try { 
       HttpURLConnection conn = (HttpURLConnection) myFileUrl 
         .openConnection(); 
       conn.setDoInput(true); 
       conn.connect(); 
       InputStream is = conn.getInputStream(); 

       bmImg = BitmapFactory.decodeStream(is); 

      } catch (IOException e) { 
       // TODO Auto-generated catch block 
       e.printStackTrace(); 
      } 

      return null; 
     } 
     protected void onPostExecute(String unused) { 
      imageView.setImageBitmap(bmImg); 
     } 
    }; 
    task.execute(fileUrl); 

}} 

AndroidGridLayoutActivity.java

public class AndroidGridLayoutActivity extends Activity { 

@Override 
public void onCreate(Bundle savedInstanceState) { 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.grid_layout); 

    GridView gridView = (GridView) findViewById(R.id.grid_view); 

    // Instance of ImageAdapter Class 
    gridView.setAdapter(new ImageAdapter(this)); 
}} 

grid_layout。 xml

<?xml version="1.0" encoding="utf-8"?> 
<GridView xmlns:android="http://schemas.android.com/apk/res/android" 
android:id="@+id/grid_view" 
android:layout_width="fill_parent" 
android:layout_height="fill_parent" 
android:numColumns="auto_fit" 
android:columnWidth="90dp" 
android:horizontalSpacing="10dp" 
android:verticalSpacing="10dp" 
android:gravity="center" 
android:stretchMode="columnWidth" > 

</GridView> 

而在我的清单中，我添加了这两行：

<uses-permission android:name="android.permission.INTERNET"></uses-permission> 
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" /> 

有没有人有想法，我在这里错过了什么？

在此先感谢您的帮助，

Germain。

Answer 1

您应该为网格的每个单元格创建一个单独的ImageView对象，而不是使用全局变量。要做到这一点，这样类似的东西：

public View getView(int position, View convertView, ViewGroup parent) { 
    ImageView imageView = new ImageView(mContext); 
    downloadFile(imageView, mThumbIds[position]); 
    imageView.setScaleType(ImageView.ScaleType.CENTER_CROP); 
    imageView.setLayoutParams(new GridView.LayoutParams(135, 135)); 
    imageView.setPadding(0, 0, 1, 0); 
    return imageView; 
} 

void downloadFile(ImageView imageView, String fileUrl) { 
    // ... 
    // change your AsyncTask so you can pass imageView as well 
    // instead of setting loaded bitmap to global mImageView, set it to the passed in imageView. 
    // ... 
} 

要更改您的AsyncTask，可以更改为params到Object的类型，所以你可以通过imageView和fileUrl到doInBackground()或创建一个类为您AyncTask与构造，你可以传递两个对象并将它们存储在AsyncTask本地的变量中，以便您可以从doInBackground()方法访问它们。

另请注意，如果网格中包含更多项目，则当它具有非空值时，应重新使用在convertView中传递的值。