我正在尝试在Gridview中显示来自互联网的图像。当我运行我的程序时,catlog中没有错误,但我的屏幕上没有显示图像..
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>
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步骤?
先谢谢你的帮助,
Germain的。
答案 0 :(得分: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,所以:
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AsyncTask,请将params
的类型更改为Object
,以便将imageView
和fileUrl
传递给doInBackground()
或创建一个类使用构造函数的AyncTask,您可以传递两个对象并将它们存储在AsyncTask的本地变量中,以便您可以从doInBackground()
方法访问它们。
另请注意,如果您的网格中有更多项目,那么当它具有非空值时,您应该重用传入的convertView
。