我试图从我的在线数据库中获取图片,在我的表中的数据字段“imagelink”中,我将我上传的图片的 url 放在那里,但不幸的是它给了我这个错误。
02-08 15:05:29.432 14364-14364/com.example.jithea.testlogin E/BitmapFactory﹕ Unable to decode stream: java.io.FileNotFoundException: /http:/agustiniancampusevents.site40.net/newsDB/images/Visual%20Report%20Filipino%20Final-12%20copy.JPG: open failed: ENOENT (No such file or directory)
这是我在 onPostExecute 中的代码:
protected void onPostExecute(String file_url) {
// dismiss the dialog after getting all products
pDialog.dismiss();
// updating UI from Background Thread
runOnUiThread(new Runnable() {
public void run() {
/**
* Updating parsed JSON data into ListView
* */
ListAdapter adapter = new SimpleAdapter(
NewsActivity.this, productsList,
R.layout.news_list_item, new String[]{TAG_PID, TAG_IMAGELINK,
TAG_NEWSTITLE, TAG_DESCRIPTION},
new int[]{R.id.pid, R.id.imageView, R.id.newstitle, R.id.description});
// updating listview
setListAdapter(adapter);
}
});
}
最佳答案
使用 BitmapFactory.decodeStream
而不是 BitmapFactory.decodeFile
。
try ( InputStream is = new URL( file_url ).openStream() ) {
Bitmap bitmap = BitmapFactory.decodeStream( is );
}
关于java - BitmapFactory 无法在 Android 中解码流 : java. io.FileNotFoundException,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28391597/