2015-02-08 101 views
1

我试图从我的数据库在线获取图片,在我的“图片链接”这是我的表中的数据字段,我把那里我上传的图片的网址,但不幸的是它给了我这个错误。BitmapFactory无法解码流:Android中的java.io.FileNotFoundException

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); 
      } 
     }); 

    } 
+0

请显示一些代码! – AADProgramming 2015-02-08 07:22:17

+2

这将是因为文件“/http:/agustiniancampusevents.site40.net/newsDB/images/Visual%20Report%20Filipino%20Final-12%20copy.JPG”不存在。您是否将URL传递给从文件读取的方法? (URL不是文件) – immibis 2015-02-08 07:22:59

+0

如何将url传递给读取文件的方法? – 2015-02-08 07:25:15

回答

11

使用BitmapFactory.decodeStream而不是BitmapFactory.decodeFile

try (InputStream is = new URL(file_url).openStream()) { 
    Bitmap bitmap = BitmapFactory.decodeStream(is); 
} 
+0

“试用资源需要API级别19” – ibrahimyilmaz 2016-04-27 19:36:27

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