2015-02-08 105 views
1

我試圖從我的數據庫在線獲取圖片,在我的「圖片鏈接」這是我的表中的數據字段,我把那裏我上傳的圖片的網址,但不幸的是它給了我這個錯誤。BitmapFactory無法解碼流: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) 

這是我在onPostExecute代碼:

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

請顯示一些代碼! – AADProgramming 2015-02-08 07:22:17

+2

這將是因爲文件「/http:/agustiniancampusevents.site40.net/newsDB/images/Visual%20Report%20Filipino%20Final-12%20copy.JPG」不存在。您是否將URL傳遞給從文件讀取的方法? (URL不是文件) – immibis 2015-02-08 07:22:59

+0

如何將url傳遞給讀取文件的方法? – 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

「試用資源需要API級別19」 – ibrahimyilmaz 2016-04-27 19:36:27

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