1

我在 phpmyadmin 中创建了一个名为 article 的数据库，其中包含有关文章和图像的信息（存储为 blob）。我能够从数据库中获取数据并将其显示在 android 列表视图中，但我不知道如何从数据库中检索 blob 图像并将其与相应的文章信息一起显示在 android 列表视图中。请帮助我..我做了很多谷歌搜索，但我没有做..任何指向教程的链接或发布的任何代码都会非常棒......我的项目需要这个..提前谢谢:)

我的php代码是-

<?php
header('Content-type: application/json');
mysql_query('SET CHARACTER SET utf8');
mysql_connect("localhost","root","");
mysql_select_db("reader");
$id=$_REQUEST['keyword'];
$id1=(int)$id;
$sql=mysql_query("SELECT * FROM article a WHERE a.a_id='{$id}'");
while($row=mysql_fetch_assoc($sql))
{
$row['a_thumbnail']=base64_encode($row['a_thumbnail']);
$row1= array_slice($row, 0, 2);
$row_slice=array_slice($row,2);
$output2=array_map('utf8_encode', $row_slice);
$output[]=array_merge((array)$row1, (array)$output2);
}
print(json_encode($output));
mysql_close();
?>

我为检索图像而编写的 java 代码是-

public class SearchByLikeCount extends ListActivity {
ArrayList<HashMap<String, String>> mylist = new ArrayList<HashMap<String, String>>();

@Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main2);
        String result = "";
        InputStream is= null;
        try{
                HttpClient httpclient = new DefaultHttpClient();
                HttpGet httpget = new HttpGet("http://10.0.2.2/likeCountsearch.php");
                HttpResponse response = httpclient.execute(httpget);
                HttpEntity entity = response.getEntity();
                is = entity.getContent();
        }catch(Exception e){
                Log.e("log_tag", "Error in http connection "+e.toString());
        }
        try{
                BufferedReader reader = new BufferedReader(new InputStreamReader(is,"utf-8"),8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                        sb.append(line + "\n");
                }
                is.close();

                result=sb.toString();
        }catch(Exception e){
                Log.e("log_tag", "Error converting result "+e.toString());
        }
        try{
                JSONArray jArray = new JSONArray(result);
                for(int i=0;i<jArray.length();i++){

                    HashMap<String,String> map = new HashMap<String, String>();
                    JSONObject e = jArray.getJSONObject(i);
                    Log.i("shruthi", "jason object length = " + e.length());
                    map.put("id",  e.getString("a_id"));
                    map.put("title",  e.getString("a_title"));
                    map.put("author",  e.getString("a_author"));
                    map.put("image", e.getString("a_thumbnail"));
                    mylist.add(map);
            }
               }catch(JSONException e)        {
                 Log.e("log_tag", "Error parsing data "+e.toString());
               }

        ListAdapter adapter = new SimpleAdapter(SearchByLikeCount.this, mylist , R.layout.main2,
        new String[] { "title", "author","image"},
        new int[] { R.id.item_title ,R.id.item_author,R.id.list_image});

        setListAdapter(adapter);

        final ListView lv = getListView();
        lv.setTextFilterEnabled(true);
        lv.setOnItemClickListener(new OnItemClickListener() {
        public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
        @SuppressWarnings("unchecked")
        HashMap<String, String> o = (HashMap<String, String>) lv.getItemAtPosition(position);
        Object v=o.get("id");
        String id1=v.toString();
        Intent myintent= new Intent(SearchByLikeCount.this,ViewArticle.class);
        myintent.putExtra("articleId", id1);
        startActivity(myintent);
        }
    }); 
 }
}

这没有用:(谁能告诉我我犯了什么错误？？

4

2 回答 2

0

也许您可以将图像存储在文件夹
中并将文件夹链接存储在数据库中，然后通过链接检索图像 

   try 
{
    Bitmap bitmap = BitmapFactory.decodeStream((InputStream)new URL(imageUrl).getContent());
    imageView.setImageBitmap(bitmap); 
} 
catch (MalformedURLException e) 
{
      e.printStackTrace();
} 
catch (IOException e)
 {
       e.printStackTrace();
 }
于 2013-05-22T09:40:58.347 回答
0

如果您从数据库中检索图像，那么您想使用哪个设备并不重要，图像就是图像

您可以执行以下操作

$im = imagecreatefromstring($imageContent);
if ($im !== false) {
    header('Content-Type: image/png'); // Change to what you want
    imagepng($im);
    imagedestroy($im);
}
else {
    echo 'An error occurred.';
}
于 2012-04-21T14:24:16.453 回答