如何将android捕获的图像发送到服务器并将其保存为blob？

Question

我正在尝试创建一个可以拍照并将其发送到服务器的应用，然后将其作为blob保存到数据库。

以错误的方式输入图像数据。我正在使用assynchttp post方法 和这段代码：

PictureCallback mPicture = new PictureCallback() {

    @Override
    public void onPictureTaken(byte[] data, Camera camera) {
        Bitmap m_bitmap = BitmapFactory.decodeByteArray(data , 0, data .length);
        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        m_bitmap.compress(Bitmap.CompressFormat.JPEG, 50, stream);
        byte[] byteArray = stream.toByteArray();
        String encodedImage = Base64.encodeToString(byteArray,Base64.DEFAULT);
        try {
            RequestParams params = new RequestParams();

            params.put("city", city2);
            params.put("country", country);
            params.put("langi", langi);
            params.put("lati", lati);
            params.put("details", details);
            params.put("type", type);
            params.put("ID", myID);
            params.put("image", encodedImage);
            AsyncHttpClient client = new AsyncHttpClient();
            client.post(getResources().getString(R.string.app_server_Name)
             + "addnewplace.php", params, new TextHttpResponseHandler() {


                @Override
                public void onStart() {
                     Toast.makeText(getApplicationContext(),"saving...", Toast.LENGTH_SHORT).show();
                }

                @Override
                public void onFailure(int statusCode, Header[] headers, String responseString, Throwable throwable) {

                        Toast.makeText(getApplicationContext(), responseString, Toast.LENGTH_SHORT).show();
                }

                @Override
                public void onSuccess(int statusCode, Header[] headers, String responseString) {

                        Toast.makeText(getApplicationContext(), responseString, Toast.LENGTH_SHORT).show();
                }                        

                @Override
                public void onFinish() {

                }
            });

        } catch (Exception e) {
            // TODO: handle exception
        } finally {
            frameLayout.destroyDrawingCache();
        }     
    }
}; 

这是我的PHP代码：

header("Content-Type: application/json;charset=utf-8");

$image     = base64_decode($_POST['image']); 
$city      = $_POST['city']; 
$country   = $_POST['country'];
$langi     = $_POST['langi'];
$lati      = $_POST['lati'];
$details   = $_POST['details'];
$type      = $_POST['type'];
$ID        = $_POST['ID']; 
$latFrom   = deg2rad($lati);
$lonFrom   = deg2rad($langi);
$user_name = "id1015576_placeitnowapp"; 
$password  = "place311311place";
$database  = "id1015576_placeitnow"; 
$server    = "localhost"; $conn = new mysqli($server , $user_name , $password,$database ); 
$conn2     = new mysqli($server , $user_name , $password,$database ); 

$sql1 = "INSERT 
        INTO `places`(`PlaceName`, `PlaceLocationLongitude`, `PlaceLocationLatitude`, `Type`, `image`, `VisitorID`)
        VALUES('".$details."','".$langi."','".$lati."','".$type."','".$image."','".$ID."')";
$result2 = $conn2->query($sql1);

if ($result2) {
    $response = array('status' => true );
} else {
    $response = array('status' => false );
}

$myJSN = json_encode($response);
echo $myJSN;