JSON並將圖像上傳到服務器

Question

我想將圖像從android上傳到我的sql數據庫，並且我有這樣的代碼：

private void uploadFile() {
    // TODO Auto-generated method stub
    Bitmap bitmapOrg= BitmapFactory.decodeFile(Environment.getExternalStorageDirectory().getAbsolutePath() +"/Chart1.png");
    ByteArrayOutputStream bao = new ByteArrayOutputStream();
    bitmapOrg.compress(Bitmap.CompressFormat.JPEG, 90, bao);
    byte [] ba = bao.toByteArray();
    String ba1=Base64.encodeBytes(ba);
    ArrayList nameValuePairs = new
    ArrayList();
    nameValuePairs.add(new BasicNameValuePair("image",ba1));
    try{
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new
    HttpPost("http://ipadress/base.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    is = entity.getContent();
    }catch(Exception e){
    Log.e("log_tag", "Error in http connection "+e.toString());
    }
}

但是同時我也想將我的用戶名也上傳到我的數據庫（假設我使用edittext檢索用戶名），有人知道該怎么做嗎？ 我應該添加哪種代碼？ 之前感謝

我的數據庫表應該是這樣的：

ID | 用戶名| 文件

我可以用來上傳字符串數據的JSON代碼是這樣的：

 private void uploadFile() {
    // TODO Auto-generated method stub
    String nama = getIntent().getStringExtra("user");
    Bitmap bitmapOrg= BitmapFactory.decodeFile(Environment.getExternalStorageDirectory().getAbsolutePath() +"/Chart1.png");
    ByteArrayOutputStream bao = new ByteArrayOutputStream();
    bitmapOrg.compress(Bitmap.CompressFormat.JPEG, 90, bao);
    byte [] ba = bao.toByteArray();
    String ba1=Base64.encodeBytes(ba);
    ArrayList nameValuePairs = new ArrayList();
    nameValuePairs.add(new BasicNameValuePair("image",ba1));
    nameValuePairs.add(new BasicNameValuePair("username",nama));
    try{
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new
    HttpPost("http://139.195.144.67/BloodGlucose/base2.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    is = entity.getContent();

    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse httpRespose = httpclient.execute(httppost);
   HttpEntity httpEntity = httpRespose.getEntity();
   InputStream in = httpEntity.getContent();
   BufferedReader read = new BufferedReader(new InputStreamReader(in));

   String isi= "";
   String baris= "";

   while((baris = read.readLine())!=null){
      isi+= baris;
   }

       //Jika isi tidak sama dengan "null " maka akan tampil Toast "Register Success" sebaliknya akan tampil "Register Failure"
       if(!isi.equals("null")){                  
           Toast.makeText(this, "Register Success", Toast.LENGTH_LONG).show();
       }else{
           Toast.makeText(this, "Register Failure", Toast.LENGTH_LONG).show();
       }

    }catch(Exception e){
    Log.e("log_tag", "Error in http connection "+e.toString());
    }

我可以合並這些代碼嗎？ 還是有另一種方式同時從android上傳文件和字符串？ 之前感謝

我的PHP代碼：

<?php
include_once("koneksi.php");

$username = $_REQUEST['username'];

$hasil = mysql_query("select (max(ID)+1)as newid  from userownfile"); 
$row = mysql_fetch_row($hasil); 

$base = $_REQUEST['image'];
$filename = $row[0] . ".jpg";
$buffer=base64_decode($base);
$path = "img/".$filename.".jpg";
$handle = fopen($path, 'wb');
$numbytes = fwrite($handle, $buffer);
fclose($handle);
$conn=mysql_connect("localhost","root","");
mysql_select_db("db_bloodglucose");


$sql = "insert into userownfile(username,file) values('$username','" . $path . "')";
mysql_query($sql);


$string= "select * from userownfile";
$my_string= mysql_query($string);
if($my_string){
   while($object= mysql_fetch_assoc($my_string)){
      $output[] = $object;
   }

   echo json_encode($output);

?>

Answer 1

在我的方法中，我使用了org.apache.http.entity.mime.MultipartEntity並將圖像文件名作為FileBody添加

entity.addPart("image_" + photo_count, new FileBody(
                        new File(failed.getFilenames()[i])));

然后將MultiPartEntity傳遞給HttpPost。 我沒有發布完整的代碼，因為它包含與您的問題無關的注釋和代碼。 通過將圖像作為FileBody傳遞，可以使用標准php文件處理代碼來獲取圖像（請參見下文）。

  if ((!empty($_FILES[$im])) && ($_FILES[$im]['error'] == 0)) {
              $newname = dirname(__FILE__) . '/../photo/' . $campaign . '/' . $fn;
              if (!file_exists($newname)) {
                  if (move_uploaded_file($_FILES[$im]['tmp_name'], $newname)) {
                      //$resp = "The file " . $fn . " has been uploaded";
                      //printf("%s", $resp);
                  } else {
                    $error = $error + 1;      
                  } 
              }else{
                //image file already exists
                $error = $error + 1;
              }
          } else {
              $error = $error +1;
          }

出於我的目的，上面的代碼處於循環中，因為我要處理多個圖像

$im = 'image_' . $i;

指實體中圖像的名稱。

對不起，我趕時間來了。

忘記提及我之所以不使用Base64字符串方法的原因是它限制了您可以發送的圖像大小。 我發現實體中的FileBody方法是最好的方法。

您可以使用以下方式傳遞字符串：

entity.addPart("address", new StringBody(failed[0].getAddress()));

HttpClient client = new DefaultHttpClient();
HttpConnectionParams.setConnectionTimeout(client.getParams(), 20000); // Timeout

MultipartEntity entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
entity.addPart("address", new StringBody("my address example"));
entity.addPart("image_0", new FileBody(new File("filename of image")));

HttpPost post = new HttpPost("server address");
post.setEntity(entity);

HttpResponse response  = client.execute(post);

Answer 2

是的，您可以並且應該最大程度地減少對服務器的呼叫次數。 只需將帶有適當數據的另一個參數添加到您的nameValuePairs 。

nameValuePairs.add(new BasicNameValuePair("image", image));
nameValuePairs.add(new BasicNameValuePair("username", username));

這很簡單。 您真正應該看的是服務器端代碼，因為這需要能夠處理不同的數據。