Question

我想将一个图像从android上传到我的sql数据库，我有一个像这样的代码：

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());
    }
}

但与此同时我想将我的用户名也上传到我的数据库（假设我使用edittext检索用户名），有谁知道怎么做？我应该添加什么样的代码？

之前谢谢

我在数据库中的表应该是这样的：

ID |用户名|档案|

我可以用来上传字符串数据的JSON代码是这样的：

 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());
    }

我可以合并这些代码吗？还是有另一种方法从android同时上传文件和字符串？

之前谢谢

我的PHP代码：

<?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并添加了传递图像文件名作为FileBody

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

然后将MultiPartEntity传递给HttpPost。我没有发布完整的代码，因为它的大量评论和代码与您的问题无关。通过将图像作为FileBody传递，可以使用stand php文件处理代码获取图像（参见下文）。

  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;
          }

出于我的目的，上面的代码处于一个循环中，因为我处理了多个图像

$im = 'image_' . $i;

是指实体中图像的名称。

对不起，这篇短文我很忙。

忘了提到我没有使用Base64字符串方法的原因是它限制了你可以发送的图像的大小。实体中的FileBody方法是我发现的最佳方法。

您可以使用以下方式传递字符串：

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，而且您应该尽量减少对服务器的呼叫次数。只需使用适当的数据向您的nameValuePairs添加其他参数即可。

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

这很直截了当。您应该关注的是服务器端代码，因为它需要能够处理不同的数据。

JSON并将图像上传到服务器

2 个答案: