Question

我想将我的图像数据（文件）和字符串数据（用户名）从android上传到服务器上的mysql数据库

我的JSON代码：

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("file",ba1));
        nameValuePairs.add(new BasicNameValuePair("username",nama));
        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();

        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());
        }

  }

和我的PHP代码：

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

$username = $_REQUEST['username'];
$file = $_REQUEST['file'];

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

$base = $_REQUEST['file'];
$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");

if($username && $file){
$sql = "insert into userownfile(username,file) values('$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);

?>

我在数据库上的表应该是这样的：

ID |用户名|文件

但是当我尝试上传时，图像和用户名没有插入我的数据库，我错过了什么？有谁可以帮我修复代码？

之前谢谢

Answer 1

使用类似Fiddler的东西来确保您发送到服务器的数据包是服务器所期望的数据包。

如果您想将文本格式化为JSON，我建议您使用内置类并执行以下操作：

JSONObject json = new JSONObject();
json.put("file", ba1);
json.put("username", nama);

用于将数据上载到服务器的JSON代码

1 个答案: