我需要通过在c#中调用restful webservice从我的android应用程序上传图像,但是当我通过向JSONObject添加byte []来尝试它时,它将byte []转换为字符串并且c#service抛出Bad请求(& #34;从命名空间反序列化Object.Element时出错'期望。发现文本' [B @ 22b6cc7f'。
Android Code:
Bitmap bitmap = BitmapFactory.decodeResource(getResources(),R.drawable.ttulips);
ByteArrayOutputStream stream = new ByteArrayOutputStream();
bitmap.compress(Bitmap.CompressFormat.PNG, 100, stream);
bitMapData = stream.toByteArray();
JSONObject jsonParam = new JSONObject();
try {
jsonParam.put("IncomingFile",bitMapData);
jsonParam.put("FileName", "name.jpg");
Log.d("Json",jsonParam+"");
} catch (JSONException e) {
e.printStackTrace();
}
JSON请求的日志即将发布 {" IncomingFile":" [B @ 22b67f""文件名":" name.jpg"}
甚至尝试将字节数组转换为Base64编码的字节数组,但在将base64字节数组添加到jsonobject时,它被视为字符串。
我应该如何解决这个问题?提前致谢。
答案 0 :(得分:1)
尝试将此位图转换为字符串并将此字符串传递给c#server
if(fileUri1 != null) {
bitmap1 = BitmapFactory.decodeFile(fileUri1.getPath(),
options);
ByteArrayOutputStream baos1 = new ByteArrayOutputStream();
if(bitmap1 != null) {
bitmap1.compress(Bitmap.CompressFormat.PNG, 50, baos1);
byte[] b1 = baos1.toByteArray();
bitmapstring1 = Base64.encodeToString(b1,
Base64.DEFAULT);
}
}
webservice电话:
public class CallWebService extends AsyncTask<Void, Void, Void> {
@Override
protected Void doInBackground(Void... params) {
// Call Webservice for Get Menus
WebServiceCall webServiceCall = new WebServiceCall(); // Custom class for call webservice
BitmapFactory.Options options = new BitmapFactory.Options();
// downsizing image as it throws OutOfMemory Exception for larger
// images
options.inSampleSize = 8;
parameters = new ArrayList<NameValuePair>();
parameters.add(new BasicNameValuePair("Name",uname12));
parameters.add(new BasicNameValuePair("Address", uaddr12));
parameters.add(new BasicNameValuePair("Email", en));
parameters.add(new BasicNameValuePair("Qualification", uquali12));
parameters.add(new BasicNameValuePair("Phoneno", ucontactno12));
parameters.add(new BasicNameValuePair("Appliedfor", uappfor12));
parameters.add(new BasicNameValuePair("Image", bitmapstring));
parameters.add(new BasicNameValuePair("Resumeimage", bitmapstring1));
parameters.add(new BasicNameValuePair("Operation", "i"));
Log.i("param::",parameters.toString());
response = webServiceCall.makeServiceCall(mUrlWebServiceLogin, parameters);
Log.d("ResponseLogin:", response);
return null;
}
@Override
protected void onPostExecute(Void result) {
if (progressDialog.isShowing())
progressDialog.dismiss();
if(response.contains("\"success\"")){
session.createLoginSession(uname12);
Toast.makeText(getApplicationContext(),"Successfully inserted",Toast.LENGTH_SHORT).show();
Intent in = new Intent(getApplicationContext(),InterView.class);
in.putExtra("Name",uname12);
startActivity(in);
finish();
}else{
Toast.makeText(getApplicationContext(),"data not inserted",Toast.LENGTH_SHORT).show();
}
}
@Override
protected void onPreExecute() {
progressDialog = new ProgressDialog(MainActivity.this);
progressDialog.setMessage("Loading...");
progressDialog.show();
progressDialog.setCanceledOnTouchOutside(false);
super.onPreExecute();
}
}
答案 1 :(得分:0)
双方使用Base64
,实际上将byte[]
转换为String
不是主要问题
read here
答案 2 :(得分:0)
请尝试以下创建JSON:
String json = "{\"IncomingFile\":\""+ new String(bytesEncoded) +"\",\"FileName\":\""+name.jpg+"\"}";
其中
bytesEncoded
应为Base64编码图像。
希望它有所帮助!
答案 3 :(得分:-1)
这里对象bitMapData
是一个bytearray,必须转换为String然后你必须在Json对象中使用
jsonParam.put("IncomingFile",new String(bitMapData));