How to upload Image on Android?
Posted
by
Mattiah85
on Stack Overflow
See other posts from Stack Overflow
or by Mattiah85
Published on 2014-06-04T06:55:13Z
Indexed on
2014/06/04
9:25 UTC
Read the original article
Hit count: 234
I havve to upload image from my SD card to PHP server. I have read a lot of articles and topics but I have some problems...
First I have use that code:
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
//DataInputStream inputStream = null;
String urlServer = hostName+"Upload";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
String serverResponseMessage;
//int serverResponseCode;
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1*1024*1024;
try
{
showLog("uploading file: " + file);
FileInputStream fileInputStream = new FileInputStream(new File(pictureFileDir+"/"+file) );
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs.
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// Set HTTP method to POST.
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type", "multipart/form-data;boundary="+boundary);
outputStream = new DataOutputStream( connection.getOutputStream() );
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\"" + file +"\"" + lineEnd);
outputStream.writeBytes(lineEnd);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// Read file
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0)
{
outputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
//serverResponseCode = connection.getResponseCode();
serverResponseMessage = connection.getResponseMessage();
showLog("server response: " + serverResponseMessage);
fileInputStream.close();
outputStream.flush();
outputStream.close();
}
catch (Exception ex)
{
ex.printStackTrace();
}
but server response 200/OK and no file was on destination server...
After i have read about Multipart:
try {
HttpParams params = new BasicHttpParams();
params.setParameter(CoreProtocolPNames.PROTOCOL_VERSION, HttpVersion.HTTP_1_1);
DefaultHttpClient mHttpClient = new DefaultHttpClient(params);
File image = new File(pictureFileDir + "/" + filename);
HttpPost httppost = new HttpPost(hostName+"Upload");
MultipartEntity multipartEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
multipartEntity.addPart("Image", new FileBody(image));
httppost.setEntity(multipartEntity);
mHttpClient.execute(httppost, new PhotoUploadResponseHandler());
} catch (Exception e) {
e.printStackTrace();
}
but then a i have such LOG in LogCat and nothing else...
06-04 06:50:52.277: D/dalvikvm(1584): DexOpt: couldn't find static field Lorg/apache/http/message/BasicHeaderValueParser;.INSTANCE 06-04 06:50:52.277: W/dalvikvm(1584): VFY: unable to resolve static field 6688 (INSTANCE) in Lorg/apache/http/message/BasicHeaderValueParser; 06-04 06:50:52.277: D/dalvikvm(1584): VFY: replacing opcode 0x62 at 0x001b
ServerSide Script:
$target_path = "uploads";
$target_path = $target_path . basename( $_FILES['Image']);
if(move_uploaded_file($_FILES['tmp_name'], $file_path)) {
echo "success";
} else{
echo "fail";
}
why? What is the simplest way to upload image?
© Stack Overflow or respective owner