We Keep Coding

JAVA POST Request MultiFormData with File and Parameters

I am trying to make a POST request using "multipart/form-data" , i need to post a file (Code Below) and 4 parameters(Name,category ...) all Strings.
I can already send the File using code below but never with parameters.
// open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("fileToUpload", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"fileToUpload\";filename=" + fileName + "" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
Log.i("uploadFile", "HTTP Response is : " + serverResponseMessage + ": " + serverResponseCode);
if (serverResponseCode == 200) {
runOnUiThread(new Runnable() {
public void run() {
Toast.makeText(PerdidosEAchados.this, "File Upload Complete.",
Toast.LENGTH_SHORT).show();
}
});
}
//close the streams //
fileInputStream.close();
dos.flush();
dos.close();
The server Code
<?php
echo $_POST["Name"]) ;
echo $_POST["category "]) ;
?>
I have tryed adding
dos.writeBytes("Content-Disposition: form-data; name=\"Name\";" + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(Variable);
But the server never registers the parameters, how can i solve this?

Maybe you shuold build the POST request like the forrowing demo code? Hope it helps.
### Send a form with the text and file fields
POST https://httpbin.org/post
Content-Type: multipart/form-data; boundary=WebAppBoundary
--WebAppBoundary
Content-Disposition: form-data; name="Name"
myName
--WebAppBoundary
Content-Disposition: form-data; name="category"
myCategory
--WebAppBoundary
Content-Disposition: form-data; name="data"; filename=".gitignore"
Content-Type: application/json
< ./.gitignore
--WebAppBoundary--
<> 2019-09-23T045805.200.json
###

Android multi-part request to web api not getting text associated with request

Hello I am working with android . I want to send a multi-part data to asp.net web api. I have a json data and multiple images.I used the following method to post data to api
public int uploadFile(Context context,ArrayList<String> sourceFileUri,String json,String url1) {
// Toast.makeText(this,"uploading...", Toast.LENGTH_SHORT).show();
StrictMode.ThreadPolicy policy = new
StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
Log.i("uploadimage", "" + sourceFileUri.get(0));
HttpURLConnection conn = null;
DataOutputStream dos = null;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
int bytesRead, bytesAvailable, bufferSize;
byte[] buffer;
int maxBufferSize = 1 * 1024 * 1024;
File sourceFile = new File(android.os.Environment.getExternalStorageDirectory() + "/TrackonPro", sourceFileUri.get(0));
if (!sourceFile.isFile()) {
Log.i("uploadFile", "Source File not exist :" + sourceFileUri);
return 0;
}
else
{
// Toast.makeText(this,"found Path : "+imagePath , Toast.LENGTH_LONG).show();
try {
// Toast.makeText(this,"try....", Toast.LENGTH_LONG).show();
// open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(url1);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
// Toast.makeText(this,"url : "+ url, Toast.LENGTH_LONG).show();
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
// JSON STRING
dos.writeBytes("Content-Disposition: form-data; name=\"sanu\"");
dos.writeBytes(lineEnd);
dos.writeBytes(json);
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
Log.i("immmm11",""+dos);
// IMAGE
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename="+ sourceFileUri + "" + lineEnd);
dos.writeBytes(lineEnd);
Log.i("immmm22", "" + dos);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
// Toast.makeText(this,": while", Toast.LENGTH_LONG).show();
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
Log.i("immmm33", "" + dos);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
Log.i("uploadFile", "HTTP Response is : "+ serverResponseMessage + ": " + serverResponseCode);
// Toast.makeText(getApplicationContext(), "HTTP Response is : "+ serverResponseMessage + ": " + serverResponseCode,
// Toast.LENGTH_SHORT).show();
if(serverResponseCode == 200){
// Toast.makeText(context, "File Upload Complete.",
// Toast.LENGTH_SHORT).show();
}
//close the streams //
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
ex.printStackTrace();
// Toast.makeText(context, "MalformedURLException",
// Toast.LENGTH_SHORT).show();
Log.e("Upload file to server", "error: " + ex.getMessage(), ex);
} catch (Exception e) {
e.printStackTrace();
// Toast.makeText(context, " Exception : try again ! ",
// Toast.LENGTH_SHORT).show();
// Log.i("Upload file to server Exception", "Exception : " + e.getMessage());
}
return serverResponseCode;
} // End else block
}
by using this I can able to get upload images to server.But text not received at server side. But http post request using Fidler app to web API get uploaded both images and text ? Is there any problem with this sending data. Please help me thanks in advance
UPDATE
now I got the error at server
System.IO.IOException: Error writing MIME multipart body part to output stream. ---> System.InvalidOperationException: The stream provider of type 'MultipartFormDataStreamProvider' threw an exception. ---> System.InvalidOperationException: Did not find required 'Content-Disposition' header field in MIME multipart body part.
at System.Net.Http.MultipartFormDataStreamProviderHelper.IsFileContent(HttpContent parent, HttpContentHeaders headers)
at System.Net.Http.MultipartFormDataStreamProvider.GetStream(HttpContent parent, HttpContentHeaders headers)
at System.Net.Http.MimeBodyPart.GetOutputStream()
--- End of inner exception stack trace ---
at System.Net.Http.MimeBodyPart.GetOutputStream()
at System.Net.Http.MimeBodyPart.<WriteSegment>d__0.MoveNext()
--- End of stack trace from previous location where exception was thrown ---
at System.Runtime.CompilerServices.TaskAwaiter.ThrowForNonSuccess(Task task)
at System.Runtime.CompilerServices.TaskAwaiter.HandleNonSuccessAndDebuggerNotification(Task task)
at System.Net.Http.HttpContentMultipartExtensions.<MultipartReadAsync>d__8.MoveNext()
--- End of inner exception stack trace ---
at System.Net.Http.HttpContentMultipartExtensions.<MultipartReadAsync>d__8.MoveNext()
--- End of stack trace from previous location where exception was thrown ---
at System.Runtime.CompilerServices.TaskAwaiter.ThrowForNonSuccess(Task task)
at System.Runtime.CompilerServices.TaskAwaiter.HandleNonSuccessAndDebuggerNotification(Task task)
at System.Net.Http.HttpContentMultipartExtensions.<ReadAsMultipartAsync>d__0`1.MoveNext()
--- End of stack trace from previous location where exception was thrown ---
at System.Runtime.CompilerServices.TaskAwaiter.ThrowForNonSuccess(Task task)
at System.Runtime.CompilerServices.TaskAwaiter.HandleNonSuccessAndDebuggerNotification(Task task)
at System.Runtime.CompilerServices.TaskAwaiter.ValidateEnd(Task task)
at WebApplication.Areas.Json.Controllers.TravelClaimsController.<PostFormData>d__2.MoveNext()

I've used this method to upload multipart request (binary data and text)
public class HttpClient {
....
where the methods are:
public void connectForMultipart() throws Exception {
con = (HttpURLConnection) ( new URL(url)).openConnection();
con.setRequestMethod("POST");
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestProperty("Connection", "Keep-Alive");
con.setRequestProperty("Content-Type", "multipart/form-data; boundary=" + boundary);
con.connect();
os = con.getOutputStream();
}
and
public void addFormPart(String paramName, String value) throws Exception {
writeParamData(paramName, value);
}
private void writeParamData(String paramName, String value) throws Exception {
os.write( (delimiter + boundary + "\r\n").getBytes());
os.write( "Content-Type: text/plain\r\n".getBytes());
os.write( ("Content-Disposition: form-data; name=\"" + paramName + "\"\r\n").getBytes());;
os.write( ("\r\n" + value + "\r\n").getBytes());
}
where
private String delimiter = "--";
private String boundary = "SwA"+Long.toString(System.currentTimeMillis())+"SwA";
The binary data (the image or something like that)
public void addFilePart(String paramName, String fileName, byte[] data) throws Exception {
os.write( (delimiter + boundary + "\r\n").getBytes());
os.write( ("Content-Disposition: form-data; name=\"" + paramName + "\"; filename=\"" + fileName + "\"\r\n" ).getBytes());
os.write( ("Content-Type: application/octet-stream\r\n" ).getBytes());
os.write( ("Content-Transfer-Encoding: binary\r\n" ).getBytes());
os.write("\r\n".getBytes());
os.write(data);
os.write("\r\n".getBytes());
}
When you call the HttpClient you add the parts:
HttpClient client = new HttpClient(url);
client.connectForMultipart();
client.addFormPart("param1", param1);
client.addFormPart("param2", param2);
client.addFilePart("file", "logo.png", baos.toByteArray());
client.finishMultipart();
and
public void finishMultipart() throws Exception {
os.write( (delimiter + boundary + delimiter + "\r\n").getBytes());
}
Hope it helps you. To have more info give a look at my blog post

Android to Php server communicaton using HttpURLConnection

My project is to upload the image, audio files with some parameter(like description and date).
Though Google announced to use HttpURLConnection instead of httpclient. I am using HttpURLConnection.
I have an code which upload the image and audio in server folder.
But the description which I send is not received in the server.
Like this question many in Stackover flow. But I did not get exact solution.
My android code is:
FileInputStream fileInputStream = new FileInputStream(sourceFile_image);
URL url = new URL(upLoadServerUri);
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type","multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
//adding parameter
String description = ""+"Desceiption about the image";
// Send parameter #name
dos.writeBytes("Content-Disposition: form-data; name=\"description\"" + lineEnd);
dos.writeBytes("Content-Type: text/plain; charset=UTF-8" + lineEnd);
dos.writeBytes("Content-Length: " + description.length() + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(description + lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
// Send #image
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
And Php Code:
$description= $_POST['description'];
$file_path = $file_path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
echo "success";
} else{
echo "fail";
}
Image and audio updating successfully.
But parameter not received or I dont know how to receive the parameter in php.
Is my android and php code to send and receive parameter is correct?
Is Any other solution.
I am trying lot but not works and not getting idea too.

check this link
Android code:
String description = ""+"Desceiption about the image";
dos.writeBytes("Content-Disposition: form-data; name=\"description\"" + lineEnd);
//dos.writeBytes("Content-Type: text/plain; charset=UTF-8" + lineEnd);
//dos.writeBytes("Content-Length: " + description.length() + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(description); // mobile_no is String variable
dos.writeBytes(lineEnd);
Php code:
$description =$_POST['description'];

// Send parameter #name
dos.writeBytes("Content-Disposition: form-data; name=\"name\"" + lineEnd);
That should be:
// Send parameter #description
dos.writeBytes("Content-Disposition: form-data; name=\"description\"" + lineEnd);

Error in moving variables from android class to php side [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
I am currently developing an app and i have a problem in transferring the variables from android to the php script. However the connection is established. Any help will be appreciated.
Here is my Java code.
// open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""
+ fileName + "\"" + lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";name=\""
+ username + "\"" + lineEnd);
dos.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dos.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
And this is my php script
<?php
require("config.inc.php");
$file_path = "uploads/";
$file_path = $file_path . basename( $_FILES['uploaded_file']['filename']);
$user = $_FILES['username'];
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
$query = "INSERT INTO users ( username, password, picture ) VALUES ( '1234567', $user, $file_path ) ";
$stmt = $db->prepare($query);
$stmt->execute();
echo "success";
} else{
echo "fail";
}
?>
The problem is in the section of setting the values and the variables are moved empty. In the database nothing is being added.

The reason why it fails is because you did not prepare your query properly:
$query = "INSERT INTO users ( username, password, picture ) VALUES ( ?, ?, ? ) ";
$stmt = $db->prepare($query);
$stmt->execute(array('1234567', $user, $file_path));

Android file uploader with server-side php

I've been looking for a solution to this for hours, but I can't find any.
Basically, I want to upload, from my android device, files to an http website. However, I have no clue whatsoever how to do this. I'm using java on the device, and I would like to use PHP on the server-side of things. I just want to upload the files, not do anything fancy with them on the server.
Can anyone provide code and/or a good link to what I need? I have little to no experience in this, and I am at a loss.
Thanks,
NS
PS. I have no experience in PHP coding.

Yeah, so I found the java-side of things. This works, so... yeah.
public class Uploader extends Activity {
private String Tag = "UPLOADER";
private String urlString = "YOUR_ONLINE_PHP";
HttpURLConnection conn;
/** Called when the activity is first created. */
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
String exsistingFileName = "/sdcard/uploader/data/testfile";
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
try {
// ------------------ CLIENT REQUEST
Log.e(Tag, "Inside second Method");
FileInputStream fileInputStream = new FileInputStream(new File(
exsistingFileName));
// open a URL connection to the Servlet
URL url = new URL(urlString);
// Open a HTTP connection to the URL
conn = (HttpURLConnection) url.openConnection();
// Allow Inputs
conn.setDoInput(true);
// Allow Outputs
conn.setDoOutput(true);
// Don't use a cached copy.
conn.setUseCaches(false);
// Use a post method.
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("Content-Type",
"multipart/form-data;boundary=" + boundary);
DataOutputStream dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos
.writeBytes("Content-Disposition: post-data; name=uploadedfile;filename="
+ exsistingFileName + "" + lineEnd);
dos.writeBytes(lineEnd);
Log.e(Tag, "Headers are written");
// create a buffer of maximum size
int bytesAvailable = fileInputStream.available();
int maxBufferSize = 1000;
// int bufferSize = Math.min(bytesAvailable, maxBufferSize);
byte[] buffer = new byte[bytesAvailable];
// read file and write it into form...
int bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
while (bytesRead > 0) {
dos.write(buffer, 0, bytesAvailable);
bytesAvailable = fileInputStream.available();
bytesAvailable = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bytesAvailable);
}
// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
// close streams
Log.e(Tag, "File is written");
fileInputStream.close();
dos.flush();
dos.close();
} catch (MalformedURLException ex) {
Log.e(Tag, "error: " + ex.getMessage(), ex);
}
catch (IOException ioe) {
Log.e(Tag, "error: " + ioe.getMessage(), ioe);
}
try {
BufferedReader rd = new BufferedReader(new InputStreamReader(conn
.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
Log.e("Dialoge Box", "Message: " + line);
}
rd.close();
} catch (IOException ioex) {
Log.e("MediaPlayer", "error: " + ioex.getMessage(), ioex);
}
}
}

If you want to upload files that is larger than 2MB you need to edit your PHP configuraton
; Maximum allowed size for uploaded files.
; http://php.net/upload-max-filesize
upload_max_filesize = 2M
Eg:
; Maximum allowed size for uploaded files.
; http://php.net/upload-max-filesize
upload_max_filesize = 10M
That would allow you to upload files up to 10MB of size

For the PHP side of things take a look at
move_uploaded_file and also the global $_FILES
http://php.net/manual/en/function.move-uploaded-file.php
http://php.net/manual/en/features.file-upload.php