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
###
Related
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
I am trying to upload an image file from an android app using java to an aspnet webapi with no success. I always get a response code of 500. Below is the basic code for preparing the sending of data from my android app. This code works when calling a php script but not when I call a web api controller shown below.
FileInputStream fileInputStream = new FileInputStream(sourceFile);
url = new URL("http://www.myserver.com/webapi/api/fileupload/post/thisfile");
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=\"" + sourceFile +"\"" + 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();
Below is the basic web api controller code to do the file upload:
[HttpPost]
[AcceptVerbs("GET", "POST")]
public Task<HttpResponseMessage> Post()
{
// Check if the request contains multipart/form-data.
if (!Request.Content.IsMimeMultipartContent())
{
// return "Unsupported media type";
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = "e:/web/myserver/htdocs/cemetery/pics/";
var provider = new MultipartFormDataStreamProvider(root);
return task;
}
I changed the code above to send back a string and it says that my request does not contain multipart/form-data. Below is my webapiconfig.cs file route for this controller.
config.Routes.MapHttpRoute(
name: "FileUploadApi",
routeTemplate: "api/{controller}/{action}/{name}",
defaults: new { action = "post", name = RouteParameter.Optional }
);
I appreciate any help in determining what is wrong with my code so I can upload a file from an android java app to a webapi controller.
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);
I'm trying to post a file and data to my server from an Android java script (to my php), it seems I'm pretty much there, but someone PLEASE help me, as I can't seem to format the name/value part (the FILE uploads perfect, but it doesn't send the name value :( )
JAVA:
try{
int serverResponseCode = 0;
final String upLoadServerUri = "http://myUrl/upload_file_functions.php";
String fileName = "/mnt/sdcard/myFile.dat";
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("/mnt/sdcard/myFile.dat");
// 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("file", fileName);
conn.setRequestProperty("gmail", names[0]);
conn.setRequestProperty("phn", phn);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"file\";filename=\""
+ fileName + "\"" + lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
String data = URLEncoder.encode("gmail", "UTF-8") + "=" + URLEncoder.encode(names[0], "UTF-8");
dos.writeBytes(data);
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);
dos.writeBytes("Content-Disposition: form-data; name=\"" + names[0] + "\"" + lineEnd);
dos.writeBytes("Content-Type: text/plain"+lineEnd);
dos.writeBytes(lineEnd);
dos.writeBytes(names[0]);
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){
// it worked !
}
//close the streams //
fileInputStream.close();
dos.flush();
dos.close();
}catch (Exception e){
}
It doesn't work, the FILE sends fine, but I can't get a freaking key/name to send ("gmail:"names[0]) I've also tried:
// 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("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("gmail", names[0]);
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"file\";filename=\""
+ fileName + "\"" + lineEnd);
dos.writeBytes(twoHyphens + boundary + lineEnd);
DOESN'T WORK. I've tried:
dos.writeBytes("Content-Disposition: form-data; name=\"gmail\";filename=\""+ names[0] + "\"" + lineEnd);
Doesn't WORK! WTF! I've programmed for years in C++ and python, it's a simple thing! But I can't figure this out I need help, if you know how to do it PLEASE DO TELL because I've spent two days banging my head against the wall. I'm not lazy I spent 32+ f'n hours on this please I beg you..
What I WANT to happen: send the file for upload, along with the value (name=gmail value=names[0]; name=phn value=phn), so that the email is associated to the file on my server.
What IS happening: file is uploading fine, but data is not passed (the name/value pairs are not sent)
PHP:
<?php
set_time_limit(100);
//need to get email also (gmail address of user)
//************************************************
if ($_FILES["file"]["error"] > 0)
{
echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Stored in: " . $_FILES["file"]["tmp_name"];
move_uploaded_file($_FILES["file"]["tmp_name"],
"upload/" . $_FILES["file"]["name"]);
echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
}
}
function Three(){
$to = 'me#email.com';
$subject = $_POST['phn'] . " " . $_POST['gmail'];
$bound_text = "file";
$bound = "--".$bound_text."\r\n";
$bound_last = "--".$bound_text."--\r\n";
$headers = "From: me#email.com\r\n";
$headers .= "MIME-Version: 1.0\r\n"
."Content-Type: multipart/mixed; boundary=\"$bound_text\"";
$message .= "If you can see this MIME than your client doesn't accept MIME types!\r\n"
.$bound;
$message .= "Content-Type: text/html; charset=\"iso-8859-1\"\r\n"
."Content-Transfer-Encoding: 7bit\r\n\r\n"
."hey my <b>good</b> friend here is a picture of regal beagle\r\n"
.$bound;
$file = file_get_contents("http://myURL/upload/myFile.dat");
$message .= "Content-Type: image/jpg; name=\"myFile.dat\"\r\n"
."Content-Transfer-Encoding: base64\r\n"
."Content-disposition: attachment; file=\"myFile.dat"\r\n"
."\r\n"
.chunk_split(base64_encode($file))
.$bound_last;
#mail($to, $subject, $message, $headers);
//delete files
$fileArray=array($_FILES["file"]["name"],"myfile.dat","myfile.dat");
foreach($fileArray as $value){
if(file_exists($value)){
unlink($value);
}
}
chdir($old_path);
}
function runAll(){
One();
Two();
Three();
}
runAll();
$randx=null;
unset($randx);
?>
PLEASE HELP! The JAVA is not sending the name='gmail' value=names[0], nor the name='phn' value=phn ..
You should really read up on a few things: HTTP (and the distinction between request header fields and the post body), and the structure of a multipart/form-data post body.
This:
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("file", fileName);
conn.setRequestProperty("gmail", names[0]);
conn.setRequestProperty("phn", phn);
sends a few request headers, which is fine for Content-Type and such, but not necessarily for the data you're posting. Lose all but the Content-Type line.
This:
dos.writeBytes(twoHyphens + boundary + lineEnd);
is the right way to start a post field (or file), you should output this for every posted field.
This:
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
signals that all fields have been sent, you should send this as the very last line.
Either use something like Wireshark to see what your final request looks like (either side will do; the device doing the request or the server handling it), or log your request so you can inspect it, and see if it is perfect. It has to be nearly perfect for the webserver/php to process it correctly.
Well, I never figured out how to send a simple string parameter with the file upload, but my workaround was to simply append the filename of the upload file to INCLUDE the string I wanted to send:
#SuppressWarnings("deprecation")
#Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
addPreferencesFromResource(R.xml.preferences);
try{
Account[] accounts=AccountManager.get(this).getAccountsByType("com.google");
String myEmailid=accounts[0].toString(); Log.d("My email id that i want", myEmailid);
String[] names = new String[accounts.length];
for (int i = 0; i < names.length; i++) {
names[i] = accounts[i].name;
}
// THE DEVICE EMAIL ADDRESS WAS ONE OF THE DATA STRINGS I NEEDED TO SEND
File from = new File("/mnt/sdcard/","UPLOADFILE.DAT");
File to = new File("/mnt/sdcard/",names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2");
from.renameTo(to);
// DATASTRING2 is the SECOND piece of DATA I wanted to send
// SO YOU SEE I'M SIMPLY APPENDING THE UPLOAD FILE WITH THE DATA I WANT
// TO SEND WITH THE FILE, AND WHEN MY SERVER RECEIVES IT, I USE SOME SIMPLE
// PHP TO PARSE OUT WHATS BEFORE .BLOCK1. AND THEN .BLOCK2
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
try{
int serverResponseCode = 0;
final String upLoadServerUri = "http://MYURL/upload_file_functions.php";
String fileName = "/mnt/sdcard/"+names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2";
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("/mnt/sdcard/"+names[0]+".BLOCK1."+"DATASTRING2"+".BLOCK2"");
// 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("Content-Type", "multipart/form-data;boundary=" + boundary);
dos = new DataOutputStream(conn.getOutputStream());
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"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);
// Responses from the server (code and message)
serverResponseCode = conn.getResponseCode();
String serverResponseMessage = conn.getResponseMessage();
Log.i("uploadFile", "HTTP Response is : "
+ serverResponseMessage + ": " + serverResponseCode);
//close the streams //
fileInputStream.close();
dos.flush();
dos.close();
}catch (Exception e){
}
}catch (Exception e){
}
}
Thanks for the help! (THICK WITH SARCASM..) Seriously, I have to code in python, C++, java, PHP, on linux, Android, iPhone, Windows, MAC and Ubuntu... my work has me building Triple-Boot OS box's, building apps in Android and iPhone to support my business needs, and thus I must know PHP html and the like, needed to configure my own server because I needed email-notification services, so I needed to run my own email server (Exim on Ubuntu Server), and I've had to learn all this in the past 6 months.
Forgive me if my code's not pretty, but I don't have the luxury of time, as I'm going INSANE trying to keep up in the land of AI, Object recognition, and trying to make the rent (though I've learned enough to do so..)
DC:)
This can be same and can be seen as duplicate since there are many questions about uploading an image. But I want to know how to upload MANY images at a time to a servlet.That said, if there are 6 images in SD card, all should be uploaded within one request,not one by one. Most of samples over the internet are regarding one image or one file. I want to know if images are stored in a ArrayList, how they can be uploaded ?
In Servlet
List<FileItem> multiparts = new ServletFileUpload(
new DiskFileItemFactory()).parseRequest(request);
for(FileItem item : multiparts){
if(!item.isFormField()){
String name = new File(item.getName()).getName();
item.write( new File(UPLOAD_DIRECTORY + File.separator + name));
}
}
This part works fine as I tested with normal JSP multiplart many images uploads.
In android for one image upload (took from another source):
FileInputStream fileInputStream = new FileInputStream(new File(exsistingFileName) );
URL url = new URL(urlString);
conn = (HttpURLConnection) url.openConnection();
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);
dos = new DataOutputStream( conn.getOutputStream() );
dos.writeBytes(twoHyphens + boundary + lineEnd);
dos.writeBytes("Content-Disposition: form-data; name=\"uploadedfile\";filename=\"" + exsistingFileName +"\"" + lineEnd);
dos.writeBytes(lineEnd);
Log.e("MediaPlayer","Headers are written");
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
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);
}
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);
BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
tv.append(inputLine);
// close streams
Log.e("MediaPlayer","File is written");
fileInputStream.close();
dos.flush();
dos.close();
This uploads only one image. How this can be used to send many with one POST request.? any other sample code or tutorial please.