Best approach: HTTP POST (multi-part) from Android to GAE - java

I would like to capture an image from the camera on Android, and send it to Google App Engine, which will store the image in the blob store. Sounds simple enough, and I can get the multi-part POST to GAE happening, but storing to the Blob store requires the servlet return an HTTP redirect (302). So, I need a connection that can follow redirects after doing an HTTP POST. Here is the code I WISH would work:
public static String sendPhoto(String myUrl, byte[] imageData) {
HttpURLConnection connection = null;
DataOutputStream outputStream = null;
String pathToOurFile = "/data/file_to_send.jpg";
String urlServer = myUrl;
String lineEnd = "\r\n";
String twoHyphens = "--";
String boundary = "*****";
// Please follow redirects
HttpURLConnection.setFollowRedirects(true);
BufferedReader rd = null;
StringBuilder sb = null;
String line = null;
try {
URL url = new URL(urlServer);
connection = (HttpURLConnection) url.openConnection();
// Allow Inputs & Outputs
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setUseCaches(false);
// I really want you to follow redirects
connection.setInstanceFollowRedirects(true);
connection.setConnectTimeout(10000); // 10 sec
connection.setReadTimeout(10000); // 10 sec
// Enable POST method
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type",
"multipart/form-data;boundary=" + boundary);
connection.connect();
outputStream = new DataOutputStream(connection.getOutputStream());
outputStream.writeBytes(twoHyphens + boundary + lineEnd);
outputStream
.writeBytes("Content-Disposition: form-data; name="
+ "\"file1\";filename=\""
+ pathToOurFile + "\"" + lineEnd);
outputStream.writeBytes(lineEnd);
outputStream.write(imageData);
outputStream.writeBytes(lineEnd);
outputStream.writeBytes(twoHyphens + boundary + twoHyphens
+ lineEnd);
outputStream.flush();
outputStream.close();
rd = new BufferedReader(new InputStreamReader(
connection.getInputStream()));
sb = new StringBuilder();
while ((line = rd.readLine()) != null) {
sb.append(line + '\n');
}
Log.i("Response: ", sb.toString());
// Responses from the server (code and message)
int serverResponseCode = connection.getResponseCode();
String serverResponseMessage = connection.getResponseMessage();
Log.i("Response: serverResponseCode:",
String.valueOf(serverResponseCode));
Log.i("Response: serverResponseMessage:", serverResponseMessage);
} catch (Exception ex) {
ex.printStackTrace();
}
}
As much as I try, this code will not follow redirects. The input steam is always empty, and the response is always 302. The image is uploaded nicely though. If someone could tell me what I can do to make it follow the redirect so I can read the response, I would really appreciate it.
Alternatively, if there is a better approach, I would love to hear it. I know there are libraries out there like Apache HTTP Client, but it requires so many dependencies that is seems like too much bloat for a simple task.
Thanks

A few months ago I was trying to do the exact same thing!
Basically, don't use HttpURLConnection. Instead, use HttpClient and HttpGet in the org.apache.http package which are part of the Android SDK.
Unfortunately I don't have my source code to provide an example, but hopefully that'll set you in the right direction.

not sure what went wrong, but you can follow redirect yourself. take the Location header out and make a new connection to it.

Related

How to write a file to an HTTP Request Java

I am new to using HTTP and I have questions about writing a file and another value to an HTTP Post request in Java. I am using an public API provided by a company called Mojang to write what is known as a "skin" (a png file) to the game Minecraft for player character modles. Here is the documentation of how to use this public API for reference:https://wiki.vg/Mojang_API#Upload_Skin
Here is the code I have written. When ran I get the 415 HTTP Response code (which I assume is "unsupported media type"). Any suggestions on what I am doing wrong and how I can fix this? I found other stack overflow issues for uploading files but I need to also add a value called "variant={classic or slim}". I am a little lost on how to make all of this work. Any help is much appreciated.
(I could not get the code to properally format in the code sample using ' ', it is in a javascript snippet)
public static void uploadSkin(String accessToken, String variant, File file) throws IOException {
URL url = new URL("https://api.minecraftservices.com/minecraft/profile/skins");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Authorization", "Bearer " + accessToken); // The access token is provided after an
// authentication request has been send, I
// have done this sucessfully in another
// method and am passing it in here
con.addRequestProperty("variant", variant);
OutputStream outputStream = con.getOutputStream();
PrintWriter writer = new PrintWriter(new OutputStreamWriter(con.getOutputStream(), "utf-8"), true);
String boundary = "===" + System.currentTimeMillis() + "===";
String fileName = file.getName();
String LINE_FEED = "\r\n";
String fieldName = "file";
writer.append("--" + boundary).append(LINE_FEED);
writer.append("Content-Disposition: form-data; name=\"" + fieldName + "\"; filename=\"" + fileName + "\"")
.append(LINE_FEED);
writer.append("Content-Type: " + URLConnection.guessContentTypeFromName(fileName)).append(LINE_FEED);
writer.append("Content-Transfer-Encoding: binary").append(LINE_FEED);
writer.append(LINE_FEED);
writer.flush();
FileInputStream inputStream = new FileInputStream(file);
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = inputStream.read(buffer)) != -1) {
outputStream.write(buffer, 0, bytesRead);
}
outputStream.flush();
inputStream.close();
writer.append(LINE_FEED);
writer.flush();
}
Alright, found a solution to the problem. Using this maven dependency:
<!-- https://mvnrepository.com/artifact/org.jodd/jodd-http -->
<dependency>
<groupId>org.jodd</groupId>
<artifactId>jodd-http</artifactId>
<version>5.0.2</version>
</dependency>
And then this:
HttpResponse response = HttpRequest.post("https://api.minecraftservices.com/minecraft/profile/skins")
.header("Authorization", "Bearer " + accessToken).header("Content-Type", "multipart/form-data")
.form("variant", variant).form("file", file).send();
I was able to get it to work. Hope this is helpful to anyone that needs to upload a Skin Png file to Minecraft.

Send Image with Parameters To Server Using HttpURLConnection in android

I am trying to upload two images along with some parameters to a server through my android application. After having searched online and following the instructions from here and here, as well as other sources, I have the following code:
String boundary = "***" + System.currentTimeMillis() + "***";
String twoHyphens = "--";
String crlf = "\r\n";
String output = "";
try {
HttpURLConnection httpUrlConnection = null;
URL url = new URL(myUrl);
httpUrlConnection = (HttpURLConnection) url.openConnection();
httpUrlConnection.setUseCaches(false);
httpUrlConnection.setDoInput(true);
httpUrlConnection.setDoOutput(true);
httpUrlConnection.setRequestMethod("POST");
httpUrlConnection.setRequestProperty("Connection", "Keep-Alive");
httpUrlConnection.setRequestProperty("Cache-Control", "no-cache");
httpUrlConnection.setRequestProperty("ENCTYPE", "multipart/form-data");
httpUrlConnection.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
DataOutputStream request = new DataOutputStream(httpUrlConnection.getOutputStream());
request.writeBytes(twoHyphens + boundary + crlf);
// Convert and add first image
ByteArrayOutputStream bao1 = new ByteArrayOutputStream();
params[0].compress(Bitmap.CompressFormat.JPEG, 100, bao1);
byte[] ba1 = bao1.toByteArray();
request.writeBytes("Content-Disposition: form-data; name=\"image1\";filename=\"image1\"" + crlf);
request.writeBytes(crlf);
request.write(ba1);
request.writeBytes(crlf);
request.writeBytes(twoHyphens + boundary + crlf);
// Convert and add second image
ByteArrayOutputStream bao2 = new ByteArrayOutputStream();
params[1].compress(Bitmap.CompressFormat.JPEG, 100, bao2);
byte[] ba2 = bao2.toByteArray();
request.writeBytes("Content-Disposition: form-data; name=\"image2\";filename=\"image2\"" + crlf);
request.writeBytes(crlf);
request.write(ba2);
request.writeBytes(crlf);
request.writeBytes(twoHyphens + boundary + crlf);
request.writeBytes("Content-Disposition: form-data; name=\"username\"" + crlf);
request.writeBytes(crlf);
request.writeBytes(username);
request.writeBytes(crlf);
request.writeBytes(twoHyphens + boundary + twoHyphens);
request.writeBytes("Content-Disposition: form-data; name=\"datestr\"" + crlf);
request.writeBytes(crlf);
request.writeBytes(timeStampString);
request.writeBytes(crlf);
request.writeBytes(twoHyphens + boundary + twoHyphens);
request.flush();
request.close();
int responseCode = httpUrlConnection.getResponseCode();
if (responseCode == HttpsURLConnection.HTTP_OK) {
InputStream responseStream = new BufferedInputStream(httpUrlConnection.getInputStream());
BufferedReader responseStreamReader = new BufferedReader(new InputStreamReader(responseStream, Charset.forName("UTF-8")));
String line;
while ((line = responseStreamReader.readLine()) != null) {
output = line;
Log.d(TAG, line);
}
responseStreamReader.close();
}
httpUrlConnection.disconnect();
if (output == "") {
httpResultsReturned = false;
} else {
httpResultsReturned = true;
}
} catch (ProtocolException e) {
e.printStackTrace();
return "failed";
} catch (MalformedURLException e) {
e.printStackTrace();
return "failed";
} catch (IOException e) {
e.printStackTrace();
return "failed";
}
On the server side I try to access the data as follows:
<?php
if($_SERVER['REQUEST_METHOD'] === 'POST'){
$image1 = $_FILES['image1']['name'];
$image2 = $_FILES['image2']['name'];
$datestr= $_POST['datestr'];
$username= $_POST['username'];
}
?>
Eventually, both images are successfully transmitted, however I cannot send/receive the extra parameters. I receive the responses correctly, but throughout all the php code (there are parts that I omitted in this question), it seems like none of the parameters are sent/received.
In this question, AndroSco shared the solution that worked for him, but in his php file it looks like he only accesses the image and not the parameter...
Since I don't have a lot of experience on this field, I believe that there is probably something very obvious that I do incorrectly/don't do at all!
Any suggestions will be appreciated!
Thanks!
After much frustration, I found the bug in my code. After having imported the two images in the transmitted message and when I wanted to import the other parameters, I wrote the boundary incorrectly. Instead of adding this:
request.writeBytes(twoHyphens + boundary + crlf);
having a new line at the end, I wrote this:
request.writeBytes(twoHyphens + boundary + twoHyphens);
adding two hyphens at the end of the line.
After replacing the twoHyphens with crlf, everything worked nicely!

Uploading a bitmap from android to a server results in an unopenable file

So I've used some code from stackoverflow that uses HttpURLConnection to send a bitmap from Android to my local WAMP server. The only issue I have is that the data being sent can't be open by an image viewer. I have a feeling it has to do with the method that I am sending the data.
HttpURLConnection httpUrlConnection = null;
URL url = new URL(ip+urls[0]);
httpUrlConnection = (HttpURLConnection) url.openConnection();
httpUrlConnection.setUseCaches(false);
httpUrlConnection.setDoOutput(true);
httpUrlConnection.setRequestMethod("POST");
httpUrlConnection.setRequestProperty("Connection", "Keep-Alive");
httpUrlConnection.setRequestProperty("Cache-Control", "no-cache");
httpUrlConnection.setRequestProperty(
"Content-Type", "multipart/form-data;boundary=" + this.boundary);
DataOutputStream request = new DataOutputStream(
httpUrlConnection.getOutputStream());
request.writeBytes(this.twoHyphens + this.boundary + this.crlf);
request.writeBytes("Content-Disposition: form-data; name=\"" +
this.attachmentName + "\";filename=\"" +
this.attachmentFileName + "\"" + this.crlf);
request.writeBytes(this.crlf);
int bytes = bitmap.getByteCount();
Log.e("bitmap bytes",Integer.toString(bytes));
ByteBuffer buffer = ByteBuffer.allocate(bytes);
bitmap.copyPixelsToBuffer(buffer);
byte[] pixels = buffer.array();
request.write(pixels);
request.writeBytes(this.crlf);
request.writeBytes(this.twoHyphens + this.boundary +
this.twoHyphens + this.crlf);
request.flush();
request.close();
//Response
InputStream responseStream = new
BufferedInputStream(httpUrlConnection.getInputStream());
BufferedReader responseStreamReader =
new BufferedReader(new InputStreamReader(responseStream));
String line = "";
StringBuilder stringBuilder = new StringBuilder();
while ((line = responseStreamReader.readLine()) != null) {
stringBuilder.append(line).append("\n");
}
responseStreamReader.close();
String response = stringBuilder.toString();
responseStream.close();
httpUrlConnection.disconnect();
return response;
I have checked the file sizes and they are the same. I plan on sending the image back to my android device in the future, however I wanted to make sure the uploading part worked first. If the file is supposed to be unopenable and I'm understanding the code wrong I apologize for the wasted time.

Imgur API uploading

So there is this line of code
String data = URLEncoder.encode("image", "UTF-8") + "=" + URLEncoder.encode(Base64.encodeBase64String(baos.toByteArray()).toString(), "UTF-8");
data += "&" + URLEncoder.encode("key", "UTF-8") + "=" + URLEncoder.encode(YOUR API KEY GOES HERE, "UTF-8");
and when I registered for the Imgur API I was given a client_id and a client_secret and was wondering which one I use for where it says "YOUR API KEY GOES HERE" also in the first part in the second line where it says "key" what do I enter there? Also is the site to upload it http://imgur.com/api/upload because I have seen a few different ones.
try this out:
public static String getImgurContent(String clientID) throws Exception {
URL url;
url = new URL("https://api.imgur.com/3/image");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
String data = URLEncoder.encode("image", "UTF-8") + "="
+ URLEncoder.encode(IMAGE_URL, "UTF-8");
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestMethod("POST");
conn.setRequestProperty("Authorization", "Client-ID " + clientID);
conn.setRequestMethod("POST");
conn.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded");
conn.connect();
StringBuilder stb = new StringBuilder();
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data);
wr.flush();
// Get the response
BufferedReader rd = new BufferedReader(
new InputStreamReader(conn.getInputStream()));
String line;
while ((line = rd.readLine()) != null) {
stb.append(line).append("\n");
}
wr.close();
rd.close();
return stb.toString();
}
was almost like humpty dumpty, getting every piece back together, codes from everywhere, at least it worked as expected, its a shame they don't have examples...
enjoy.
ps: ou can also make with FILES (haven't tried yet) but you need to convert an image to base64 and then to utf8 (to replace the url)
edit, use this instead of the URL, so you can upload files:
//create base64 image
BufferedImage image = null;
File file = new File(imageDir);
//read image
image = ImageIO.read(file);
ByteArrayOutputStream byteArray = new ByteArrayOutputStream();
ImageIO.write(image, "png", byteArray);
byte[] byteImage = byteArray.toByteArray();
String dataImage = Base64.encode(byteImage);
String data = URLEncoder.encode("image", "UTF-8") + "="
+ URLEncoder.encode(dataImage, "UTF-8");
The site to upload to is - https://api.imgur.com/3/image or you can alternatively use the same link with "upload" instead of image.
I am currently trying to use the Imgur API myself and although I have not got it completely right yet (I can't seem to parse the URL response) I have looked at quite a few code examples for it. Are you definitely using version 3 of the API?
Because the homepage of the API says that you should give your client ID in this format "Authorization Client-ID YOUR_CLIENT_ID", not using "key" like you are.
Have a look at http://api.imgur.com/
Edit: you might find the following useful - Anonymous Uploading File object to Imgur API (JSON) gives Authentication Error 401

Connect to web that requires user/password

I'm a bit new to Java and more to connections stuff with it. I'm trying to create a program to connect to a website ("www.buybackprofesional.com") where I would like to download pictures and get some text from cars (after the login you have to enter a plate number to access a car's file).
This is what I have right now, but it always says that the session has expired, I need a way to login using the username and password of the mainpage, am I right? can someone give me some advice? Thanks
Note: I want to do it in Java, maybe I was not clear in the question.
//URL web = new URL("http://www.buybackprofesional.com/DetallePeri.asp?mat=9073FCV&fec=27/07/2010&tipo=C&modelo=4582&Foto=0");
URL web = new URL("http://www.buybackprofesional.com/");
HttpURLConnection con = (HttpURLConnection) web.openConnection();
con.setRequestMethod("GET");
con.setRequestProperty("User-Agent", "Mozilla/4.0 (compatible; JVM)");
con.setRequestProperty("Pragma", "no-cache");
con.connect();
BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
A colleage helped me with this so I'll post the code that works:
public static URLConnection login(String _url, String _username, String _password) throws IOException, MalformedURLException {
String data = URLEncoder.encode("Usuario", "UTF-8") + "=" + URLEncoder.encode(_username, "UTF-8");
data += "&" + URLEncoder.encode("Contrase", "UTF-8") + "=" + URLEncoder.encode(_password, "UTF-8");
// Send data
URL url = new URL(_url);
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data);
wr.flush();
wr.close();
return conn;
}
This will submit the form info on the page I need and after that, using cookies I can stay connected!
To connect to a website using java consider using httpunit or httpcore (offered by apache). They handle sessions much better then you (or I) could do on your own.
Edit: Fixed the location of the link. Thanks for the correction!

Categories

Resources