I need to send a binary message which is divided to 2 parts:
The first part is 4 bytes and it has some information (say an integer)
The second part has an XMLtext stream.
I have never done something like this before, how can I do this?
My code is something like this:
public String serverCall(String link, String data){
HttpURLConnection connection;
OutputStreamWriter writer = null;
URL url = null;
String parameters = data;
try
{
url = new URL(link);
connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setRequestProperty("Content-Type", "text/xml");
connection.setRequestMethod("POST");
writer = new OutputStreamWriter(connection.getOutputStream());
writer.write(parameters);
writer.flush();
writer.close();
}
catch(IOException e)
{
e.printStackTrace();
}
}
How do I set the XML to be 4 bytes and how do I have 4 bytes of text before it?
(Info) The HTTP protocol uses method PUT to "transport a file on the server" (instead of POST).
The transport of binary data better not have a content type "text/..." but "application/bin".
You however could send the XML as "text/xml; charset=UTF-8", and use your own header
connection.setHeader("MyCode",
String.format("%02x%02x%02x%02x", bytes[0], bytes[1], bytes[2], bytes[3]));
If you send all as binary, do not use a Writer (converts bytes to some character encoding), but a Stream (BufferedOutputStream). The XML as:
byte[] xmlBytes = xml.getBytes("UTF-8");
UTF-8 if there is no other encoding mentioned in <?xml ...>.
The close() already flushes, so flush() is not needed.
If the binary portion of your message are mere 4 bytes, percent-encode it and send it as an additional url-parameter. alternatively, add it to the existing xml stream.
the java classes Uri and URLEncoder provide the necessary methods.
Related
NOTICE UPDATE!!
The problem got solved and i added my own answer in the thread
In short, I have attempted to add the parameter "scan_id" value but since it is a POST i can't add the value directly in the url path.
using the code i already have, how would i go about modifying or adding so that the url is correct, that is, so that it accepts my POST?.
somehow i have been unable to find any examples that have helped me in figuring out how i would go about doing this..
I know how to do a POST with a payload, a GET with params. but a post with Params is very confusing to me.
Appreciate any help. (i'd like to continue using HttpUrlConnection unless an other example is provided that also tells me how to send the request and not only configuring the path.
I've tried adding it to the payload.
I've tried UriBuilder but found it confusing and in contrast with the rest of my code, so wanted to ask for help with HttpUrlConnection.
URL url = new URL("http://localhost/scans/{scan_id}/launch");
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod("POST");
con.setRequestProperty("tmp_value_dont_mind_this", "432432");
con.setRequestProperty("X-Cookie", "token=" + "43432");
con.setRequestProperty("X-ApiKeys", "accessKey="+"43234;" + " secretKey="+"43234;");
con.setDoInput(true);
con.setDoOutput(true); //NOT NEEDED FOR GETS
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
//First example of writing (works when writing a payload)
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//second attemp at writing, doens't work (wanted to replace {scan_id} in the url)
DataOutputStream writer = new DataOutputStream(con.getOutputStream());
writer.writeChars("scan_id=42324"); //tried writing directly
//writer.write(payload);
writer.close();
Exception:
Exception in thread "main" java.io.IOException: Server returned HTTP response code: 400 for URL: http://localhost/scans/launch
I'd like one of the three response codes because then i know the Url is correct:
200 Returned if the scan was successfully launched.
403 Returned if the scan is disabled.
404 Returned if the scan does not exist.
I've tried several urls
localhost/scans/launch,
localhost/scans//launch,
localhost/scans/?/launch,
localhost/scans/{scan_id}/launch,
So with the help of a friend and everyone here i solved my problem.
The below code is all the code in an entire class explained bit by bit. at the bottom you have the full class with all its syntax etc, that takes parameters and returns a string.
in a HTTP request there are certain sections.
Such sections include in my case, Request headers, parameters in the Url and a Payload.
depending on the API certain variables required by the API need to go into their respective category.
My ORIGINAL URL looked like this: "http://host:port/scans/{scan_id}/export?{history_id}"
I CHANGED to: "https://host:port/scans/" + scan_Id + "/export?history_id=" + ID;
and the API i am calling required an argument in the payload called "format" with a value.
String payload = "{\"format\" : \"csv\"}";
So with my new URL i opened a connection and set the request headers i needed to set.
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
The setDoOutput should be commented out when making a GET request.
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"23243;" +"secretKey="+"45543;");
Here i write to the payload.
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
After i've written the payload i read whatever response i get back (this depends on the call, when i do a file download (GET Request) i don't have a response to read as i've already read the response through another piece of code).
I hope this helps anyone who might encounter this thread.
public String requestScan(int scan_Id, String token, String ID) throws MalformedInputException, ProtocolException, IOException {
try {
String endpoint = "https://host:port/scans/" + scan_Id + "/export?history_id=" ID;
URL url = new URL(endpoint);
String payload= "{\"format\" : \"csv\"}";
HttpsURLConnection con = (HttpsURLConnection) url.openConnection();
con.setDoInput(true);
con.setDoOutput(true);
con.setRequestMethod("POST");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Content-Type", "application/json; charset=UTF-8");
con.setRequestProperty("X-Cookie", "token=" + token);
con.setRequestProperty("X-ApiKeys", "accessKey="+"324324;" +
"secretKey="+"43242;");
//WRITING THE PAYLOAD to the http call
OutputStreamWriter writer = new OutputStreamWriter(con.getOutputStream(), "UTF-8");
writer.write(payload);
writer.close();
//READING RESPONSE
BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream()));
StringBuffer jsonString = new StringBuffer();
String line;
while ((line = br.readLine()) != null) {
jsonString.append(line);
}
br.close();
con.disconnect();
return jsonString.toString();
} catch (Exception e) {
throw new RuntimeException(e.getMessage());
}
}
As discussed here the solution would be to change the content type to application/x-www-form-urlencoded, but since you are already using application/json; charset=UTF-8 (which I am assuming is a requirement of your project) you have no choise to redesign the whole thing. I suggest you one of the following:
Add another GET service;
Add another POST service with content type application/x-www-form-urlencoded;
Replace this service with one of the above.
Do not specify the content type at all so the client will accept anything. (Don't know if possible in java)
If there are another solutions I'm not aware of, I don't know how much they would be compliant to HTTP protocol.
(More info)
Hope I helped!
Why you are not using like this. Since you need to do a POST with HttpURLConnection, you need to write the parameters to the connection after you have opened the connection.
String urlParameters = "scan_id=42324";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
Or if you have launch in the end, just change the above code to the following,
String urlParameters = "42324/launch";
byte[] postData = urlParameters.getBytes(StandardCharsets.UTF_8);
DataOutputStream dataOutputStream = new DataOutputStream(conn.getOutputStream());
dataOutputStream.write(postData);
URL url = new URL("http://localhost/scans/{scan_id}/launch");
That line looks odd to me; it seems you are trying to use a URL where you are intending the behavior of a URI Template.
The exact syntax will depend on which template implementation you choose; an implementation using the Spring libraries might look like:
import org.springframework.web.util.UriTemplate;
import java.net.url;
// Warning - UNTESTED code ahead
UriTemplate template = new UriTemplate("http://localhost/scans/{scan_id}/launch");
Map<String,String> uriVariables = Collections.singletonMap("scan_id", "42324");
URI uri = template.expand(uriVariables);
URL url = uri.toURL();
In Grails web application, I am trying to post minutiae (finger print) byte array from applet to server using rest API.
This what i tried so for
private String post(String purl,String customerId, byte[] regMin1,byte[] regMin2) throws Exception {
StringBuilder parameters = new StringBuilder();
parameters.append("customerId=");
parameters.append(customerId);
parameters.append("®Min1=");
parameters.append(URLEncoder.encode(new String(regMin1),"UTF-8"));
parameters.append("®Min2=");
parameters.append(URLEncoder.encode(new String(regMin2),"UTF-8"));
URL url = new URL(purl);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty("Content-Length",Integer.toString(parameters.toString().getBytes().length));
DataOutputStream wr = new DataOutputStream(connection.getOutputStream ());
wr.writeBytes(parameters.toString());
wr.flush();
wr.close();
BufferedReader in = new BufferedReader(new InputStreamReader(
connection.getInputStream()));
StringBuilder builder = new StringBuilder();
String aux = "";
while ((aux = in.readLine()) != null) {
builder.append(aux);
}
in.close();
connection.disconnect();
return builder.toString();
}
I can post regMin1, regMin2 successfully but fingerprint verification always failing. I doubt, am i posting correctly.
This looks like a very bad idea to me:
parameters.append(URLEncoder.encode(new String(regMin1),"UTF-8"));
...
parameters.append(URLEncoder.encode(new String(regMin2),"UTF-8"));
If regMin1 and regMin2 aren't actually UTF-8 text (and my guess is that they're not) you'll almost certainly be losing data here.
Don't treat arbitrary binary data as if it's encoded text.
Instead, convert regMin1 and regMin2 to base64 - that way you'll end up with ASCII characters which you can then decode on the server to definitely get the original binary data. You can use a URL-safe version of base64 to avoid having to worry about further encoding the result.
There's a good public domain base64 library you can use for this if you don't have anything else to hand. So for example:
parameters.append("®Min1=")
.append(Base64.encodeBytes(regMin1, Base64.URL_SAFE))
.append("®Min2=")
.append(Base64.encodeBytes(regMin2, Base64.URL_SAFE));
Note that you'd want to decode with the URL_SAFE option as well - don't just try to decode it as "normal" base64 data.
(You might still want to convert this to a POST request, and you'd definitely have an easier time if you could use a better HTTP library, but they're slightly separate concerns.)
I am currently developing an Android application and encounter the following problem.
I am making an HTTP request to a server that is supposed to send me back XML content that I then parse. I noticed recurring errors while parsing long XML strings so I decided to display the result of my requests and discovered that the string (or the stream?) that I receive is randomly truncated. Sometimes I get the whole string, sometimes half, sometimes a third, and it seems to follow a certain pattern in the amount of characters that are truncated, what I mean by that is that I sometimes get 320 characters after a request then 156 after the next then 320 twice, then 156 again (these aren't the actual numbers but it follows a pattern).
Here is my code for the request and conversion of the InputStream into a string:
private String downloadUrlGet(String myurl) throws IOException {
InputStream is = null;
// Only display the first 20000 characters of the retrieved
// web page content.
int len = 20000;
try {
URL url = new URL(myurl);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(10000 /* milliseconds */);
conn.setConnectTimeout(15000 /* milliseconds */);
conn.setRequestMethod("GET");
conn.setDoInput(true);
conn.setRequestProperty("Content-Type", "application/xml");
// Starts the query
conn.connect();
int response = conn.getResponseCode();
Log.d(DEBUG_TAG, "The response is: " + response);
is = conn.getInputStream();
// Convert the InputStream into a string
String contentAsString = readIt(is, len);
return contentAsString;
// Makes sure that the InputStream is closed after the app is
// finished using it.
} finally {
if (is != null) {
is.close();
}
}
}
// Reads an InputStream and converts it to a String.
private String readIt(InputStream stream, int len) throws IOException, UnsupportedEncodingException {
Reader reader = null;
reader = new InputStreamReader(stream, "UTF-8");
char[] buffer = new char[len];
reader.read(buffer);
return new String(buffer);
}
The length of the XML that I try to retrieve is much less than 20000.
I tried to use HttpURLConnection.setChunkedStreamingMode() with 0 and various other numbers as parameter but it didn't change anything.
Thanks in advance for any suggestions.
You are making the usual mistake of assuming that read() fills the buffer. See the Javadoc. It isn't obliged to do that. It isn't obliged to transfer more than one byte as a matter of fact. You need to read in a loop until you have encountered end of stream (read() returns -1).
I am trying to download a portion of a PDF file (just for testing "Range" header). I requested the server for the bytes (0-24) in Range but still, instead of getting first 25 bytes (a portion) out of the content, I am getting the full length content. Moreover, instead of getting response code as 206 (partial content), I'm getting response code as 200.
Here's my code:
public static void main(String a[]) {
try {
URL url = new URL("http://download.oracle.com/otn-pub/java/jdk/7u21-b11/jdk-7u21-windows-x64.exe?AuthParam=1372502269_599691fc0025a1f2da7723b644f44ece");
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setRequestProperty("Range", "Bytes=0-24");
urlConnection.connect();
System.out.println("Respnse Code: " + urlConnection.getResponseCode());
System.out.println("Content-Length: " + urlConnection.getContentLengthLong());
InputStream inputStream = urlConnection.getInputStream();
long size = 0;
while(inputStream.read() != -1 )
size++;
System.out.println("Downloaded Size: " + size);
}catch(MalformedURLException mue) {
mue.printStackTrace();
}catch(IOException ioe) {
ioe.printStackTrace();
}
}
Here's the output:
Respnse Code: 200
Content-Length: 94973848
Downloaded Size: 94973848
Thanks in Advance.
Try changing following:
urlConnection.setRequestProperty("Range", "Bytes=0-24");
with:
urlConnection.setRequestProperty("Range", "bytes=0-24");
as per the spec 14.35.1 Byte Ranges
Similarly, as per the spec 14.5 Accept-Ranges, you can also check whether your server actually supports partial content retrieval or not using following:
boolean support = urlConnection.getHeaderField("Accept-Ranges").equals("bytes");
System.out.println("Partial content retrieval support = " + (support ? "Yes" : "No));
If the server supports it (and HTTP 1.1 servers should), only then you can use range requests...
and if all you want to do is check, then just send a HEAD request instead of a GET request. Same headers, same everything, just "HEAD" instead of "GET". If you receive a 206 response, you'll know Range is supported, and otherwise you'll get a 200 response.
You have to connect to url before setRequestProperty
Change:
urlConnection.setRequestProperty("Range", "Bytes=0-24");
urlConnection.connect();
To:
urlConnection.connect();
urlConnection.setRequestProperty("Range", "Bytes=0-24");
I think the correct header is "Content-Range", not "Range" as you are using.
I want to post among other data a String variable to a PHP file by using the HttpConnection stuff. The first data is the byte[] data returned from a recordstore. So it should be posted alone. So how to post the String variable also ?
You can pass the data to a PHP file using GET or POST methods.
Get method is the easy way to pass simple data. Using GET you can add the variable to the URL
Example:
192.168.1.123/myproject/uploads/treatphoto.php?myVariable1=MyContent&myVariable2=MyContent2
And in PHP:
$content1 = $_GET['myVariable1'];
$content2 = $_GET['myVariable2'];
Also the content of "MyContent" needs to be an string encoded. using any UrlEncoder.
To pass a byte[] array using this method you need to convert the byte array to an string encoded in some printable encoding like base64
The GET method also has a sort limit of data that can be passed safely (usually 2048 bytes)
The other method "POST" is more complex (but not a lot), way to add more data.
You need to prepare the HttpConnection to pass the data as POST.
Also the data stored in urlParamenters need to be according to the url enconding.
Passing the data using post is similar to GET but instead of adding all the variables next to the url the varuiables are added in the Stream of the httpConnection request.
example of the java code:
String urlParameters = "myVariable1=myValue1&myVariable2=myValue2";
HttpURLConnection connection = null;
try {
url = new URL(targetURL);
connection = (HttpURLConnection)url.openConnection();
// Use post and add the type of post data as URLENCODED
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
// Optinally add the language and the data content
connection.setRequestProperty("Content-Length", "" + Integer.toString(urlParameters.getBytes().length));
connection.setRequestProperty("Content-Language", "en-US");
// Set the mode as output and disable cache.
connection.setUseCaches (false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
DataOutputStream wr = new DataOutputStream (connection.getOutputStream ());
wr.writeBytes (urlParameters);
wr.flush ();
wr.close ();
// Get Response
// Optionally you can get the response of php call.
InputStream is = connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
String line;
StringBuffer response = new StringBuffer();
while((line = rd.readLine()) != null) {
response.append(line);
response.append('\r');
}
rd.close();
return response.toString();
The php is similar, you only need to replace $_GET by $_POST:
$content1 = $_POST['myVariable1'];
$content2 = $_POST['myVariable2'];