Develop Reference

How to write / convert CURL for Android java

I am trying to implement the MOT history API https://dvsa.github.io/mot-history-api-documentation/ and they give an example using CURL which works with the supplied api key successfully when using an online CURL tool.
I am trying to implement this in Android and realise I have to use something like HttpPost rather than CURL, this is my code:
//Tried with full URL and by adding the registration as a header.
//HttpPost httpPost = new HttpPost("https://beta.check-mot.service.gov.uk/trade/vehicles/mot-tests?registration=" + reg_selected);
HttpPost httpPost = new HttpPost("https://beta.check-mot.service.gov.uk/trade/vehicles/mot-tests");
httpPost.addHeader("Content-Type", "application/json");
httpPost.addHeader("Accept", "application/json+v6");
httpPost.addHeader("x-api-key", "abcdefgh123456");
httpPost.addHeader("registration", reg_selected);
StringEntity entity = new StringEntity(jsonObj.toString(), HTTP.UTF_8);
httpPost.setEntity(entity);
HttpClient client = new DefaultHttpClient();
try {
HttpResponse response = client.execute(httpPost);
if (response.getStatusLine().getStatusCode() == 200) {
InputStream inputStream = response.getEntity().getContent();
bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
String readLine = bufferedReader.readLine();
String jsonStr = readLine;
JSONObject myJsonObj = new JSONObject(jsonStr);
}else if (response.getStatusLine().getStatusCode() == 400){
//Bad Request Invalid data in the request. Check your URL and parameters
error_text = "Bad Request";
}else if (response.getStatusLine().getStatusCode() == 403){
//Unauthorised – The x-api-key is missing or invalid in the header
error_text = "Authentication error"; //<<<< FAILS HERE 403
}
response.getStatusLine().getStatusCode() returns • "403 – Unauthorised – The x-api-key is missing or invalid in the header".
However the x-api-key that I use works correctly with the online CURL test so the actual key is correct but how I am adding it to my android code request must be invalid or similar.
Can anyone throw any light as to the correct way to convert the CURL into Android java so that the server does not return 403?
Thanks

It's easy to do with Jsoup:
// CREATE CONNECTION
Connection conn=Jsoup.connect("URL_GOES_HERE");
// ADD POST/FORM DATA
conn.data("KEY", "VALUE");
// ADD HEADERS HERE
conn.header("KEY", "VALUE");
// SET METHOD AS POST
conn.method(Connection.Method.POST);
// ACCEPT RESPONDING CONTENT TYPE
conn.ignoreContentType(true);
try
{
// GET RESPONSE
String response = conn.execute().body();
// USE RESPONSE HERE
// CREATE JSON OBJECT OR ANYTHING...
} catch(HttpStatusException e)
{
int status = e.getStatusCode();
// HANDLE HTTP ERROR HERE
} catch (IOException e)
{
// HANDLE IO ERRORS HERE
}
Ps: I guess you are confused with Header and Post Data. The key etc (Credentials) must be used as Post Data and Content Type etc as Header.

Read InputStream from file via URL query string

Is it possible to use the java URL.openStream() method to read the file into an input stream when the URL is a query string rather than a direct link to a file? E.g. the code I have is:
URL myURL = new URL("http://www.test.com/myFile.doc");
InputStream is = myURL.openStream();
This works fine for a direct file link. But what if the URL was http://www.test.com?file=myFile.doc ? Would I still be able to obtain the file stream from the server response?
Thanks!

Generally YES, it will work.
But note that URL.openStream() method doesn't follow redirects and not so agile with specifying some additional HTTP behaviours: request type, headers, etc.
I'd recommend to use Apache HTTP Client instead:
final CloseableHttpClient httpclient = HttpClients.createDefault();
final HttpGet request = new HttpGet("http://any-url");
try (CloseableHttpResponse response = httpclient.execute(request)) {
final int status = response.getStatusLine().getStatusCode();
if (status == 200) {
final InputStream is = response.getEntity().getContent();
} else {
throw new IOException("Got " + status + " from server!");
}
}
finally {
request.reset();
}

The URL class works for any url, including:
new URL("http://www.example.com/");
new URL("file://C/windows/system32/cmd.exe");
new URL("ftp://user:password#example.com/filename;type=i");
Its up to the application to do something with the data, for example download the data, or treat it as plain text.

How to post a JSON object to an URL?

I'm trying to post a JSON to a web service so that I can get a JSON in response as return, I have searched in Google but most of the response I found is for Android but not for core Java, this question is for a Swing application I'll give the code the code I used below.
Connection class
public class Connection extends Thread {
private String url1;
private JSONObject data;
String line;
//Constuctor to initialize the variables.
public Connection(String url1, JSONObject data) {
this.url1 = url1;
this.data = data;
start();
}
public void run() {
ConnectionReaderWriter();
}
//To fetch the data from the input stream
public String getResult() {
return line;
}
public String ConnectionReaderWriter() {
URL url;
HttpURLConnection connection = null;
ObjectOutputStream out;
try {
/*URL url = new URL(Url.server_url + url1); //Creating the URL.
URLConnection conn = url.openConnection(); //Opening the connection.
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());
wr.write(data); //Posting the data to the ouput stream.
wr.flush();
BufferedReader rd = new BufferedReader(new InputStreamReader(conn.getInputStream()));
line=rd.readLine(); //Reading the data from the input stream.
wr.close();
rd.close();*/
url = new URL(Url.server_url + url1); //Creating the URL.
connection = (HttpURLConnection) url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/json");
connection.setRequestProperty("Accept", "application/json");
connection.setRequestProperty("api_key", "123456");
connection.setUseCaches(false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
out = new ObjectOutputStream(connection.getOutputStream());
out.writeObject(data);
out.flush();
out.close();
} catch (MalformedURLException ex) {
Logger.getLogger(Connection.class.getName()).log(Level.SEVERE, null, ex);
String nonet = "No Network Connection";
line = nonet;
} catch (IOException ex) {
Logger.getLogger(Connection.class.getName()).log(Level.SEVERE, null, ex);
String nonet = "No Server Connection";
line = nonet;
}
return line; //Return te stream recived from the input stream.
}
}
The commented code is the one I used before when I was passing as text encoded to the URL. The function call is given below
JSONObject json = new JSONObject();
json.put("username", username);
json.put("password", passwordenc);
Connection conn = new Connection(Url.login, json);
conn.join();
On execution I get the exception shown below
Jan 20, 2014 1:18:32 PM SupportingClass.Connection ConnectionReaderWriter
SEVERE: null
java.io.NotSerializableException: org.json.JSONObject
at java.io.ObjectOutputStream.writeObject0(ObjectOutputStream.java:1180)
at java.io.ObjectOutputStream.writeObject(ObjectOutputStream.java:346)
at SupportingClass.Connection.ConnectionReaderWriter(Connection.java:74)
at SupportingClass.Connection.run(Connection.java:40)
Please tell me the problem in this code or an alternative to this method.

I'll give the answer:
replace out.writeObject(data); with out.write(data.toString().getBytes());
You were trying to write the JSONObject object, and the writeObject() method will attempt to serialize the object, but it will fail since the JSONObject class does not implement Serializable.

JSon is text, so you cannot use an ObjectOutputStream. The POST method uses the first line of the content as the arguments, so you would need a blank line before the actual content:
OutputStream stream = connection.getOutputStream();
stream.write('\r');
stream.write('\n');
out = new OutputStreamWriter(stream, "UTF-8");
out.write(data.toString());
out.flush();
out.close();
EDIT: actually we need CR LF, so println() may not work.

If you find it awkward to use HttpURLConnection natively, you might use an abstraction library. There are many powerful out there. One of them is DavidWebb. You can find a long list of alternatives at the end of that page.
With this library your code would be shorter and more readable:
JSONObject nameAndPassword = new JSONObject();
// set name and password
Webb webb = Webb.create();
JSONObject result = webb.post("your_serverUrl")
.header("api_key", "123456")
.useCaches(false)
.body(nameAndPassword)
.ensureSuccess()
.asJsonObject()
.getBody();
The code shows only the part that would run in your run() method of the Thread. It is not necessary to set the Content-Type header, because this is done automatically by detecting the type of object you set as the body. The same is true for the Accept header.
I assumed that you receive a JSON Object from your REST service, since you set the "Accept" header to "application/json". For receiving a plain String one could write String result = ... .asString().getBody().
BTW the library has been developed with Android in mind, but it can be used with Swing or server-side as well. Maybe you choose another library (e.g. Jersey Client), because you have no restriction about the size. DavidWebb weighs about 20 kilobytes while most other libraries add several hundred kilobytes to your deployment artifacts.
Another thing: you use JSONObject. For small web services, this is not a problem, but if you have to marshal many and/or big objects, you could think about using JAXB + Jackson or Gson. Less code, less errors, more fun!

FileNotFoundException while getting the InputStream object from HttpURLConnection

I am trying to send a post request to a url using HttpURLConnection (for using cUrl in java).
The content of the request is xml and at the end point, the application processes the xml and stores a record to the database and then sends back a response in form of xml string. The app is hosted on apache-tomcat locally.
When I execute this code from the terminal, a row gets added to the db as expected. But an exception is thrown as follows while getting the InputStream from the connection
java.io.FileNotFoundException: http://localhost:8080/myapp/service/generate
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1401)
at org.kodeplay.helloworld.HttpCurl.main(HttpCurl.java:30)
Here is the code
public class HttpCurl {
public static void main(String [] args) {
HttpURLConnection con;
try {
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
con.setRequestMethod("POST");
con.setDoOutput(true);
con.setDoInput(true);
File xmlFile = new File("test.xml");
String xml = ReadWriteTextFile.getContents(xmlFile);
con.getOutputStream().write(xml.getBytes("UTF-8"));
InputStream response = con.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(response));
for (String line ; (line = reader.readLine()) != null;) {
System.out.println(line);
}
reader.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
}
Its confusing because the exception is traced to the line InputStream response = con.getInputStream(); and there doesn't seem to be any file involved for a FileNotFoundException.
When I try to open a connection to an xml file directly, it doesn't throw this exception.
The service app uses spring framework and Jaxb2Marshaller to create the response xml.
The class ReadWriteTextFile is taken from here
Thanks.
Edit:
Well it saves the data in the DB and sends back a 404 response status code at the same time.
I also tried doing a curl using php and print out the CURLINFO_HTTP_CODE which turns out to be 200.
Any ideas on how do I go about debugging this ? Both service and client are on the local server.
Resolved:
I could solve the problem after referring to an answer on SO itself.
It seems HttpURLConnection always returns 404 response when connecting to a url with a non standard port.
Adding these lines solved it
con.setRequestProperty("User-Agent","Mozilla/5.0 ( compatible ) ");
con.setRequestProperty("Accept","*/*");

I don't know about your Spring/JAXB combination, but the average REST webservice won't return a response body on POST/PUT, just a response status. You'd like to determine it instead of the body.
Replace
InputStream response = con.getInputStream();
by
int status = con.getResponseCode();
All available status codes and their meaning are available in the HTTP spec, as linked before. The webservice itself should also come along with some documentation which overviews all status codes supported by the webservice and their special meaning, if any.
If the status starts with 4nn or 5nn, you'd like to use getErrorStream() instead to read the response body which may contain the error details.
InputStream error = con.getErrorStream();

FileNotFound is just an unfortunate exception used to indicate that the web server returned a 404.

To anyone with this problem in the future, the reason is because the status code was a 404 (or in my case was a 500). It appears the InpuStream function will throw an error when the status code is not 200.
In my case I control my own server and was returning a 500 status code to indicate an error occurred. Despite me also sending a body with a string message detailing the error, the inputstream threw an error regardless of the body being completely readable.
If you control your server I suppose this can be handled by sending yourself a 200 status code and then handling whatever the string error response was.

For anybody else stumbling over this, the same happened to me while trying to send a SOAP request header to a SOAP service. The issue was a wrong order in the code, I requested the input stream first before sending the XML body. In the code snipped below, the line InputStream in = conn.getInputStream(); came immediately after ByteArrayOutputStream out = new ByteArrayOutputStream(); which is the incorrect order of things.
ByteArrayOutputStream out = new ByteArrayOutputStream();
// send SOAP request as part of HTTP body
byte[] data = request.getHttpBody().getBytes("UTF-8");
conn.getOutputStream().write(data);
if (conn.getResponseCode() != HttpURLConnection.HTTP_OK) {
Log.d(TAG, "http response code is " + conn.getResponseCode());
return null;
}
InputStream in = conn.getInputStream();
FileNotFound in this case was an unfortunate way to encode HTTP response code 400.

FileNotFound in this case means you got a 404 from your server - could it be that the server does not like "POST" requests?

FileNotFound in this case means you got a 404 from your server
You Have to Set the Request Content-Type Header Parameter
Set “content-type” request header to “application/json” to send the request content in JSON form.
This parameter has to be set to send the request body in JSON format.
Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
Full Script ->
public class SendDeviceDetails extends AsyncTask<String, Void, String> {
#Override
protected String doInBackground(String... params) {
String data = "";
String url = "";
HttpURLConnection con = null;
try {
// From the above URL object,
// we can invoke the openConnection method to get the HttpURLConnection object.
// We can't instantiate HttpURLConnection directly, as it's an abstract class:
con = (HttpURLConnection)new URL(url).openConnection();
//To send a POST request, we'll have to set the request method property to POST:
con.setRequestMethod("POST");
// Set the Request Content-Type Header Parameter
// Set “content-type” request header to “application/json” to send the request content in JSON form.
// This parameter has to be set to send the request body in JSON format.
//Failing to do so, the server returns HTTP status code “400-bad request”.
con.setRequestProperty("Content-Type", "application/json; utf-8");
//Set Response Format Type
//Set the “Accept” request header to “application/json” to read the response in the desired format:
con.setRequestProperty("Accept", "application/json");
//To send request content, let's enable the URLConnection object's doOutput property to true.
//Otherwise, we'll not be able to write content to the connection output stream:
con.setDoOutput(true);
//JSON String need to be constructed for the specific resource.
//We may construct complex JSON using any third-party JSON libraries such as jackson or org.json
String jsonInputString = params[0];
try(OutputStream os = con.getOutputStream()){
byte[] input = jsonInputString.getBytes("utf-8");
os.write(input, 0, input.length);
}
int code = con.getResponseCode();
System.out.println(code);
//Get the input stream to read the response content.
// Remember to use try-with-resources to close the response stream automatically.
try(BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream(), "utf-8"))){
StringBuilder response = new StringBuilder();
String responseLine = null;
while ((responseLine = br.readLine()) != null) {
response.append(responseLine.trim());
}
System.out.println(response.toString());
}
} catch (Exception e) {
e.printStackTrace();
} finally {
if (con != null) {
con.disconnect();
}
}
return data;
}
#Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
Log.e("TAG", result); // this is expecting a response code to be sent from your server upon receiving the POST data
}
and call it
new SendDeviceDetails().execute("");
you can find more details in this tutorial
https://www.baeldung.com/httpurlconnection-post

The solution:
just change localhost for the IP of your PC
if you want to know this: Windows+r > cmd > ipconfig
example: http://192.168.0.107/directory/service/program.php?action=sendSomething
just replace 192.168.0.107 for your own IP (don't try 127.0.0.1 because it's same as localhost)

Please change
con = (HttpURLConnection) new URL("http://localhost:8080/myapp/service/generate").openConnection();
To
con = (HttpURLConnection) new URL("http://YOUR_IP:8080/myapp/service/generate").openConnection();

twitter streaming api - set filter in http post request (apache httpcomponents)

I'm trying out the twitter streaming api. I could succesfully filter tweets by using curl, as stated here:
curl -d #tracking http://stream.twitter.com/1/statuses/filter.json -u <user>:<pass>
where tracking is a plain file with the content:
track=Berlin
Now I tried to do the same thing in JavaSE, using Apache's HTTPComponents:
UsernamePasswordCredentials creds = new UsernamePasswordCredentials(<user>, <pass>);
DefaultHttpClient httpClient = new DefaultHttpClient();
httpClient.getCredentialsProvider().setCredentials(AuthScope.ANY, creds);
HttpPost httpPost = new HttpPost("http://stream.twitter.com/1/statuses/filter.json");
HttpParams params = new BasicHttpParams();
params = params.setParameter("track", "Berlin");
httpPost.setParams(params);
try {
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity entity = httpResponse.getEntity();
if (entity != null) {
InputStream instream = entity.getContent();
String t;
BufferedReader br = new BufferedReader(new InputStreamReader(instream));
while(true) {
t = br.readLine();
if(t != null) {
linkedQueue.offer(t);
}
}
}
} catch (IOException ioe) {
System.out.println(ioe.getMessage());
}
finally{
httpClient.getConnectionManager().shutdown();
}
When I run that, I get:
No filter parameters found. Expect at least one parameter: follow track
as a single entry in my linkedQueue. Seems the api wants the parameter in a different form, but cannot find any hint in the documentation. Can somebody share some experiences with the api or see any other problem with the code? Thanks!
EDIT
Putting the filter parameter into the params was a bad idea. As it's post data, it needs to be defined as an Entity before the request is being made:
StringEntity postEntity = new StringEntity("track=Berlin", "UTF-8");
postEntity.setContentType("application/x-www-form-urlencoded");
httpPost.setEntity(postEntity);
That's what I was doing wrong. Thanks Brian!

I suspect you need to post the data as the contents of your HTTP post. The man page for curl -d says:
(HTTP) Sends the specified data in a
POST request to the HTTP server, in
the same way that a browser does when
a user has filled in an HTML form and
presses the submit button. This will
cause curl to pass the data to the
server using the content-type
application/x-www-form-urlencoded.
so I believe you have to set that content type and put the contents of the tracking file in the body of your post.