We Keep Coding

Rest Api to Decrypt GZIP Body

An API is sending a large amount of data in the body in form of GZIP, I need to create rest API to decrypt and save it in the database, but I am not able to decrypt the data.
`#GetMapping
public void hello() throws IOException {
String payload = "{\n" +
" \"name1\": \"shrikant\",\n" +
" \"date\": \"Fri Apr 05 15:48:59 IST 2019\"\n" +
"}";
String urlStr = "http://localhost:8080/hello";
URL url = new URL(urlStr);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(60000);
conn.setConnectTimeout(60000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
conn.setRequestProperty("Accept", "application/json");
conn.setRequestProperty("Content-Type", "application/json");
conn.addRequestProperty("Content-Encoding", "gzip");
OutputStream os = conn.getOutputStream();
GZIPOutputStream gos = new GZIPOutputStream(os);
gos.write(payload.getBytes(StandardCharsets.UTF_8));
System.out.println("payload " +
Arrays.toString(payload.getBytes(StandardCharsets.UTF_8)));
os.close();
int responseCode = conn.getResponseCode();
System.out.println("\nSending 'POST' request to URL : " + url);
System.out.println("Response Code : " + responseCode);
}`
API to receive the data.
#PostMapping("hello")
public byte[] hello1(HttpServletRequest request) throws IOException {
System.out.println("hi");
ByteArrayInputStream bis = new ByteArrayInputStream();
GZIPInputStream gis = new GZIPInputStream(bis);
BufferedReader br = new BufferedReader(new InputStreamReader(gis, "UTF-8"));
StringBuilder sb = new StringBuilder();
String line;
while((line = br.readLine()) != null) {
sb.append(line);
}
}
but not able to decrypt data.
how to decrypt request.

On the client side you should close GZIPOutputStream before closing OutputStream.
gos.close();
os.close();
On the server side you should use InputStream from request
ServletInputStream inputStream = request.getInputStream();
GZIPInputStream gis = new GZIPInputStream(inputStream);
BufferedReader br = new BufferedReader(new InputStreamReader(gis, "UTF-8"));
StringBuilder sb = new StringBuilder();
String line;
while((line = br.readLine()) != null) {
sb.append(line);
}
System.out.println(sb.toString());
For streams better use try-with-resources blok you won't have to remember about closing streams.
try (OutputStream os = conn.getOutputStream()) {
try (GZIPOutputStream gos = new GZIPOutputStream(os)) {
gos.write(payload.getBytes(StandardCharsets.UTF_8));
}
}

Sending a JSON formatted string through HttpUrlConnection

I've done some research on using HttpUrlConnect and most examples I've seen uses either
a) a params string which looks like this:
paramString = "param1=someParam&param2=2ndparam&param3=3rdparam";
b) uses a put method to place the parameters:
JSONObject json = new JSONObject();
json.put("param1", "Parameter");
json.put("param2", "Parameter2");
json.put("param3", "Parameter3");
The format I want to send looks like this:
{
"grant_type":"password",
"username":"testuser#someid.com",
"password":"testPwd123$"
}
Is there a way for me to send a formatted JSON string instead of setting parameters or using a param string? The code I'm using to send my POST request looks like the following:
public static String PostRequest(String urlString, String token, String jsonString) throws IOException {
byte[] postData = jsonString.getBytes(StandardCharsets.UTF_8);
URL url = new URL(urlString);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setDoOutput(true);
conn.setInstanceFollowRedirects(false);
conn.setRequestMethod("POST");
conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
conn.setRequestProperty ("Authorization", "Bearer " + token);
conn.setUseCaches(false);
try( DataOutputStream wr = new DataOutputStream(conn.getOutputStream())) {
wr.write(postData);
}
int responseCode = conn.getResponseCode();
System.out.println("POST response code: " + responseCode);
BufferedReader in = new BufferedReader(
new InputStreamReader(conn.getInputStream()));
String line;
StringBuffer response = new StringBuffer();
while ((line = in.readLine()) != null) {
response.append(line);
}
in.close();
return response.toString();
}
I'm open to suggestions whether to use a different library, or if there are any code changes that I should make in order to take a JSON formatted string.

In HttpURLConnection Why don't JSONObject as Params work but String as Params are working

I am using HttpUrlConnection to post some data to my server here is the function:
private String register(String myurl) throws IOException {
String resp = null;
try {
JSONObject parameters = new JSONObject();
// parameters.put("jsonArray", ((makeJSON())));
parameters.put("key", "key");//getencryptkey());
URL url = new URL(myurl);
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
// conn.setReadTimeout(10000 /* milliseconds *///);
// conn.setConnectTimeout(15000 /* milliseconds */);
conn.setRequestProperty("Content-Type", "application/json");
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestMethod("POST");
OutputStream out = new BufferedOutputStream(conn.getOutputStream());
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(out, "UTF-8"));
writer.write(parameters.toString());
writer.close();
out.close();
int responseCode = conn.getResponseCode();
System.out.println("\nSending 'POST' request to URL : " + url);
System.out.println("Response Code : " + responseCode);
BufferedReader in = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
System.out.println("strngbuffr" + response.toString());
resp = response.toString();
} catch (Exception exception) {
System.out.println("Exception: " + exception);
}
System.out.println("rsp"+ resp.toString());
return resp.toString();
}
I get the response code as 200, which means connection is okay however I get empty variables on PHP side, what can be wrong here?
Earlier I was sending a JSON array too but just to test functonality I commented that out now I am only sending one variable key as "key"
Its amazing to see, this sample code works - sans the JSON array and the key value pairs:
private String sendPost(String url) throws Exception {
String USER_AGENT = "Mozilla/5.0";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
//add reuqest header
con.setRequestMethod("POST");
con.setRequestProperty("User-Agent", USER_AGENT);
con.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
String urlParameters ="sn=C02G8416DRJM&cn=&locale=&caller=&num=12345";
// Send post request
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
wr.close();
int responseCode = con.getResponseCode();
System.out.println("\nSending 'POST' request to URL : " + url);
System.out.println("Post parameters : " + urlParameters);
System.out.println("Response Code : " + responseCode);
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
//print result
System.out.println("rvsp"+response.toString());
return response.toString();
}
So it boils down to replacing this:
JSONObject parameters = new JSONObject();
parameters.put("jsonArray", new JSONArray(Arrays.asList(makeJSON())));
parameters.put("key", getencryptkey());
by this:
String urlParameters ="jArr="+makeJSON()+"Key="+getencryptkey();
and I am still curious.

I reckon the problem here is not at the Java side, If the parameters is of fixed type like in json in your case, the JSON Object as POST params method will work if collected this way on the php side:
<?php
$json = file_get_contents('php://input');
$obj = json_decode($json);
print_r($obj);
print_r("this is a test response");
?>

The problem here was not with the Java side, it was with the php side, the JSON Object as POST params method will work if collected this way on the php side:
<?php
$json = file_get_contents('php://input');
$obj = json_decode($json);
print_r($obj);
print_r("this is a test");
?>

How to set parameters in a GET request in Java

So I want to send a GET request with parameters. But it only seems to have conventions for the url you send the request to. Unlike the POST request, I see no way to pass parameters in it.
How I send the GET request now, without parameters (might be wrong):
String url = "http://api.netatmo.net/api/getuser";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
// optional default is GET
con.setRequestMethod("GET");
//add request header
con.setRequestProperty("User-Agent", USER_AGENT);
int responseCode = con.getResponseCode();
Log.v(TAG, ("\nSending 'GET' request to URL : " + url));
Log.v(TAG, ("Response Code : " + responseCode));
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
//print result
Log.v(TAG, (response.toString()));
How I send the POST request with parameters:
String url = "https://api.netatmo.net/oauth2/token";
URL obj = new URL(url);
HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();
//add request header
con.setRequestMethod("POST");
con.setRequestProperty("User-Agent", USER_AGENT);
con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
String urlParameters = "grant_type=password&client_id=myid&client_secret=mysecret&username=myusername&password=mypass";
// Send post request
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
wr.close();
int responseCode = con.getResponseCode();
Log.v(TAG, "\nSending 'POST' request to URL : " + url);
Log.v(TAG, "Post parameters : " + urlParameters);
Log.v(TAG, "Response Code : " + responseCode);
BufferedReader in = new BufferedReader(
new InputStreamReader(con.getInputStream()));
String inputLine;
StringBuffer response = new StringBuffer();
while ((inputLine = in.readLine()) != null) {
response.append(inputLine);
}
in.close();
//print result
Log.v(TAG, response.toString());
access_token = response.substring(17, 74);
refresh_token = response.substring(93,150);
getRequest = "/api/getuser?access_token=" + access_token + " HTTP/1.1";
Log.v(TAG, access_token);
Log.v(TAG, refresh_token);
Log.v(TAG, getRequest);

As per the HTTP specification GET supports only path params or url params and hence you cannot put the params in HTTP request body as you do in POST request.
As Sotirios mentioned in the comments, technically you can still push params in the GET body, but if the APIs are respecting the specs, they will not provide you a way to do it.

Have you tried to add the query params to the request java.net.URL?
String url = "http://api.netatmo.net/api/getuser?access_token=" + access_token;
URL obj = new URL(url);

I was encountering the same problem, trying this:
String bla = "http://api.netatmo.net/api/devicelist?access_token=" + AUTH_TOKEN;
URL url = new URL(bla);
BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream()));
String line = "";
String message = "";
while ((line = reader.readLine()) != null)
{
message += line;
}
I got an exception that the syntax was not correct. When I changed the syntax (by for example encoding with UTF 8) the API would just return errors (like 404 not found...).
I finally got it working using this:
try
{
System.out.println("Access Token: " + AUTH_TOKEN);
String url = "http://api.netatmo.net/api/devicelist";
String query = "access_token=" + URLEncoder.encode(AUTH_TOKEN, CHARSET);
URLConnection connection = new URL(url + "?" + query).openConnection();
connection.setRequestProperty("Accept-Charset", CHARSET);
InputStream response = connection.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(response));
String line = "";
String message = "";
while ((line = reader.readLine()) != null)
{
message += line;
}
return message;
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
Note: CHARSET = "UTF-8"

Turns out the url the API provided confused me greatly. I fixed the url and it works now.

how to pass the parameters to the urlconnection in java/android?

i can establish a connection using HttpUrlConnection. my code below.
client = new DefaultHttpClient();
URL action_url = new URL(actionUrl);
conn = (HttpURLConnection) action_url.openConnection();
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestProperty("userType", "2");
conn.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded");
conn.setRequestMethod(HttpPost.METHOD_NAME);
DataOutputStream ds = new DataOutputStream(conn.getOutputStream());
String content = "username=username1&password=password11";
Log.v(TAG, "content: " + content);
ds.writeBytes(content);
ds.flush();
ds.close();
InputStream in = conn.getInputStream();//**getting filenotfound exception here.**
BufferedReader reader = new BufferedReader(
new InputStreamReader(in));
StringBuilder str1 = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
str1.append(line);
Log.v(TAG, "line:" + line);
}
in.close();
s = str1.toString();
getting filenotfound exception. dont know why?
else give me some suggestion to pass username and passwrod parameter to the url by code..

The HTTPClient offers a much simpler way to access http resources, when all you want to do is fetch the repsonse body:
HttpGet httpGet = new HttpGet("http://domain.com/path?var1=bla&var2=foo");
HTTPResponse reponse = httpClient.execute(httpGet);
String responseBody = EntityUtils.toString(response.getEntity());