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();
I need to obtain the input stream to a HTTPS URL eg. https://baseurl.com/mypdfgenerated.php?param=somevalue. In order to access this URL I need to get through the login page (eg. https://baseurl.com/login.php) by supplying BODY parameters:
user_name, web_pwd and submit_login
I'm assuming the only way to successfully access the first URL is by a POST to the /login.php followed by storing the cookies and then reusing the cookie-session-ID in the next GET request; if this is the correct approach then could someone please share a solution with the correct/recent libraries?
Not sure which is the best way but what helped me achieve this is the CloseableHttpClient class which along with BasicCookieStore retains cookies for subsequent requests once logged in, implemented below:
BasicCookieStore cookieStore = new BasicCookieStore();
CloseableHttpClient httpclient = HttpClients.custom().setDefaultCookieStore(cookieStore).build();
HttpUriRequest login = RequestBuilder.post()
.setUri(new URI(url_login))
.addParameter("login", "loginuname")
.addParameter("password", "pwd")
.addParameter("submit", "sub_mit");
CloseableHttpResponse response = httpclient.execute(login);
List<Cookie> cookies = cookieStore.getCookies();
response.close();
HttpGet httpget2 = new HttpGet(url_to_get_after_login);
CloseableHttpResponse response2 = httpclient.execute(httpget2);
response2.close();
Sample code snippet from Java Samples
try {
System.setProperty("java.protocol.handler.pkgs", "com.sun.net.ssl.internal.www.protocol");
java.security.Security.addProvider(new com.sun.net.ssl.internal.ssl.Provider());
URL url = new URL("https://www.yourwebsite.com/"); // Some URL
HttpsURLConnection connection = (HttpsURLConnection) url.openConnection();
connection.setDoInput(true);
connection.setDoOutput(true);
connection.setRequestMethod("POST");
connection.setFollowRedirects(true);
String query = "UserID=" + URLEncoder.encode("username");
query += "&";
query += "password=" + URLEncoder.encode("password");
query += "&";
// open up the output stream of the connection
DataOutputStream output = new DataOutputStream( connection.getOutputStream() );
// write out the data
output.writeBytes( query );
}catch(Exception err){
err.printStackTrace();
}
Have a look at Usage of cookies
You should use a library which handles cookies for you, such as Apache HTTPClient.
I am newbie developer in Java. STEP 1 I have already done the the following:
Logged in to REST API server (with login&password)
Received a token in XML format which i parsed with SAX parser so now i
am in a position of a token. Below is the sample code for Login:
Java code:
String url1 = "https://api4.liverail.com/login";
URL obj = new URL(url1);
HttpsURLConnection con1 = (HttpsURLConnection) obj.openConnection();
String urlParameters ="username=paania#gmail.com&password=d372a15b714bd250e";
con1.setDoOutput(true);
con1.setRequestMethod("POST");
DataOutputStream wr = new DataOutputStream(con1.getOutputStream());
wr.writeBytes(urlParameters);
STEP 2: I want to pass the token to REST API to obtain some information e.g a list from category but when i send the request via GET method , i get a response in XML saying [CDATA[You need to be logged in]] This is the code in Java:
String url = "http://api4.liverail.com/advertising/category/list/?token="72938howdwoi";
URL obj = new URL(url);
HttpURLConnection con = (HttpURLConnection) obj.openConnection();
con.setRequestMethod("GET");
BufferedReader in = new BufferedReader(newInputStreamReader(con.getInputStream()));
in.close();
con.disconnect();
I am not sure what i am missing here.
Any suggestions?
Just changed your url for request of data :
String url = "http://api4.liverail.com/advertising/category/list/?token=72938howdwoi";
When I am sending POST request to an url with authentication from GAE, I am getting MalformedURLException on urlConn.getResponseCode() function.
This problem arises only on the deployment server and does not arise when I run the code on local server.
Invalid URL specified: https://user:passtoken#twilix.exotel.in/v1/Accounts/user/Sms/send
java.net.MalformedURLException: Invalid URL specified: https://user:passtoken#twilix.exotel.in/v1/Accounts/user/Sms/send
at com.google.appengine.api.urlfetch.URLFetchServiceImpl.convertApplicationException(URLFetchServiceImpl.java:120)
at com.google.appengine.api.urlfetch.URLFetchServiceImpl.fetch(URLFetchServiceImpl.java:43)
at com.google.apphosting.utils.security.urlfetch.URLFetchServiceStreamHandler$Connection.fetchResponse(URLFetchServiceStreamHandler.java:417)
at com.google.apphosting.utils.security.urlfetch.URLFetchServiceStreamHandler$Connection.getInputStream(URLFetchServiceStreamHandler.java:296)
My code:
String urlString =
"https://user:passtoken#twilix.exotel.in/v1/Accounts/user/Sms/send";
URL url = new URL(urlString.toString());
String userpass = "user:passtoken";
String basicAuth = "Basic " +
javax.xml.bind.DatatypeConverter.printBase64Binary(userpass.getBytes());
urlConn = (HttpURLConnection) url.openConnection();
urlConn.setRequestMethod("POST");
urlConn.setRequestProperty ("Authorization", basicAuth);
urlConn.connect();
responseCode = urlConn.getResponseCode(); //Throws Exception
Everything works fine when I try to set my url as https://httpbin.org/post
Is this a bug of GAE or some problem with my code?
It looks like the problem is with the ampersand symbol. Since you are already putting the credentials in the HTTP Request, they should not be needed in the URL as well. Try changing the first line to this instead.
String urlString = "https://twilix.exotel.in/v1/Accounts/user/Sms/send";
String album = "http://picasaweb.google.com/data/feed/api/user/"+email;
HttpURLConnection con = (HttpURLConnection) new URL(albumUrl).openConnection();
// request method, timeout and headers
con.setRequestMethod("GET") ;
con.setReadTimeout(15000);
con.setRequestProperty("Authorization", "GoogleLogin auth="+auth);
con.setRequestProperty("GData-Version", "2");
// set timeout and that we will process output
con.setReadTimeout(15000);
con.setDoOutput(true);
// connnect to url
con.connect();
// read output returned for url
BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));
Problem : Everytime i call con.getInputStream() it gives me file not found exception.
But when i load the same url in the desktop browser then it is displaying correct data.
I am confused why on android it is throwing exception.
Thanks in advance.
Did you get this? Maybe you just missed the https
below example uses default for authenticated user and the experimental fields list.
url = "https://picasaweb.google.com/data/feed/api/user/default?kind=album&access=public&fields="
+ URLEncoder
.encode("entry(title,id,gphoto:numphotosremaining,gphoto:numphotos,media:group/media:thumbnail)",
"UTF-8");
https://developers.google.com/picasa-web/docs/2.0/developers_guide_protocol#ListAlbums