I'm trying to use the API from Web Of Knowledge(WoK) to obtain some data. The documentation explain that you have to do POST Requests through HTTPS, sending a XML which contains the queries. But I only get the error 400 form server. (Bad Request)
Here is my code, I found it in Google and I make some fixes for my case.
public static void main(String[] args) throws Exception {
// Get target URL
String strURL = /*Here the Server URL*/;
// Get file to be posted
String strXMLFilename = "src/main/resources/xml/wosdata.xml";
File input = new File(strXMLFilename);
// Prepare HTTP post
PostMethod post = new PostMethod(strURL);
// Request content will be retrieved directly
// from the input stream
// Per default, the request content needs to be buffered
// in order to determine its length.
// Request body buffering can be avoided when
// content length is explicitly specified
post.setRequestEntity(new InputStreamRequestEntity(
new FileInputStream(input), input.length()));
// Specify content type and encoding
// If content encoding is not explicitly specified
// ISO-8859-1 is assumed
post.setRequestHeader(
"Content-type", "text/xml; charset=ISO-8859-1");
// Get HTTP client
HttpClient httpclient = new HttpClient();
// Execute request
try {
int result = httpclient.executeMethod(post);
// Display status code
System.out.println("Response status code: " + result);
// Display response
System.out.println("Response body: ");
System.out.println(post.getResponseBodyAsString());
}catch (Exception e) {
e.printStackTrace();
} finally {
// Release current connection to the connection pool
// once you are done
post.releaseConnection();
}
}
There is something wrong with the XML you are sending. You will have to look at server logs to find out exactly what, as 400 deliberately tells you as little as possible.
You should do it like this. First read the contents of the xml to String and do post using a StringRequestEntity.
// Get file to be posted
String strXMLFilename = "src/main/resources/xml/wosdata.xml";
StringBuilder contents = new StringBuilder();
try {
BufferedReader input = new BufferedReader(new FileReader(new File(strXMLFilename)));
try {
while (( line = input.readLine()) != null){
contents.append(line);
contents.append(System.getProperty("line.separator"));
}
}
finally {
input.close();
}
StringEntity requestEntity = new StringEntity(contents.toString());
post.setEntity(requestEntity);
Related
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.
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.
I have a problem with a WebService on Android. I am getting a 400 error but there is no information on the ErrorStream.
What I am trying to do is a POST request on a WCF Webservice using JSON.
I must add that I have includeExceptionDetailInFaults Enabled on my Service. The last time I got a 400 error, it was because I hadn't defined the RequestProperty. Now I don't get any error in the stream.
Here is the code:
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
// In my last error I had not included these lines. Maybe they are still wrong?
urlConnection.setRequestProperty("Content-Type", "application/json");
urlConnection.setRequestProperty("Accept", "application/json");
urlConnection.setRequestMethod("POST");
urlConnection.setDoOutput(true);
urlConnection.setChunkedStreamingMode(0);
OutputStream out = new BufferedOutputStream(urlConnection.getOutputStream());
OutputStreamWriter outputStreamWriter = new OutputStreamWriter(out);
outputStreamWriter.write(jsonObject.toString(), 0, jsonObject.length());
outputStreamWriter.flush();
//outputStreamWriter.close();
int code = urlConnection.getResponseCode();
System.out.println(code);
if(code == 400) {
BufferedInputStream errorStream = new BufferedInputStream(urlConnection.getErrorStream());
InputStreamReader errorStreamReader = new InputStreamReader(errorStream);
BufferedReader bufferedReader = new BufferedReader(errorStreamReader);
StringBuilder builder = new StringBuilder();
String aux = "";
while ((aux = bufferedReader.readLine()) != null) {
builder.append(aux);
}
String output = builder.toString(); // The output is empty.
System.out.print(output);
}
Check Retrofit library from Square it's more easy and thin for GET/POST request and especially for REST. I suggest you to try it. It will make your life easy.
You can use different JSON parsers, error handlers, etc. Very flexible.
POST request definition using retrofit it's simple like this:
An object can be specified for use as an HTTP request body with the #Body annotation.
#POST("/users/new")
void createUser(#Body User user, Callback<User> cb);
Methods can also be declared to send form-encoded and multipart data.
Form-encoded data is sent when #FormUrlEncoded is present on the method. Each key-value pair is annotated with #Field containing the name and the object providing the value.
#FormUrlEncoded
#POST("/user/edit")
User updateUser(#Field("first_name") String first, #Field("last_name") String last);
After you define method inside your Java interface like shown above instantiate it:
RestAdapter restAdapter = new RestAdapter.Builder()
.setEndpoint("https://api.soundcloud.com")
.build();
MyInterface service = restAdapter.create(MyInterface.class);
And then you can call your method synchronously or asynchronously (in case you pass Callback instance).
service.myapi(requestBody);
See Retrofit documentation (http://square.github.io/retrofit/javadoc/index.html) and samples on GitHub for more details.
A 400 error might be occuring (and usually occurs in my case) because of incorrect URL or bad JSON format in post. please check those two
Using an HttpPost object will make your job a lot easier in my opinion
HttpPost post = new HttpPost(url);
if(payload != null){
try {
StringEntity entity = new StringEntity(payload,HTTP.UTF_8);
entity.setContentType(contentType);
post.setEntity(entity);
} catch (UnsupportedEncodingException e) {
LOG.d(TAG, "post err url : " + url);
LOG.e(TAG, "post err url" , e);
throw new Exception(1, e);
}
}
HttpResponse response=executeRequest(owner, post);
I'm trying to login to a webpage, but even before that, I'm loading the page using HttpGet, and this is one the lines that's being returned,
ÓA;
That's all I could put, won't let me paste any other characters. But they are all like that, like I'm somehow getting the wrong encoding? Here is the code I am using to GET
HttpGet httpget = new HttpGet(url);
if(headers == null) {
headers = getDefaultHeaders();
}
for(String s : headers.keySet()) {
httpget.addHeader(s, headers.get(s));
}
HttpResponse response = getClient().execute(httpget);
HttpEntity entity = response.getEntity();
System.out.println("Status Line: " + response.getStatusLine());
if (entity != null) {
InputStream input = entity.getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(input));
String ln = "";
while((ln = reader.readLine()) != null) {
System.out.println("During Get - " + ln);
}
}
What am I doing wrong?
Thanks for any help.
If you need any more information like headers, just ask.
The following line is possibly the cause of your problems:
BufferedReader reader = new BufferedReader(new InputStreamReader(input));
You are creating a reader using the default characterset of your platform, and completely ignoring any character set that may be specified in the HTTP response headers.
If you are getting the same problem when reading the content the correct way, then it is possible that the server is at fault for not setting the response header correctly.
DO the entity reading like this:
String content = org.apache.http.util.EntityUtils.toString( entity );
System.out.println(content);
This is going to read it all for you so you can check what's being really returned.
Make sure that you didn't accidentally go to port 443 with a simple HTTP connection. Because in that case you will get back the SSL handshake instead of an HTTP response.
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();