Tomcat returns http response code 400 - http

I was googling for a long time, but still can't find a solution to my case.
My Tomcat sometimes returns an exception :
Error in postRequest(): Server returned HTTP response code: 400 for URL: http://localhost:80/CITIUS2/webresources/entities.personainterna/
Sometimes it works and sometimes it returns this exception, so I really don't know what is the reason...
Connection function:
public static String excutePost(String targetURL, String urlParameters) throws UnsupportedEncodingException {
URL url;
HttpURLConnection connection = null;
String responseXML = null;
try {
//Create connection
url = new URL(targetURL);
connection = (HttpURLConnection) url.openConnection();
byte[] requestXML = urlParameters.getBytes();
connection.setRequestProperty("Content-Length", String.valueOf(requestXML.length));
connection.setRequestProperty("Content-Type", "application/xml; charset=utf-8");
connection.setRequestMethod("POST");
connection.setDoOutput(true);
connection.setDoInput(true);
// Send the String that was read into postByte.
OutputStream out = connection.getOutputStream();
out.write(requestXML);
out.close();
// Read the response and write it to standard out.
InputStreamReader isr = new InputStreamReader(connection.getInputStream());
BufferedReader br = new BufferedReader(isr);
String temp;
String tempResponse = "";
//Create a string using response from web services
while ((temp = br.readLine()) != null) {
tempResponse = tempResponse + temp;
}
responseXML = tempResponse;
br.close();
isr.close();
} catch (java.net.MalformedURLException e) {
System.out.println("Error in postRequest(): Secure Service Required");
} catch (Exception e) {
System.out.println("Error in postRequest(): " + e.getMessage());
}
return responseXML;
}
# Edit:
In general build is successful, there are no errors, only this one in the Apache Tomcat's output window.
Rest method:
#POST
#Consumes({"application/xml", "application/json"})
public Response create(Personainterna entity) {
try {
getJpaController().create(entity);
return Response.created(URI.create(entity.getPersonaId().toString())).build();
} catch (Exception ex) {
return Response.notModified(ex.getMessage()).build();
}
}

Related

Problem to execute https connection from a servlet: http 404 error occours

From my Tomcat's servlet I execute an https connection to an external servlet.
This is the code:
HttpsURLConnection hpcon = null;
try {
URL url = new URL(surl);
hpcon = (HttpsURLConnection) url.openConnection();
hpcon.setRequestMethod("POST");
hpcon.setDoInput(true);
hpcon.setDoOutput(true);
hpcon.setUseCaches(false);
hpcon.setRequestProperty("Content-type", "application/x-www-form-urlencoded");
OutputStreamWriter wr = new OutputStreamWriter(hpcon.getOutputStream());
String params = "user=" + URLEncoder.encode(user, "UTF-8");
params += "&psswd=" + URLEncoder.encode(pssw, "UTF-8");
params += "&metodo=" + URLEncoder.encode(metodo, "UTF-8");
wr.write(params);
wr.flush();
wr.close();
hpcon.connect();
int respCode = hpcon.getResponseCode();
if (respCode == 200) {
BufferedReader br = new BufferedReader(new InputStreamReader(hpcon.getInputStream()));
String response = "";
String output;
while ((output = br.readLine()) != null) {
response += output;
}
if (response.indexOf("-") > 0) {
response = "-12";
System.out.println("ret = -12 - response = " + response);
}
br.close();
} else {
ret = "-11";
System.out.println("ret = -11 - respCode = " + respCode);
}
} catch (Exception e) {
e.printStackTrace();
ret = "-10";
System.out.println("ret = -10");
} finally {
if (hpcon != null) {
hpcon.disconnect();
}
}
Where surl is the full url of a servlet present in a different domain and the three parameters are read from a db table (the third really is fixed and is the operation that is make by the external servlet).
The result is:
ret = -11 - respCode = 404
Before make the connection I turn off the certificate's verify using the above code:
try {
TrustManager[] trustAllCerts = new TrustManager[]{new X509TrustManager() {
public java.security.cert.X509Certificate[] getAcceptedIssuers() {
return null;
}
public void checkClientTrusted(X509Certificate[] certs, String authType) {
}
public void checkServerTrusted(X509Certificate[] certs, String authType) {
}
}};
SSLContext sc = SSLContext.getInstance("SSL");
sc.init(null, trustAllCerts, new java.security.SecureRandom());
HttpsURLConnection.setDefaultSSLSocketFactory(sc.getSocketFactory());
HostnameVerifier allHostsValid = new HostnameVerifier() {
public boolean verify(String hostname, SSLSession session) {
return true;
}
};
HttpsURLConnection.setDefaultHostnameVerifier(allHostsValid);
} catch (Exception e) {
e.printStackTrace();
}
If I execute the same servlet manually from a browser with parameters in get mode all run correctly.
I tried to execute it on my code using the get mode and passing the three parameters in query string, but the result is the same.
How can I do to resolve the problem?

Upload file from applet to servlet using apache fileupload

To accomplish:
Upload a file from my local to server using an applet and servlet using apache fileupload jar.
Tried:
I have used a simple jsp, with a browse button and posted the action to my servlet (where I used apache fileupload). I was successful in uploading the file to the server.
Issue:
I am trying to upload a file, from my local machine, using an applet. I do not want to manually select files, instead upload files that are present in a specific folder. For now I have hardcoded the folder. I am able to look at the folder and get the list of files I want to upload.
Also, I have successfully established a connection from my applet to servlet.
Issue arises at the upload.parseRequest(request) in the servlet. I'm thinking its because the applet cannot post to servlet's request object.
Also, I have set the request type to multipart/form-data in my applet.
Right now, I am trying to pass the absolute path of the file to servlet and upload.
I have seen other posts where byte stream data is passed from applet to servlet, but the servlet uses the traditional File.write.
For me, it is mandatory to achieve this using apache fileupload.
Please suggest on how to pass a file/file path from applet to servlet, where the upload is handled by apache fileupload.
Below are my FileUploadHandler (where the HTTP requests are handled) and FileUpload(which is my applet)
Below is my FileUpload Handler:
#WebServlet(name = "FileUploadHandler", urlPatterns = { "/upload" })
#MultipartConfig
public class FileUploadHandler extends HttpServlet {
String uploadFolder ="";
#Override
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
System.out.println("doPost-servlet URL is: "
+ request.getRequestURL());
try {
uploadFolder = fileToUpload(request);
} catch (ClassNotFoundException e) {
e.printStackTrace();
}
uploadFolder = getServletContext().getRealPath("")+ File.separator;
// Create a factory for disk-based file items
DiskFileItemFactory factory = new DiskFileItemFactory();
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
// process only if its multipart content
if (ServletFileUpload.isMultipartContent(request)) {
System.out.println("Yes, it is a multipart request...");
try {
List<FileItem> multiparts = upload.parseRequest(request);
System.out.println("Upload.parseRequest success !");
for (FileItem item : multiparts) {
if (!item.isFormField()) {
String name = new File(item.getName()).getName();
item.write(new File(uploadFolder + File.separator
+ name));
}
}
System.out.println("File uploaded to server !");
// File uploaded successfully
request.setAttribute("message", "File Uploaded Successfully");
} catch (Exception ex) {
request.setAttribute("message", "File Upload Failed due to "
+ ex);
}
} if(!ServletFileUpload.isMultipartContent(request)){
throw new ServletException("Content type is not multipart/form-data");
}
doGet(request, response);
//request.getRequestDispatcher("/result.jsp").forward(request, response);
OutputStream outputStream = response.getOutputStream();
ObjectOutputStream objectOutputStream = new ObjectOutputStream(outputStream);
objectOutputStream.writeObject("Success !");
objectOutputStream.flush();
objectOutputStream.close();
}
private String fileToUpload(HttpServletRequest request) throws IOException,
ClassNotFoundException {
ServletInputStream servletIn = request.getInputStream();
ObjectInputStream in = new ObjectInputStream(servletIn);
String uploadFile = (String) in.readObject();
System.out.println("Value in uploadFolder is: " + uploadFile);
return uploadFile;
}
Below is the fileupload applet:
public class FileUpload extends Applet {
private JButton capture;
private JTextField textField;
private final String pathDirectory = "C:\\";
private final String captureConfirmMessage = "Are you sure you want to continue?";
private final String confirmDialogTitle = "Confirm upload";
final File folder = new File(pathDirectory);
public void init() {
upload= new JButton("Upload");
textField = new JTextField();
capture.addActionListener(new ActionListener() {
#Override
public void actionPerformed(ActionEvent e) {
int selection = JOptionPane.showConfirmDialog(upload,
uploadConfirmMessage, confirmDialogTitle,
JOptionPane.YES_NO_OPTION);
if (selection == JOptionPane.OK_OPTION) {
listFilesForFolder(folder);
} else if (selection == JOptionPane.CANCEL_OPTION) {
JOptionPane.showMessageDialog(upload,
"You have aborted upload", "Upload Cancelled", 2);
}
}
});
add(upload);
add(textField);
}
public void listFilesForFolder(final File folder) {
for (final File fileEntry : folder.listFiles()) {
if (fileEntry.isDirectory()) {
listFilesForFolder(fileEntry);
} else {
try {
onSendData(fileEntry.getAbsolutePath());
System.out.println(fileEntry.getAbsolutePath());
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
private URLConnection getServletConnection() throws MalformedURLException,
IOException {
// Open the servlet connection
URL urlServlet = new URL("http://localhost:8081/UploadFile/upload");
HttpURLConnection servletConnection = (HttpURLConnection) urlServlet
.openConnection();
// Config
servletConnection.setDoInput(true);
servletConnection.setDoOutput(true);
servletConnection.setUseCaches(false);
servletConnection.setDefaultUseCaches(false);
servletConnection.setRequestProperty("Content-Type", "multipart/form-data;");
servletConnection.connect();
return servletConnection;
}
private void onSendData(String fileEntry) {
try {
// Send data to the servlet
HttpURLConnection servletConnection = (HttpURLConnection) getServletConnection();
OutputStream outstream = servletConnection.getOutputStream();
ObjectOutputStream objectOutputStream= new ObjectOutputStream(
outstream);
objectOutputStream.writeObject(fileEntry);
// Receive result from servlet
InputStream inputStream = servletconnection.getInputStream();
ObjectInputStream objectInputStream = new ObjectInputStream(
inputStream);
String result = (String) objectInputStream.readObject();
objectInputStream.close();
inputStream.close();
out.flush();
out.close();
// Display result on the applet
textField.setText(result);
} catch (java.net.MalformedURLException mue) {
mue.printStackTrace();
textField.setText("Invalid serlvetUrl, error: " + mue.getMessage());
} catch (java.io.IOException ioe) {
ioe.printStackTrace();
textField.setText("Couldn't open a URLConnection, error: "
+ ioe.getMessage());
} catch (Exception e) {
e.printStackTrace();
textField.setText("Exception caught, error: " + e.getMessage());
}
}
public void paint(Graphics g) {
g.drawString("Click the button above to capture", 5, 50);
}
I could finally succeed posting the request to the servlet from the applet.
It was a simple logic that I was missing. I did not add the header and trailer while posting to the servlet, which was the key, in the servlet to identify the incoming request as a multipart data.
FileInputStream fileInputStream = new FileInputStream(new File(
fileEntry));
dataOutputStream.writeBytes(twoHyphens + boundary + lineEnd);
dataOutputStream
.writeBytes("Content-Disposition: form-data; name=\"upload\";"
+ " filename=\"" + fileEntry + "\"" + lineEnd);
dataOutputStream.writeBytes(lineEnd);
// create a buffer of maximum size
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];
// read file and write it into form
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
while (bytesRead > 0) {
dataOutputStream.write(buffer, 0, bufferSize);
bytesAvailable = fileInputStream.available();
bufferSize = Math.min(bytesAvailable, maxBufferSize);
bytesRead = fileInputStream.read(buffer, 0, bufferSize);
System.out.println(fileEntry + " uploaded.");
}
// send multipart form data necesssary after file data
dataOutputStream.writeBytes(lineEnd);
dataOutputStream.writeBytes(twoHyphens + boundary + twoHyphens
+ lineEnd);
I added the header and trailer and also used this to create the URL connection.
private URLConnection getServletConnection() throws MalformedURLException,
IOException {
// Open the servlet connection
URL urlServlet = new URL("http://localhost:8083/UploadFile/upload");
HttpURLConnection servletConnection = (HttpURLConnection) urlServlet
.openConnection();
// Config
servletConnection.setDoInput(true);
servletConnection.setDoOutput(true);
servletConnection.setUseCaches(false);
servletConnection.setDefaultUseCaches(false);
servletConnection.setRequestMethod("POST");
servletConnection.setRequestProperty("Content-Type",
"multipart/form-data;boundary=" + this.boundary);
servletConnection.setRequestProperty("Connection", "Keep-Alive");
servletConnection.connect();
return servletConnection;
}
Then, in the servlet, I was just reading the data using upload.ParseRequest(request).
Thank you for the help.

The remote server returned an error: (400) Bad Request. When try to login with facebook in asp.net

I am try to login with a user with Facebook account in my website, but the applcation gives me error that The remote server returned an error: (400) Bad Request.
Below is my code:
public string WebRequest(Method method, string url, string postData)
{
HttpWebRequest webRequest = null;
StreamWriter requestWriter = null;
string responseData = "";
webRequest = System.Net.WebRequest.Create(url) as HttpWebRequest;
webRequest.Method = method.ToString();
webRequest.ServicePoint.Expect100Continue = false;
webRequest.UserAgent = "[You user agent]";
webRequest.Timeout = 50000;
if (method == Method.POST)
{
webRequest.ContentType = "application/x-www-form-urlencoded";
//POST the data.
requestWriter = new StreamWriter(webRequest.GetRequestStream());
try
{
requestWriter.Write(postData);
}
catch
{
throw;
}
finally
{
requestWriter.Close();
requestWriter = null;
}
}
responseData = WebResponseGet(webRequest);
webRequest = null;
return responseData;
}
*It gives me error in this method:*
public string WebResponseGet(HttpWebRequest webRequest)
{
StreamReader responseReader = null;
string responseData = "";
try
{
responseReader = new StreamReader(webRequest.GetResponse().GetResponseStream());
responseData = responseReader.ReadToEnd();
}
catch
{
throw;
}
finally
{
webRequest.GetResponse().GetResponseStream().Close();
responseReader.Close();
responseReader = null;
}
return responseData;
}
Ooo been a while since Iv'e played with webRequest but I think your problem might be
webRequest.GetResponse().GetResponseStream().Close();
in the finally block. Since you've already called
webRequest.GetResponse().GetResponseStream()
in the body of try block. Documentation states:
The GetResponse method sends a request to an Internet resource and
returns a WebResponse instance. If the request has already been
initiated by a call to GetRequestStream, the GetResponse method
completes the request and returns any response.
Therefore as I read it, the response had already been returned in the try block and then when you call it again in the finally block, it fails...since it's already been retrieved. Just comment out that line and see how you go. The StreamReader should close the underlying connection when you close it.
So try:
public string WebResponseGet(HttpWebRequest webRequest)
{
StreamReader responseReader = null;
string responseData = "";
try
{
responseReader = new StreamReader(webRequest.GetResponse().GetResponseStream());
responseData = responseReader.ReadToEnd();
}
catch
{
throw;
}
finally
{
responseReader.Close();
}
return responseData;
}

Problems sending and receiving JSON between ASP.net web service and ASP.Net web client

You would think with all the posts here that this would be easy to figure out. :| Well here is what should be a simple example. NOTE The web service is VB and the client is c#. The wb service sends and receives fine when called from JQuery. From .NET There is a problem,
If the service asks for a parameter as show below then the client's getresponse method gets error 500 Internal server error
The Web Service
<WebMethod()> _
<ScriptMethod(ResponseFormat:=ResponseFormat.Json, XmlSerializeString:=False)> _
Public Function Test(WebInfo As GetUserID) As Person
Dim Someone As New Person
Someone.Name = "Bob"
Someone.FavoriteColor = "Green"
Someone.ID = WebInfo.WebUserID.ToString()
Return Someone
End Function
The Web Client (set up to be send and receive JSON)
public Person Test(int UserID, string url) {
HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create(url + "test.asmx/Test");
webRequest.Method = "POST";
webRequest.ContentType = "application/json; charset=utf-8";
StreamWriter sw = new StreamWriter(webRequest.GetRequestStream());
sw.Write("{'WebInfo':{'WebUserID':1}}"); // this works from JQuery
HttpWebResponse webResponse = (HttpWebResponse)webRequest.GetResponse();
Stream responseStream = webResponse.GetResponseStream();
DataContractJsonSerializer jsonSerializer = new DataContractJsonSerializer(typeof(Person));
Person someone = (Person)jsonSerializer.ReadObject(responseStream);
return someone;
}
Has anyone out there done this successfully?
Thanks
Here is a method that makes calls to a JSON web service, allowing the developer to both send and receive complext data types. The object passed in can be any data type or class. The result is a JSON string, and or any error message the methods type is shown below
public class WebServiceCallReturn {
public string JSONResponse { get; set; }
public string SimpleResponse { get; set; }
public string Error { get; set; }
}
public WebServiceCallReturn WebServiceJSONCall(string uri, string requestType, object postData = null) {
WebServiceCallReturn result = new WebServiceCallReturn();
// create request
HttpWebRequest webRequest = (HttpWebRequest)WebRequest.Create(uri);
webRequest.ContentType = "application/json; charset=utf-8";
webRequest.Method = requestType;
webRequest.Accept = "application/json; charset=utf-8";
// add json data object to send
if (requestType == "POST") {
string json = "{ }";
if (postData != null) {
try { // the serializer is fairly robust when used this way
DataContractJsonSerializer ser = new DataContractJsonSerializer(postData.GetType());
MemoryStream ms = new MemoryStream();
ser.WriteObject(ms, postData);
json = Encoding.UTF8.GetString(ms.ToArray());
} catch {
result.Error = "Error serializing post";
}
}
webRequest.ContentLength = json.Length;
StreamWriter sw;
try {
sw = new StreamWriter(webRequest.GetRequestStream());
} catch (Exception ex) {
// the remote name could not be resolved
result.Error = ex.Message;
return result;
}
sw.Write(json);
sw.Close();
}
// read response
HttpWebResponse webResponse;
try {
webResponse = (HttpWebResponse)webRequest.GetResponse();
} catch (Exception ex) {
// The remote server returned an error...
// (400) Bad Request
// (403) Access forbidden (check the application pool)
// (404) Not Found
// (405) Method not allowed
// (415) ...not the expected type
// (500) Internal Server Error (problem with IIS or unhandled error in web service)
result.Error = ex.Message;
return result;
}
Stream responseStream = webResponse.GetResponseStream();
StreamReader sr = new StreamReader(responseStream, Encoding.UTF8);
string resultString = sr.ReadToEnd();
sr.Close();
responseStream.Close();
result.JSONResponse = resultString;
return result;
}
This method could be used as follows
public SomeCustomDataClass Getsomeinformation(int userID) {
UserInfoClass postData = new UserInfoClass();
postData.WebUserID = userID;
SomeCustomDataClass result = new SomeCustomDataClass();
string uri = URL + "SomeServices.svc/GetSomething";
WebServiceCallReturn webReturn = WebServiceJSONCall(uri, "POST", postData);
if (webReturn.Error == null) {
//resultString = CleanJSON(resultString);
JavaScriptSerializer serializer = new JavaScriptSerializer();
try {
result = serializer.Deserialize<SomeCustomDataClass>(webReturn.JSONResponse);
} catch {
result.Error = "Error deserializing";
}
} else {
result.Error = webReturn.Error;
}
return result;
}
Hope that helps someone

BlackBerry app unable to establish Http Connection with server

This is the code that I am using for an HTTP connection:
HttpConnection connection = null;
// InputStream inputstream = null;
connection = (HttpConnection) Connector.open("http://www.google.com");
//HTTP Request
connection.setRequestMethod(HttpConnection.GET);
connection.setRequestProperty("Content-Type","//text plain");
connection.setRequestProperty("Connection", "close");
add(new LabelField(""+connection.getResponseMessage()));
connection.close();
This BlackBerry Development Guide titled "Code sample: Creating a connection over HTTP by using the first available transport" worked!!
ConnectionFactory connFact = new ConnectionFactory();
ConnectionDescriptor connDesc;
connDesc = connFact.getConnection("http://www.google.com");
if (connDesc != null) {
HttpConnection httpConn;
httpConn = (HttpConnection)connDesc.getConnection();
try {
final int iResponseCode = httpConn.getResponseCode();
UiApplication.getUiApplication().invokeLater(new Runnable() {
public void run() {
Dialog.alert("Response code: " +
Integer.toString(iResponseCode));
}
});
} catch (IOException e) {
System.err.println("Caught IOException: " + e.getMessage());
}
}

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