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();
Related
I cannot comprehend why doesn't the following code does not put a packet onto wire (confirmed via wireshark). It is a fairly standard method of sending an HTTP POST request, as I believe. I don't intend to read anything just POST.
private void sendRequest() throws IOException {
String params = "param=value";
URL url = new URL(otherUrl.toString());
HttpURLConnection con = (HttpURLConnection)url.openConnection();
con.setDoOutput(true);
con.setDoInput(true); //setting this to `false` does not help
con.setRequestMethod("POST");
con.setRequestProperty("Content-Type", "text/plain");
con.setRequestProperty("Content-Length", "" + Integer.toString(params.getBytes().length));
con.setRequestProperty("Accept", "text/plain");
con.setUseCaches(false);
con.connect();
DataOutputStream wr = new DataOutputStream(con.getOutputStream());
wr.writeBytes(params);
wr.flush();
wr.close();
//Logger.getLogger("log").info("URL: "+url+", response: "+con.getResponseCode());
con.disconnect();
}
What happens is... actually nothing, unless I try to read anything. For example by uncommenting the above log line which reads the response code. Trying to read a response via con.getInputStream(); also works. There is no movement of packets. When I uncomment the getResponseCode, I can see that http POST is sent, and then 200 OK is sent back. The order is proper. I.e. I don't get some wild response before sending POST. Everything else looks exactly the same (I can attach wireshark screenshots if needed.). In the debugger the code executes (i.e. does not block anywhere).
I don't understand under what circumstances this can be happening. I belive it should be possible, to send a POST request with con.setDoInput(false);. Currently it doesn't send anything or fails (when trying to execute con.getResponseCode()) with an exception because I obviously promised I won't read anything.
It might be relevant, that before sendRequest I do request some data from the same site, but I trust I close everything properly. I.e:
public static String getData(String urlAddress) throws MalformedURLException, IOException {
URL url = new URL(urlAddress);
HttpURLConnection con = (HttpURLConnection)url.openConnection();
con.setDoOutput(false);
InputStream in = con.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
StringBuilder data = new StringBuilder();
String line;
while((line = reader.readLine()) != null) {
data.append(line);
}
reader.close();
in.close();
con.getResponseCode();
con.disconnect();
return data.toString();
}
The server for url in both cases is the same, port also, so I believe it is possible to use the same socket for communication. The above code works and retrieves the data properly.
I am not sure, maybe I don't clean something, and it gets cached, so with out an explicit read the POST gets delayed. There is no other traffic on the socket.
Unless you're using fixed-length or chunked transfer mode, HttpURLConnection will buffer all your output until you call getInputStream() or getResponseCode(), so that it can send a correct Content-length header.
If you call getResponseCode() you should have a look at its value.
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 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!
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);
I am trying to connect to a URL from a desktop app, and I get the error indicated in the Title of my question, but when I tried to connect to the same URL from servlet, all works fine. When I load the URL from browser, all works fine. I am using the same code in the servlet. The code was in a library, when it didn't work, I pulled the code out to a class in the current project, yet it didn't work.
The URL https://graph.facebook.com/me.
The Code fragment.
public static String post(String urlSpec, String data) throws Exception {
URL url = new URL(urlSpec);
URLConnection connection = url.openConnection();
connection.setDoOutput(true);
OutputStreamWriter writer = new OutputStreamWriter(connection.getOutputStream());
writer.write(data);
writer.flush();
BufferedReader reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line = "";
StringBuilder builder = new StringBuilder();
while((line = reader.readLine()) != null) {
builder.append(line);
}
return builder.toString();
}
I'm a little bit confused here, is there something that is present is a servlet that is not a normal desktop app?
Thanks.
FULL STACK TRACE
Feb 8, 2011 9:54:14 AM com.trinisoftinc.jiraffe.objects.FacebookAlbum create
SEVERE: null
java.io.IOException: Server returned HTTP response code: 400 for URL: https://graph.facebook.com/me
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1313)
at sun.net.www.protocol.https.HttpsURLConnectionImpl.getInputStream(HttpsURLConnectionImpl.java:234)
at com.jiraffe.helpers.Util.post(Util.java:49)
at com.trinisoftinc.jiraffe.objects.FacebookAlbum.create(FacebookAlbum.java:211)
at com.trinisoftinc.jiraffe.objects.FacebookAlbum.main(FacebookAlbum.java:261)
EDIT: You need to find the exact error message that facebook is sending in the response
You can modify your code to get the message from the error stream like so:
HttpURLConnection httpConn = (HttpURLConnection)connection;
InputStream is;
if (httpConn.getResponseCode() >= 400) {
is = httpConn.getErrorStream();
} else {
is = httpConn.getInputStream();
}
Take a look at how you are passing the user context
Here's some information that could help you out:
Look at the error message behind the 400 response code:
"Facebook Platform" "invalid_request" "An active access token must be used to query information about the current user*
You'll find the solution here
HTTP/1.1 400 Bad Request
...
WWW-Authenticate: OAuth "Facebook Platform" "invalid_request" "An active access token must be used to query information about the current user."
...
I finally found the problem. Of course it's my code. One part of the code I didn't post is the value of data. data must contain only name and description but I am passing more than name and description.