I am trying to send XML data with POST request to the server. But getting 500 error.
URL url = new URL(AFConstants.ServerEndPoint);
HttpURLConnection http = (HttpURLConnection) url.openConnection();
http.setRequestMethod("POST");
http.setRequestProperty("Content-Type",
"application/xml;charset=utf-8");
http.setDoOutput(true);
http.setDoInput(true);
http.connect();
DataOutputStream dataOutputStream = new DataOutputStream(http.getOutputStream());
String encoded = URLEncoder.encode(xml,"UTF-8");
dataOutputStream.writeBytes(encoded);
dataOutputStream.flush();
dataOutputStream.close();
Any ideas what is going wrong?
String xml contains this:
<?xml version=“1.0” encoding=“utf-8"?>
<AddSalesOrder revision=“8.0” environment=“Production” lang=“en-US” xmlns:xsi=“http://www.w3.org/2001/XMLSchema-instance” xsi:noNamespaceSchemaLocation=“C:\Data2\Ntp\shoprvparts\AddSalesOrder.xsd”>
<ApplicationArea>
...
</ApplicationArea>
<DataArea>
<Add confirm=“Always”/>
<SalesOrder>
<Header>
</Header>
<Line>
<LineNumber>1</LineNumber>
<OrderItem>
...
</OrderItem>
<OrderQuantity uom=“string”></OrderQuantity>
</Line>
</SalesOrder>
</DataArea>
</AddSalesOrder>
500 Internal Server Error or HTTP 500 - Internal Server Error
Cause of HTTP 500 Errors
Most of the time, "wrong" means an issue with the page or site's programming, but there's certainly a chance that the problem is on your end
so i think there will be error logs on the server and if you have access you can provide the server logs and will be easy to identify the problem
Related
I'm about to code an Android app (using A.Studio 3.5.1) that should connect to a back-end using https. I'm quite new to the techniques so I looked at https://developer.android.com/training/articles/security-ssl.html#HttpsExample
I has four lines of code:
URL url = new URL("https://wikipedia.org");
URLConnection urlConnection = url.openConnection();
InputStream in = urlConnection.getInputStream();
copyInputStreamToOutputStream(in, System.out);
When I try it I get a red text;
new URL gives MalformedURLException
url.openConnection() gives IOException,
urlConnection.getInputStream() also
copyInputStreamToOutputStream is not found.
I have read about copyInputStreamToOutputStream, that it can be solved
copyInputStreamToOutputStream(in, System.out)
and
Easy way to write contents of a Java InputStream to an OutputStream
I tried the first way, using apache commons, with no success.
I'm mostly curious about the exceptions.
I will try another walk-through on
https://codelabs.developers.google.com/codelabs/android-network-security-config/
But it would be nice to learn about this...
I have faced similar situations many times. This occur because the webpage you are trying to reach doesn't return anything as a response after connection is established. So the parsing returns IOException as there is nothing to read in the response of URL after connection.
newURL is returning MalformedURLException because your url contains -- " -- which should be escaped as these are special characters.
Hope this helps.
i'm trying to get the source code from a site using this code
conn = (HttpURLConnection)url.openConnection();
conn.setReadTimeout(10000);
conn.setConnectTimeout(20000);
conn.setRequestMethod("GET");
conn.setDoInput(true);
conn.addRequestProperty("User-Agent", "Mozilla/5.0");
conn.setRequestProperty("Accept-Encoding", "gzip, deflate");
int response = conn.getResponseCode();
if (response = 307){
String locationHeader = conn.getHeaderField("Location");
URL redirectURL = new URL(locationHeader);
...
}
when the server responds with a 307 code i create a new connection with the same parameters as above with the new url given by the server.
this code works fine while following the first 2 redirects, at the third the server gives a relative url that forces a MalformedURLException when my code executes ' new URL(locationHeader); '.
so i tried to fix that adding the ' http://servername/ ' to the ' locationHeader ' string but doing that creates a loop cause the server then redirects to the first url of his redirection chain.
since my browser gets the source code from that server with no problems is there a way to achieve that with HttpURLConnection?
if someone is interested thanks to Fiddler i worked out a solution to this issue.
first i changed the "User-Agent" property to mimic the one of Mozilla then i manually tweaked the cookie the serer was sending in its reply with the relative path.
that did the trick. thank you Fiddler.
I need to connect to server using SSL with REST. I am using HttpsURLConnection but somehow I cannot even get response code from server. The url is correct and when I try to get response code, it just throws nullPointerException.
Code snippet:
URL url = new URL("https", "xxx.com/yyy/zzz.svc", "myApi/test");
HttpsURLConnection conn = (HttpsURLConnection) url.openConnection();
Log.e("Response code", "Response from server: " + conn.getResponseCode());
What am I doing wrong here?
Note: I have to add remote server address and authorization token to header, but it have to return some response code anyway. For authorization error I have to receive error 400.
EDIT: The nullPointerException was hidden under Malformed URL.
Check out How to parse / read JSON data from the URL.
I'm trying around with the HttpURLConnection for quite some time now and I tried several solution posted here and on other places, but nothing seems to work.
I have the following architecture:
A Ruby on Rails Web Service (Rest interface with JSON)
An iPhone Client with RestKit
An Android Client with HttpURLConnection
The iPhone Client works like a charm. It connects to the web service with RestKit.
Now the Android Client is a completely different story. I always get a 401 Unauthorized message from the server (which results in a local FileNotFoundException).
The strange thing is, that the iPhone Client gets the same error, but RestKit somehow manages the handle it by sending the same request again. I tried that of course, but I just get the same error twice.
On the Rails Log Output it looks like this:
Started POST "/api/v1/login" for 127.0.0.1 at 2012-05-03 12:44:56 +0200
Processing by Api::V1::ApiController#session_login as JSON
Parameters: {"device"=>{"model"=>"Simulator", "system"=>"Android", "version"=>"Hugo", "name"=>"Android Simulator"}, "email"=>"florian.letz#simpliflow.com", "api"=>{"device"=>{"model"=>"Simulator", "system"=>"Android", "version"=>"Hugo", "name"=>"Android Simulator"}, "email"=>"florian.letz#simpliflow.com", "action"=>"session_login", "controller"=>"api/v1/api"}}
Filter chain halted as :require_login_from_http_basic rendered or redirected
Completed 401 Unauthorized in 0ms (ActiveRecord: 0.0ms)
The exact same message occurs when the iPhone Client connect's but then suddenly a magical second request occurs and it goes through.
On the Android Client I do the following:
HttpURLConnection con = (HttpURLConnection) url.openConnection();
con.setRequestMethod(HTTP_POST);
con.setRequestProperty("Content-Type", "application/json");
con.setRequestProperty("Accept", "application/json");
con.setRequestProperty("Accept-Encoding", "gzip, deflate");
String userpassword = email + ":" + password;
con.setRequestProperty("Authorization", "Basic " + new String(Base64.encodeBase64(userpassword.getBytes())));
String body = jsonLogin.toString();
byte[] body_bytes = body.getBytes();
con.setRequestProperty("Content-Length", "" + Integer.toString(body_bytes.length));
con.setRequestProperty("Content-Language", "en-US");
con.setInstanceFollowRedirects(false);
con.setUseCaches (false);
con.setDoInput(true);
con.setDoOutput(true);
DataOutputStream wr = new DataOutputStream (con.getOutputStream ());
wr.write(body_bytes);
wr.flush ();
wr.close ();
InputStream is = con.getInputStream();
And on the last line the exception occurs.
I've read some things about redirects, but there is no redirect implemented at the server and I do not receive one on the client. I just get the 401 Unauthorized. The code in the web service and the iphone client indicate a quite simple workflow. Just send the data and receive the answer. I don't know where the SECOND login call comes from when the iPhone connects.
Does anyone here have any idea what the problem could be?
Thanks a lot!
EDIT #1:
I have identified the "magical" second request. The RestKit Log shows the following:
Asked if canAuthenticateAgainstProtectionSpace: with authenticationMethod = NSURLAuthenticationMethodDefault
This then results in the second request with quite a buch of headers I cannot make any sense of.
So do you know a way to implement this in Android?
I've encountered a rather strange error. I've written an android application that uploads an simple text file to a server. The code for the connection is as follows:
try {
URL = new URL(myURL);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "multipart/form-data; boundary="+"---");
DataOutputStream output = new DataOutPutstream(connection.getOutputStream());
output.writeBytes(<my post request>);
output.flush();
output.close();
connection.connect();
}
When I run this method, the post request is never sent over to the server. Nothing shows up on wireshark and if I look in logcat, there are no errors and the connection gets made fine, the POST message is just never sent. However, if I add a simple line right after the connection.connect() such as:
connection.getResponseCode();
Suddenly the POST message gets sent over no problem. What's going on here? Am I required to get a response code in order for the message to get sent over?
Why do you have to call URLConnection#getInputStream to be able to write out to URLConnection#getOutputStream?
In short you must call getInputStream() and close it. getResponseCode() is also working because it requires an established connection.
You do not need to call
connection.connect();
which is redundant.
You can get working sample from here: http://www.xyzws.com/Javafaq/how-to-use-httpurlconnection-post-data-to-web-server/139
You set doInput to true. Maybe the url connection waits for an input because of that. Try to set it to false. But I could also be wrong. It is just a guess.