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Soap service can't write to XML file

Simple Soap service running on Axis engine on Tomcat v9.0 server needs to read and write to XML files. I developed soap service in Eclipse like a dynamic web project, so the XML files are in the WebContent->WEB-INF->resources->...
When i read the files everything works fine, but when i want to write to the files i get InvocationTargetException. Since i read files normaly, I guess that i'm not opening stream as i should when i write in the files, so can anyone guide me how to do this properly?
Here's the method for reading the file, and this WORKS:
public Station deserialization(String name, String path) {
Station s=null;
try {
URL url=getClass().getClassLoader().getResource(path+File.separator+name+".xml");
InputStream is=url.openStream();
XMLDecoder decoder = new XMLDecoder(is);
s=(Station) decoder.readObject();
decoder.close();
} catch (Exception e) {
Main.LOGGER.info("Station deserializaton was not successful!");
}
return s;
}
and here's the method for writing into the file, this DOESN'T work:
public boolean serialize(Station s, String path) {
try {
URL url=getClass().getClassLoader() .getResource(path+File.separator+s.getName()+".xml");
URLConnection con=url.openConnection();
con.setDoOutput(true);
OutputStream out=con.getOutputStream();
XMLEncoder encoder = new XMLEncoder(out);
encoder.writeObject(s);
encoder.close();
} catch (Exception e) {
Main.LOGGER.info("Station serialization was not successful!");
return false;
}
return true;
}
My real question here is how come the same principle work when reading the file and doesn't work with writing into the file? File paths are the same in both methods.

I found what was the problem. Turns out you can read files with URL, but you cant write to URL or URLConnection. I had to use FileOutputStream for writing into the files:
XMLEncoder encoder=new XMLEncoder(new FileOutputStream("file_path"));
You can also keep the files in WebContent->WEB-INF (if you develop web service in Eclipse) if you want web service to use them, becuse it then makes copies of them.
Just keep an eye on the path that you are providing, double check if it is the right one!

Cannot read json file in java

I am using eclipse for reading a jsonfile. I put the jsonfile in my src->main->java->testjson->jsonfile.json . Now I am trying to read the jsonfile. But my progam cannot find the file. I get the output "nothing". Here is the code I already implement:
JsonParser parser = new JSONParser();
try{
Object obj = parser.parse(new FileReader("jsonfile.json"));
JSONObject jsonObject = (JSONObject) obj;
String name = (String) jsonObject.get("name");
}
catch(Exception e){
System.out.println("nothing");
}

Your file within the project is called a "resource", which will be bundled in the resulting jar-file.
In maven projects such files resides in a special folder resources (like src/main/resources/testjson/jsonfile.json), in many other project types, these files are located directly beneath the java files.
Therefore you cannot read it with FileReader, because it will not be a regular file, but zipped inside the jar file.
All you have to do is to read the file with this.getClass().getResourceAsStream("/testjson/jsonfile.json").
Your parser should be capable to read from an InputStream instead of a Reader.
If not, utilize an InputStreamReader with the correct encoding (JSON files should be UTF-8, but that depends...)
Code:
try (InputStream is = this.getClass().getResourceAsStream("/testjson/jsonfile.json"); ) {
Object obj = parser.parse(is);
} catch (Exception ex) {
System.out.println("failed to read: "+ex.getMessage());
}
Code if parser does not support InputStream:
try (InputStream is = this.getClass().getResourceAsStream("/testjson/jsonfile.json");
Reader rd = new InputStreamReader(is, "UTF-8"); ) {
Object obj = parser.parse(rd);
} catch (Exception ex) {
System.out.println("failed to read: "+ex.getMessage());
}

As you're saying that you use Eclipse, i assume you also run your code via Eclipse.
As a default, the working directory when executing a Java program in Eclipse is the root folder of the project.
Therefore, I suggest to put your jsonfile.json in the root folder of your project instead of src/main/....
Furthermore, you should not catch Exception. Catch more specific like IOExceptionor JSONException and then print the exception message (e.getMessage()), then it is much easier to solve the problem.

The file "countries.geo.json" is missing or inaccessible, make sure the URL is valid or that the file has been added to your sketch and is readable.
You have to use the data folder as a source.
if Images not appear. Click on this links to see the images
Right-Click on the data folder
You then see Buid path option
Click on "Use as a source folder"

Give the full path to the JSON file instead of the file name.
If the file path is home/src/main/java/testjson/jsonfile.json
String path = "home/src/main/java/testjson/jsonfile.json";
Object obj = parser.parse(new FileReader(path));

How to open a file without extension by URL deployed at JBoss

Require to read a flat file that do not have any extension through URL, couldn't get this as browser(404) and java(FileNotFoundException) not taking the file as a resource to open.
import java.net.*;
import java.io.*;
public class URLTest {
public static void main(String[] args) {
/*Block 1: Reading a file "test.txt" (WITH extension) that works */
try{ System.out.println("Block 1");
URL url1 = new URL("http://localhost:8080/CtxPath/test.txt"); //file test.txt available at WebContent
BufferedReader in = new BufferedReader(new InputStreamReader(url1.openStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
System.out.println(inputLine);
in.close();
} catch(Exception e) { e.printStackTrace(); }
/* Block 2 : Reading the other one "test" (WITHOUT extension) that throws Exception */
try{ System.out.println("Block 2");
URL url2 = new URL("http://localhost:8080/CtxPath/test"); ////file test also available at WebContent
BufferedReader in = new BufferedReader(new InputStreamReader(url2.openStream()));
String inputLine;
while ((inputLine = in.readLine()) != null)
System.out.println(inputLine);
in.close();
} catch(Exception e) { e.printStackTrace(); }
}
}
Output
Block 1
File content from test.txt read successfully.
Block 2
java.io.FileNotFoundException: http://localhost:8080/CtxPath/test
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1624)
at java.net.URL.openStream(URL.java:1037)
at URLTest.main(URLTest.java:28)
Jboss 5.1 Java 1.7
Getting exception on open URL for the file(test) without extension. The other file with .txt extension(test.txt) that placed in the same location with similar content opens without any issues.
Is there any app specific or server specific config to do for allowing the file without extension to be served?

In the servlet container implementation, things without suffix "test" is not treat as file, but treat as servlet mapping.
You can download tomcat source code, the logic for this is similar in jboss, and take a look.
And I don't think you should make a file without suffix, if you really need to, then put it under WEB-INF/xxx/, then use a servlet to help read it.

Java is simply calling the URL. It cares not whether it has a . in the path. The server you are calling, on the other hand, /may/ have logic which determines how it attempts to fulfill the request. The FileNotFoundException indicates a 404 http status code was returned.

It may be possible that file does have extension that you may not be able to see because Operating system settings have hide the extensions or some rules on your server may causing it to return 404. Java does not give file not found exception because of extension.
Please let us know the issue, when you resolve it.

Java Applet Output File

I am trying to create a Java Applet that outputs information to a text file located in the same directory as the java applet. I understand Java Applets are not ideal, but I have spent a great deal of time on this and if possible want to solve this through applets. Here is some of my code on how I could read code from a file into a text box. I assume it would be something similar to this, but outputted.
public void readFile() {
String line;
URL url = null;
try {
url = new URL(getCodeBase(), fileToRead);
}
catch(MalformedURLException e) {
}
try {
InputStream in = url.openStream();
BufferedReader bf = new BufferedReader
(new InputStreamReader(in));
strBuff = new StringBuffer();
while((line = bf.readLine()) != null){
strBuff.append(line + "\n");
}
a1.append("File Name : " + fileToRead + "\n");
a1.append(strBuff.toString());
}
catch(IOException e) {
e.printStackTrace();
}
}

If you want an applet to store data on the local machine, from 6u10 the javax.jnlp.PersistenceService is available.
or on your local machine...
java.io.File file = new java.io.File(System.getProperty("user.home"), "yourfile.txt");
You must have it signed, otherwise...
Keep in mind that from an Applet, you cannot directly write to the server's file system. You can issue a request to the server that causes the server to write to its own file system, but an Applet does not have a way to write to a file system on a remote machine.
A signed Applet has every right to write to the local file system of the person running the Applet. If you are writing to the "current directory" (rather than an absolute full path), then make sure you know what directory the Applet is running in. Otherwise you may indeed create a file, but not be able to find it!
EDIT
Signed Applet Tutorial

Problem sending XML via HTTP

I want to have an application which parses various RSS feeds and send the information to a remote server. The information is sent in xml format via http. At first I tried to deploy this application on my own server, so I send the xml using the method shown in this tutorial by Java Tips. Here is my code which is replicated from the example:
First Method
String strURL = "http://localhost/readme/readme_xml";
String strXMLFilename = "output.xml";
File input = new File(strXMLFilename);
PostMethod post = new PostMethod(strURL);
post.setRequestEntity(new InputStreamRequestEntity(
new FileInputStream(input), input.length()));
post.setRequestHeader(
"Content-type", "text/xml; charset=ISO-8859-1");
HttpClient httpclient = new HttpClient();
try {
int result = httpclient.executeMethod(post);
System.out.println("Response status code: " + result);
System.out.println("Response body: ");
System.out.println(post.getResponseBodyAsString());
} finally {
post.releaseConnection();
}
This works perfectly (I even tested using a remote server outside the localhost). Then, somehow I cant use my own server to deploy this application, so I decided to migrate to Google Apps Engine. One thing about it, as we know it, is that not all libraries are allowed in the environment. So I try another method shown in ExampleDepot.com (I can't find where the exact url though) as below:
Second Method
try {
/* fill up this url with the remote server url */
URL url = new URL("http://localhost/readme/readme_xml");
FileReader fr = new FileReader("output.xml");
char[] buffer = new char[1024*10];
int len = 0;
if ((len = fr.read(buffer)) != -1){
/* send http request to remote server */
URLConnection conn = url.openConnection();
conn.setRequestProperty("Content-Type","text/xml;charset=ISO-8859-1"); /* need to specify the content type */
conn.setDoOutput(true);
conn.setDoOutput(true);
PrintWriter pw = new PrintWriter(conn.getOutputStream());
pw.write(buffer, 0, len);
pw.flush();
/* receive response from remote server*/
BufferedReader bf = new BufferedReader(new InputStreamReader(conn.getInputStream()));
String input = null;
while ((input = bf.readLine()) != null){
System.out.println(input);
}
}
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
The second method though, doesn't work and gives the following error (I use SimpleXMLElement (php) object to parse xml in the remote hosting):
Error message from remote server
Here's the php code from the remote server (In here, I just want the SimpleXMLElement to parse the xml without doing anything else fancy for now)
$xml = new SimpleXMLElement('php://input', NULL, TRUE);
foreach ($xml -> attributes() as $name => $val){
echo "[".$name."] = ".$val."\n";
}
I thought the cause of this problem is the malfunction xml file (because the eclipse IDE indicates there's error of "invalid byte 1 of 1-byte utf-8 sequence"). Then I use the same exact input xml file to the first method, but it still works perfectly.
So is there any adjustment that I need to make to the second method? Or is there any other method that I can use to send xml file to remote server? Let me know if I need to add some other details. Thanks for your help.
NOTE: I actually solved this problem by using the solution given in the comments. I didn't use approaches suggested in the answers, even though those answers are pretty useful. So, I didn't select the best answer out of those answers given. Nonetheless, I still appreciate all of your helps, thus deserve my upvote. Cheers!

I guess you need to change the content type to multipart/form-data. See an already answered question in detailed. The file upload is discussed at the bottom of this example

I would, as the first answer suggest, read the file with an InputStream. Converting from byte to char and back again is unnecessary and a source of error. Also, verify that the input file really is using the ISO-8859-1 encoding.
UPDATE:
When using a FileReader, you accept the default encoding (i.e. how to make chars from bytes). This encoding must match the encoding used for the input file, otherwise there's a great risk that the result is corrupted. The default Java encoding is different for different platforms, so it is generally not a good idea to rely on it.
In your second example, there's no reason to read the file as characters, since it will be sent on the wire as bytes anyway. Using byte streams all the way also avoids the encoding issue (apart from the information in the content-type header).

never read a file as chars unless you are reading a text file. xml is not text, it is a binary format. copy the file using normal InputStreams and byte[]s.
also, as #beny23 suggested in his comment, make sure you always copy streams using a loop, not a single read() (even if your buffer is big enough, it is not guaranteed that the InputStream will give you all the bytes in one call, even for a FileInputStream).