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FileInputStream FileNotFound Exception in AIX with filename with special characters

I have a small Java Application running inside IBM Integration Bus, which is installed in an AIX Server with the character encoding set to ISO-8959-1.
My application is creating a ZIP File with the filenames received as a parameter. I have a file called "Websërvícès Guide.pdf" in the filesystem which I wanted to zip but I'm unable.
This is my code:
String zipFilePath = "/tmp/EventAttachments_2018.01.25.11.39.34.zip";
// Streams buffer
int BUFFER = 2048;
// Open I/O Buffered Streams
BufferedInputStream origin = null;
FileOutputStream dest = new FileOutputStream(zipFilePath);
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(dest));
byte data[] = new byte[BUFFER];
// Oprn File Stream to my file
Path currentFilePath = Paths.get("/tmp/Websërvícès Guide.pdf");
InputStream fi = Files.newInputStream(currentFilePath, StandardOpenOption.READ);
origin = new BufferedInputStream(fi, BUFFER);
ZipEntry entry = new ZipEntry("Websërvícès Guide.pdf");
out.putNextEntry(entry);
int count;
while ((count = origin.read(data, 0, BUFFER)) != -1) {
out.write(data, 0, count);
}
origin.close();
out.close();
Which is throwing a "File Not Found" exception in the Files.newInputStream line.
I have read that Java is not working properly when checking it files with special characters exists and so on. I'm not able to perform changes in the JVM Parameters as code is executed inside a IBM JVM.
Any idea on how to solve this issue and pack the file properly in the ZIP?
Thank you

Can you try to pass following flag to while running your java program
-Dsun.jnu.encoding=UTF-8

First: In your code, you are not taking care of any Exceptions that could be thrown. I would suggest to handle the exceptions of the method or make the method throw the exception and handle it on a higher level. But somewhere you need to handle the exception.
Maybe that's already the problem. (see https://stackoverflow.com/a/155655/8896833)
Second: According to ISO-8959-1 all the characters used in your filename should be covered. Are you really sure about the path your program is working in at the moment you are trying to access the file?

Try to use URLDecoder class method decode(String string, String encoding);.
For example:
String path = URLDecoder.decode("Websërvícès Guide.pdf", "UTF-8"));

Which one is better approach so as to able to use wrapper class read() / write() method with android.content.res.Resources.openRawResource() method?

//Reading a image file from #drawable res folder and writing to a file on external sd card
//below one works no doubt but I want to imrpove it:
OutputStream os = new FileOutputStream(file); //File file.........
InputStream is =getResources().openRawResource(R.drawable.an_image);
byte[] b = new byte[is.available()];
is.read(b);
os.write(b);
is.close();
os.close();
In above code I am using basic io classes to read and write. My question is what can I do in order to able to use wrapper classes like say DataInputStream/ BufferedReaderd or PrintStream / BufferedWriter /PrintWriter.
As openRawResources(int id ) returns InputStream ;
to read a file from res I either need to typecast like this:
DataInputStream is = (DataInputStream) getResources().openRawResource(R.drawble.an_image));
or I can link the stream directly like this:
DataInputStream is = new DataInputStream(getResources().openRawResource(R.drawable.greenball));
and then I may do this to write it to a file on sd card:
PrintStream ps =new PrintStream (new FileOutputStream(file));
while(s=is.readLine()!=null){
ps.print(s);
}
So is that correct approach ? which one is better? Is there a better way?better practice..convention?
Thanks!!!

If openRawResource() is documented to return an InputStream then you cannot rely on that result to be any more specific kind of InputStream, and in particular, you cannot rely on it to be a DataInputStream. Casting does not change that; it just gives you the chance to experience interesting and exciting exceptions. If you want a DataInputStream wrapping the the result of openRawResource() then you must obtain it via the DataInputStream constructor. Similarly for any other wrapper stream.
HOWEVER, do note that DataInputStream likely is not the class you want. It is appropriate for reading back data that were originally written via a DataOutputStream, but it is inappropriate (or at least offers no advantages over any other InputStream) for reading general data.
Furthermore, your use of InputStream.available() is incorrect. That method returns the number of bytes that can currently be read from the stream without blocking, which has only a weak relationship with the total number of bytes that could be read from the stream before it is exhausted (if indeed it ever is).
Moreover, your code is also on shaky ground where it assumes that InputStream.read(byte[]) will read enough bytes to fill the array. It probably will, since that many bytes were reported available, but that's not guaranteed. To copy from one stream to another, you should instead use code along these lines:
private final static int BUFFER_SIZE = 2048;
void copyStream(InputStream in, OutputStream out) throws IOException {
byte[] buffer = new byte[BUFFER_SIZE];
int nread;
while ( (nread = in.read(buffer) != 0 ) do {
out.write(buffer, 0, nread);
}
}

Java SequenceInputStream

I try to send multiple Files from my Server (NanoHttpd) to my Client (Apache DefaultHttpClient).
My approach is to send multiple files via one Response of NanoHttpd.
For this purpose i wanted to use SequenceInputStream.
I am trying to concatenate multiple Files, send them via the Response (InputStream) and write every File again in a seperate File with my Client.
On the Serverside i call this:
List<InputStream> data = new ArrayList<InputStream>(o_file_path.size());
for (String file_name : files)
{
File file = new File(file_name);
data.add(new FileInputStream(file));
}
InputStream is = new SequenceInputStream(Collections.enumeration(data));
return new NanoHTTPD.Response(HTTP_OK, "application/octet-stream", is);
Now my Question is how to receive and split the Files correctly.
I have tried it this way on my client, but it does not work:
int read = 0;
int remaining = 0;
byte[] bytes = new byte[buffer];
// Read till the end of the Stream
while ( (read != -1) && (counter < files.size()))
{
// Create a .o file for the current file
read = 0;
remaining = is.available();
// Should open each Stream
while (remaining > 0)
{
read = is.read(bytes);
remaining = remaining - read;
os.write(bytes, 0, read);
}
os.flush();
os.close();
}
This way I want to go over all Stream (untill read == 1, or i know there is no file anymore), and read any stream into a file.
I clearly seem to understand something groundbreaking wrong, since is.available() always is 0.
Could anyone please tell me how to read properly from this SequencedInputStream, or how to solve my Problem.
Thanks in advance.

It won't work this way. SequenceInputStream will merge all input streams in one solid byte stream. There will be no separators or EOFs. I suggest to abandon the idea and look for a different approach.

Getting XML data from the byteArray of a zipFile

I'm writing a simple program that retrieves XML data from an object, and parses it dynamically, based on user criteria. I am having trouble getting the XML data from the object, due to the format it is available in.
The object containing the XML returns the data as a byteArray of a zipFile, like so.
MyObject data = getData();
byte[] byteArray = data.getPayload();
//The above returns the byteArray of a zipFile
The way I checked this, is by writing the byteArray to a String
String str = new String(byteArray);
//The above returns a string with strange characters in it.
Then I wrote the data to a file.
FileOutputStream fos = new FileOutputStream("new.txt");
fos.write(byteArray);
I renamed new.txt as new.zip. When I opened it using WinRAR, out popped the XML.
My problem is that, I don't know how to do this conversion in Java using streams, without writing the data to a zip file first, and then reading it. Writing data to disk will make the software way too slow.
Any ideas/code snippets/info you could give me would be really appreciated!! Thanks
Also, if you need a better explanation from me, I'd be happy to elaborate.
As another option, I am wondering whether an XMLReader would work with a ZipInputStream as InputSource.
ByteArrayInputStream bis = new ByteArrayInputStream(byteArray);
ZipInputStream zis = new ZipInputStream(bis);
InputSource inputSource = new InputSource(zis);

A zip archive can contain several files. You have to position the zip stream on the first entry before parsing the content:
ByteArrayInputStream bis = new ByteArrayInputStream(byteArray);
ZipInputStream zis = new ZipInputStream(bis);
ZipEntry entry = zis.getNextEntry();
InputSource inputSource = new InputSource(new BoundedInputStream(zis, entry.getCompressedSize()));
The BoundedInputStream class is taken from Apache Commons IO (http://commons.apache.org/io)

Get filename from an inputstream (Java)

if I have this code, how could I keep the filename of the original file or reassign it to the new one?:
InputStream input= assetInfo.openStream();
File t = new File("");
OutputStream out = new FileOutputStream(t);
int read=0;
byte[] bytes = new byte[1024];
while((read = input.read(bytes))!= -1){
out.write(bytes, 0, read);
}

An input stream can be created to read from a file or from any other source of data. Therefore it makes no sense to have a filename attached to an input stream. Look in assetInfo to see if that class exposes that data (you can even look inside the class using reflection). Note that the creator or assetInfo made a design mistake not exposing this information, OR you are trying to make one now.