Making use of newOutputStream method to define file location

Question

I've the following code working fine in terms of CSV file creation (only showing relevant code below):

rsCompany = pstmtCompany.executeQuery();
Path dir = Paths.get("/srv/custom_users", userName);
Files.createDirectories(dir);

Path filecompany = dir.resolve("company_custom_file_" + unixTimestamp + ".csv");
    try (CSVWriter writer = new CSVWriter(Files.newBufferedWriter(filecompany))) {
            writer.writeAll(rsCompany, true);
  }

Now, let's say I want to create a zip file for the same, how should I make use of the dir variable (that I used in the above scenario) in this line FileOutputStream fos = new FileOutputStream("your_files.zip"); of code below?

I mean, I have to define the following things:

Path dir = Paths.get("/srv/custom_users", userName);
Files.createDirectories(dir);

FileOutputStream fos = new FileOutputStream("your_files.zip");
BufferedOutputStream bos = new BufferedOutputStream(fos);
ZipOutputStream zos = new ZipOutputStream(bos); 


ZipEntry entry = new ZipEntry(file.getFileName().toString());
zos.putNextEntry(entry);
try (CSVWriter writer = new CSVWriter(new OutputStreamWriter(zos,StandardCharsets.UTF_8)))) {
       writer.writeAll(rsDemo, true);
       writer.flush();
       zos.closeEntry();
}
zos.close();

If I go ahead and use newOutputstream method of Files class, it may looks like the following :

FileOutputStream fos = new FileOutputStream(Files.newOutputStream(dir)); 

Is it a correct way to go about it?

I'm wondering, where should I place the name of the zip file , I mean, your_files.zip that I defined in my previous line of code FileOutputStream("your_files.zip");?

Answer 1

You don’t need a FileOutputStream. At all.

You only need an OutputStream, and that is what Files.newOutputStream returns:

OutputStream fos = Files.newOutputStream(dir.resolve("your_files.zip"));
BufferedOutputStream bos = new BufferedOutputStream(fos);
ZipOutputStream zos = new ZipOutputStream(bos);