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InputStream to GZIPInputStream, and I need to know the size of the GZIPInputStream

I have a MultipartFile and I need to compress inputStream as gzip and sent it, but I need to find a way to compress it and know the compressed size of it
param: MultipartFile file
try(var inputStream = file.getInputStream()) {
var outputStream = new GZIPOutputStream(OutputStream.nullOutputStream());
IOUtils.copyLarge(inputStream, outputStream);
var compressedInputStream = someConvertMerthod(outputStream);
sendCompressed(compressedInputStream, compressedSize)
}
Maybe I can do something like this Java: How do I convert InputStream to GZIPInputStream? but I am not gonna be a able to get the compressedSize
I am not finding an easy way to do it :(

CopyLarge() returns the number of bytes copied. I would assume this is true even if the output is discarded, so all you need is to capture the return value of IOUtils.copyLarge(in,out) and you should be good to go, but this does assume the return value is bytes WRITTEN and not bytes READ, which is not really documented. So it might work!
In general though, you are assuming you can turn the output stream back into an input stream, so nullOutputStream() is not going to be an option. Instead you will be creating a temp file, writing your compressed data to it, and then closing it. At that point you can simply ask the file system API how big it is, that should be iron clad.

hey I think I found the solution :)
param: MultipartFile file
try (InputStream inputStream = file.getInputStream()) {
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
GZIPOutputStream gzipOutputStream = new GZIPOutputStream(byteArrayOutputStream);
inputStream.transferTo(gzipOutputStream);
InputStream compressedInputStream = new ByteArrayInputStream(byteArrayOutputStream.toByteArray());
byteArrayOutputStream.size() // is the compressed size
}
Thanks guys!

Write byte array to ZipArchiveOutputStream

I need to create a zip file and am limited by the following conditions:
The entries come as byte[] (or ByteArrayOutputStream) and not as File.
The filenames for the entry can be non-ascii/UTF-8.
JDK 1.6 or earlier
Since java.util.zip only supports UTF-8 filenames from JDK 1.7 and onward, it seems better to use commons-compress ZipArchiveOutputStream. But how to create a ZipEntryArchive based on a byte array or ByteArrayOutputStream rather than a File?
Thank you!

The following method takes a byte[] as input, produces a zip and returns its content as another byte[]. All is done in memory. No IO operations on the disk. I stripped exception handling for a better overview.
byte[] zip(byte[] data, String filename) {
ByteArrayOutputStream bos = new ByteArrayOutputStream();
ZipArchiveOutputStream zos = new ZipArchiveOutputStream(bos);
ZipArchiveEntry entry = new ZipArchiveEntry(filename);
entry.setSize(data.length);
zos.putArchiveEntry(entry);
zos.write(data);
zos.closeArchiveEntry();
zos.close();
bos.close();
return bos.toByteArray();
}
Does this solve your problem?

Java write byte in file

I am doing in Java a File Comprimir to a Binary File. The problem is the next, how can I write a Byte in a new File that the total size only occups 1 byte? I am doing the next:
FileOutputStream saveFile=new FileOutputStream("SaveObj3.sav");
// Create an ObjectOutputStream to put objects into save file.
ObjectOutputStream save = new ObjectOutputStream(saveFile);
save.writeByte(0);
save.close();
saveFile.close();
That, only must to write a only byte in the file, but when I look the size,it occups 7 bytes. Anyone knows how can I write a only byte? Is there another way better?

Don't use ObjectOutputStream. Use the FileOutputStream directly:
FileOutputStream out=new FileOutputStream("SaveObj3.sav");
out.write(0);
out.close();

As JB Nizet noticed documentation of ObjectOutputStream constructor states that this object also
writes the serialization stream header to the underlying stream
which explains additional bytes.
To prevent this behaviour you can just use other streams like FileOutputStream or maybe DataOutputStream
FileOutputStream saveFile = new FileOutputStream("c:/SaveObj3.sav");
DataOutputStream save = new DataOutputStream(saveFile);
save.writeByte(0);
save.close();

You can use Files class provided by Java 7. It's more easy than expected.
It can be performed in one line:
byte[] bytes = new String("message output to be written in file").getBytes();
Files.write(Paths.get("outputpath.txt"), bytes);
If you have a File class, you can just replace:
Paths.get("outputpath.txt")
to:
yourOutputFile.toPath()
To write only one byte, as you want, you can do the following:
Files.write(Paths.get("outputpath.txt"), new byte[1]);
in file properties:
size: 1 bytes

Reading the contents of a zip entry from a zip that is in the form of bytes

I'm calling onto some code that returns me an HTTP response. I can get the contents of the response which returns me a byte array. The bytes represent a zip file that I would like to extract and get the contents of a single file (the zip only contains one file).
Currently I have some messy code (I'll need to clean it up if I keep it) that seems to work:
byte[] bytes = response.out.toByteArray();
ByteArrayInputStream input = new ByteArrayInputStream(bytes);
ZipInputStream zip = new ZipInputStream(input);
ByteArrayOutputStream output = new ByteArrayOutputStream();
zip.getNextEntry();
int data;
while ((data = zip.read()) != -1) output.write(data);
output.close();
zip.close();
input.close();
byte[] kmlBytes = output.toByteArray();
String contents = new String(kmlBytes, "UTF-8");
but was wondering whether there was a cleaner way to do the same thing because the above looks incredibly ugly.

In order to avoid "reinventing the wheel" and reduce clutter and to focus on your app's business logic rather than low-level code like this, you can use Apache Commons Compress.

Byte[] to InputStream or OutputStream

I have a blob column in my database table, for which I have to use byte[] in my Java program as a mapping and to use this data I have to convert it to InputStream or OutputStream. But I don't know what happens internally when I do so. Can anyone briefly explain me what's happening when I do this conversion?

You create and use byte array I/O streams as follows:
byte[] source = ...;
ByteArrayInputStream bis = new ByteArrayInputStream(source);
// read bytes from bis ...
ByteArrayOutputStream bos = new ByteArrayOutputStream();
// write bytes to bos ...
byte[] sink = bos.toByteArray();
Assuming that you are using a JDBC driver that implements the standard JDBC Blob interface (not all do), you can also connect a InputStream or OutputStream to a blob using the getBinaryStream and setBinaryStream methods1, and you can also get and set the bytes directly.
(In general, you should take appropriate steps to handle any exceptions, and close streams. However, closing bis and bos in the example above is unnecessary, since they aren't associated with any external resources; e.g. file descriptors, sockets, database connections.)
1 - The setBinaryStream method is really a getter. Go figure.

I'm assuming you mean that 'use' means read, but what i'll explain for the read case can be basically reversed for the write case.
so you end up with a byte[]. this could represent any kind of data which may need special types of conversions (character, encrypted, etc). let's pretend you want to write this data as is to a file.
firstly you could create a ByteArrayInputStream which is basically a mechanism to supply the bytes to something in sequence.
then you could create a FileOutputStream for the file you want to create. there are many types of InputStreams and OutputStreams for different data sources and destinations.
lastly you would write the InputStream to the OutputStream. in this case, the array of bytes would be sent in sequence to the FileOutputStream for writing. For this i recommend using IOUtils
byte[] bytes = ...;//
ByteArrayInputStream in = new ByteArrayInputStream(bytes);
FileOutputStream out = new FileOutputStream(new File(...));
IOUtils.copy(in, out);
IOUtils.closeQuietly(in);
IOUtils.closeQuietly(out);
and in reverse
FileInputStream in = new FileInputStream(new File(...));
ByteArrayOutputStream out = new ByteArrayOutputStream();
IOUtils.copy(in, out);
IOUtils.closeQuietly(in);
IOUtils.closeQuietly(out);
byte[] bytes = out.toByteArray();
if you use the above code snippets you'll need to handle exceptions and i recommend you do the 'closes' in a finally block.

we can convert byte[] array into input stream by using ByteArrayInputStream
String str = "Welcome to awesome Java World";
byte[] content = str.getBytes();
int size = content.length;
InputStream is = null;
byte[] b = new byte[size];
is = new ByteArrayInputStream(content);
For full example please check here http://www.onlinecodegeek.com/2015/09/how-to-convert-byte-into-inputstream.html

There is no conversion between InputStream/OutputStream and the bytes they are working with. They are made for binary data, and just read (or write) the bytes one by one as is.
A conversion needs to happen when you want to go from byte to char. Then you need to convert using a character set. This happens when you make String or Reader from bytes, which are made for character data.

output = new ByteArrayOutputStream();
...
input = new ByteArrayInputStream( output.toByteArray() )

I do realize that my answer is way late for this question but I think the community would like a newer approach to this issue.

byte[] data = dbEntity.getBlobData();
response.getOutputStream().write();
I think this is better since you already have an existing OutputStream in the response object.
no need to create a new OutputStream.