Develop Reference

Java: how to compress a byte[] using ZipOutputStream without intermediate file

Requirement: compress a byte[] to get another byte[] using java.util.zip.ZipOutputStream BUT without using any files on disk or in-memory(like here https://stackoverflow.com/a/18406927/9132186). Is this even possible?
All the examples I found online read from a file(.txt) and write to a file(.zip). ZipOutputStream needs a ZipEntry to work with and that ZipEntry needs a file.
However, my use case is as follows: I need to compress a chunk (say 10MB) of a file at a time using a zip format and append all these compressed chunks to make a .zip file. But, when I unzip the .zip file then it is corrupted.
I am using in-memory files as suggested in https://stackoverflow.com/a/18406927/9132186 to avoid files on disk but need a solution without these files also.
public void testZipBytes() {
String infile = "test.txt";
FileInputStream in = new FileInputStream(infile);
String outfile = "test.txt.zip";
FileOutputStream out = new FileOutputStream(outfile);
byte[] buf = new byte[10];
int len;
while ((len = in.read(buf)) > 0) {
out.write(zipBytes(buf));
}
in.close();
out.close();
}
// ACTUAL function that compresses byte[]
public static class MemoryFile {
public String fileName;
public byte[] contents;
}
public byte[] zipBytesMemoryFileWORKS(byte[] input) {
MemoryFile memoryFile = new MemoryFile();
memoryFile.fileName = "try.txt";
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
ZipEntry entry = new ZipEntry(memoryFile.fileName);
entry.setSize(input.length);
zos.putNextEntry(entry);
zos.write(input);
zos.finish();
zos.closeEntry();
zos.close();
return baos.toByteArray();
}
Scenario 1:
if test.txt has small amount of data (less than 10 bytes) like "this" then unzip test.txt.zip yeilds try.txt with "this" in it.
Scenario 2:
if test.txt has larger amount of data (more than 10 bytes) like "this is a test for zip output stream and it is not working" then unzip test.txt.zip yields try.txt with broken pieces of data and is incomplete.
this 10 bytes is the buffer size in testZipBytes and is the amount of data that is compressed at a time by zipBytes
Expected (or rather desired):
1. unzip test.txt.zip does not use the "try.txt" filename i gave in the MemoryFile but rather unzips to filename test.txt itself.
2. unzipped data is not broken and yields the input data as is.
3. I have done the same with GzipOutputStream and it works perfectly fine.

Requirement: compress a byte[] to get another byte[] using java.util.zip.ZipOutputStream BUT without using any files on disk or in-memory(like here https://stackoverflow.com/a/18406927/9132186). Is this even possible?
Yes, you've already done it. You don't actually need MemoryFile in your example; just delete it from your implementation and write ZipEntry entry = new ZipEntry("try.txt") instead.
But you can't concatenate the zips of 10MB chunks of file and get a valid zip file for the combined file. Zipping doesn't work like that. You could have a solution which minimizes how much is in memory at once, perhaps. But breaking the original file up into chunks seems unworkable.

Writing a File object to a file in java

For a class, I have to send a file of any type from my client to a server. I have to handle each packet individually and use UDP. I have managed to transfer the file from the client to the server, and I now have a file object which I cannot figure out how to save to a user specified directory.
f = new File(path + '\\' + filename);//path and filename are user specified.
FileOutputStream foutput = new FileOutputStream(f);
ObjectOutputStream output = new ObjectOutputStream(foutput);
output.writeObject(result);//result is a File
output.flush();
output.close();
Any time I run this code, it writes a new file with the appropriate name, but the text file I am testing ends up just containing gibberish. Is there any way to convert the File object to a file in the appropriate directory?
EDIT: As it turns out, I was misunderstanding what, exactly, a file is. I have not been transferring the data, but rather the path. How do I transfer an actual file?

ObjectOutputStream is a class that outputs a specific format of data to a text file. Only ObjectInputStream's readObject() can decoding that text file.
If you open the text file , it is just gibberish ,as you have seen.
you want this:
FileOutputStream fos = new FileOutputStream(path + '\\' + filename);
FileInputStream fis = new FileInputStream(result);
byte[] buf = new byte[1024];
int hasRead = 0;
while((hasRead = fis.read(buf)) > 0){
fos.write(buf, 0, hasRead);
}
fis.close();
fos.close();

If I understand your question, how about using a FileWriter?
File result = new File("result.txt");
result.createNewFile();
FileWriter writer = new FileWriter(result);
writer.write("Hello user3821496\n"); //just an example how you can write a String to it
writer.flush();
writer.close();

File exists but when I perform operation on that file, system shows FileNotFoundException. Works when file is in dir where source code is

public void decrypt() throws Exception
{
//opening streams
//Error is in the line below When i try to read file from directory
//other than the one which has .java and .class files.
FileInputStream fis1 =new FileInputStream(file);
File dir=new File("C:/Crypt-R/Decrypted");
dir.mkdirs();
file=new File(dir,file.getName() +".dec");
FileOutputStream fos1 =new FileOutputStream(file);
//generating same key
byte k[] = keyRecv.getBytes();
SecretKeySpec key = new SecretKeySpec(k,"AES");
//creating and initialising cipher and cipher streams
Cipher decrypt = Cipher.getInstance(algorithm);
decrypt.init(Cipher.DECRYPT_MODE, key);
CipherInputStream cin=new CipherInputStream(fis1, decrypt);
byte[] buf = new byte[1024];
int read=0;
while((read=cin.read(buf))!=-1) //reading encrypted data from file
{
fos1.write(buf,0,read); //writing decrypted data to file
}
//closing streams
cin.close();
fos1.flush();
fos1.close();
JOptionPane.showMessageDialog (null,
"File Decrypted",
"Success..!!",
JOptionPane.INFORMATION_MESSAGE);
}
There is a Text Editor attached to this program, file gets displayed in that Editor but when i am trying to decrypt it and it does not exist in the directory where my source code is kept then it shows file not found exception. Can you please help me out with this?

You need to check the following things
1) Make sure you are mentioning filename that really exists in the location
Check Reasons for getting FileNotFoundException
2) Then you have to mention path like "C:\\Users\...". Make sure you use right way to mention file names on windows system
3) You have to again check similarly for FileOutputStream.

You need to separate your directories with
File.separator
Which will hold the value of /, : or \\ depending on which platform you are on.

How to read content of the Zipped file without extracting in java

I have file with names like ex.zip. In this example, the Zip file contains only one file with the same name(ie. `ex.txt'), which is quite large. I don't want to extract the zip file every time.Hence I need to read the content of the file(ex.txt) without extracting the zip file. I tried some code like below But i can only read the name of the file in the variable.
How do I read the content of the file and stores it in the variable?
Thank you in Advance
fis=new FileInputStream("C:/Documents and Settings/satheesh/Desktop/ex.zip");
ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));
ZipEntry entry;
while((entry = zis.getNextEntry()) != null) {
i=i+1;
System.out.println(entry);
System.out.println(i);
//read from zis until available
}

Your idea is to read the zip file as it is into a byte array and store it in a variable.
Later when you need the zip you extract it on demand, saving memory:
First read the content of the Zip file in a byte array zipFileBytes
If you have Java 1.7:
Path path = Paths.get("path/to/file");
byte[] zipFileBytes= Files.readAllBytes(path);
otherwise use Appache.commons lib
byte[] zipFileBytes;
zipFileBytes = IOUtils.toByteArray(InputStream input);
Now your Zip file is stored in a variable zipFileBytes, still in compressed form.
Then when you need to extract something use
ByteArrayInputStream bis = new ByteArrayInputStream(zipFileBytes));
ZipInputStream zis = new ZipInputStream(bis);

Try this:
String zipFile = "ex.zip";
try (ZipFile zip = new ZipFile(zipFile)) {
int i = 0;
for (Enumeration<? extends ZipEntry> e = zip.entries(); e.hasMoreElements(); ) {
ZipEntry entry = (ZipEntry) e.nextElement();
System.out.println(entry);
System.out.println(i);
InputStream in = zip.getInputStream(entry);
}
}
For example, if the file contains text, and you want to print it as a String, you can read the InputStream like this: How do I read / convert an InputStream into a String in Java?

I think that in your case the fact that a zipfile is a container that can hold many files (and thus forces you to navigate to the right contained file each time you open it) seriously complicates things, as you state that each zipfile only contains one textfile. Maybe it's a lot easier to just gzip the text file (gzip is not a container, just a compressed version of your data). And it's very simple to use:
GZIPInputStream gis = new GZIPInputStream(new FileInputStream("file.txt.gz"));
// and a BufferedReader on top to comfortably read the file
BufferedReader in = new BufferedReader(new InputStreamReader(gis) );
Producing them is equally simple:
GZIPOutputStream gos = new GZIPOutputStream(new FileOutputStream("file.txt.gz"));

How to create ZIP file for a list of "virtual files" and output to httpservletresponse

My goal is to put multiple java.io.File objects into a zip file and print to HttpServletResponse for the user to download.
The files were created by the JAXB marshaller. It's a java.io.File object, but it's not actually on the file system (it's only in memory), so I can't create a FileInputStream.
All resources I've seen use the OutputStream to print zip file contents. But, all those resources use FileInputStream (which I can't use).
Anyone know how I can accomplish this?

Have a look at the Apache Commons Compress library, it provides the functionality you need.
Of course "erickson" is right with his comment to your question. You will need the file content and not the java.io.File object. In my example I assume that you have a method
byte[] getTheContentFormSomewhere(int fileNummer) which returns the file content (in memory) for the fileNummer-th file. -- Of course this function is poor design, but it is only for illustration.
It should work a bit like this:
void compress(final OutputStream out) {
ZipOutputStream zipOutputStream = new ZipOutputStream(out);
zipOutputStream.setLevel(ZipOutputStream.STORED);
for(int i = 0; i < 10; i++) {
//of course you need the file content of the i-th file
byte[] oneFileContent = getTheContentFormSomewhere(i);
addOneFileToZipArchive(zipOutputStream, "file"+i+"."txt", oneFileContent);
}
zipOutputStream.close();
}
void addOneFileToZipArchive(final ZipOutputStream zipStream,
String fileName,
byte[] content) {
ZipArchiveEntry zipEntry = new ZipArchiveEntry(fileName);
zipStream.putNextEntry(zipEntry);
zipStream.write(pdfBytes);
zipStream.closeEntry();
}
Snipets of your http controller:
HttpServletResponse response
...
response.setContentType("application/zip");
response.addHeader("Content-Disposition", "attachment; filename=\"compress.zip\"");
response.addHeader("Content-Transfer-Encoding", "binary");
ByteArrayOutputStream outputBuffer = new ByteArrayOutputStream();
compress(outputBuffer);
response.getOutputStream().write(outputBuffer.toByteArray());
response.getOutputStream().flush();
outputBuffer.close();

Turns out I'm an idiot :) The file that was being "created" was saving to invalid path and swallowing the exception, so I thought it was being "created" ok. When I tried to to instantiate a new FileInputStream, however, it complained that file didn't exist (rightly so). I had a brainfart and assumed that the java.io.File object actually contained file information in it somewhere. But as erickson pointed out, that was false.
Thanks Ralph for the code, I used it after I solved the invalid pathing issue.
My code:
ZipOutputStream out = new ZipOutputStream(response.getOutputStream());
byte[] buf = new byte[1024];
File file;
InputStream in;
// Loop through entities
for (TitleProductAccountApproval tpAccountApproval : tpAccountApprovals) {
// Generate the file
file = xmlManager.getXML(
tpAccountApproval.getTitleProduct().getTitleProductId(),
tpAccountApproval.getAccount().getAccountId(),
username);
// Write to zip file
in = new FileInputStream(file);
out.putNextEntry(new ZipEntry(file.getName()));
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
out.closeEntry();
in.close();
}
out.close();