I have a function which creates a Zip file from a list of files. Is it possible to return the Zip file without it being saved on the disk? I need the file as I have to use the zip file as a parameter for another function. I am not sure of the ByteStream would be an option for me.
public File compressFileList(List<File> fileList,String fileName) {
FileOutputStream fileOutputStream=null;
ZipOutputStream zipOutputStream=null;
FileInputStream fileInputStream=null;
String compressedFileName=fileName +".zip";
if(fileList.isEmpty())
return null;
try
{
fileOutputStream = new FileOutputStream(compressedFileName);
zipOutputStream = new ZipOutputStream(new BufferedOutputStream(fileOutputStream));
for (File file: fileList) {
fileInputStream = new FileInputStream(file);
ZipEntry zipEntry = new ZipEntry(file.getName());
zipOutputStream.putNextEntry(zipEntry);
byte[] tmp = new byte[4*1024];
int size = 0;
while((size = fileInputStream.read(tmp)) != -1){
zipOutputStream.write(tmp, 0, size);
}
zipOutputStream.flush();
fileInputStream.close();
}
zipOutputStream.close();
return compressedFile; //This is what I am missing
}
catch (FileNotFoundException e)
{
} catch (IOException e) {
e.printStackTrace();
}
return null;
}
EDIT : adding the use case
The idea is to create a zip file and use the CreateClassifierOptions method of VisualRecognition Service of Watson.
classifierOptions = new CreateClassifierOptions.Builder()
.classifierName("Santa")
.addClass("Santa", new File("C:\\app\\GitRepo\\images\\beagle.zip"))
.negativeExamples(new File("C:\\app\\GitRepo\\images\\nosport.zip"))
.build();
The builder accepts the zip file as the parameter.
Understanding
Based on the explanation from Alexandre Dupriez, I think it is better to store the file at some place on the hard disk.
You should be able to use a ByteArrayOutputStream instead of a FileOutputStream:
zipOutputStream = new ZipOutputStream(new ByteArrayOutputStream());
The difficulty here is to provide a File to the method consuming the zip file. The java.io.File does not provide an abstraction which allows you to manipulate in-memory files.
The java.io.File abstraction and java.io.FileInputStream implementation
To simplify, if we had to boil down what the File abstraction is, we would see it as a URI. And therefore, to be able to build an in-memory File, or at least mimic it, we would need to provide an URI which would then be used by the consumer of the File to read its content.
If we look at the FileInputStream which the consumer is likely to use, we can see that it always ends up with a native call which gives us to possibility whatsoever to abstract a FileSystem for in-memory files:
// class java.io.FileInputStream
/**
* Opens the specified file for reading.
* #param name the name of the file
*/
private native void open0(String name) throws FileNotFoundException;
It would be easier if there was a possibility to adapt the consumer to accept an InputStream, but from your problem statement I guess this is not possible.
API call
Your requirement is to provide a File to the Watson Visual API.
Could you please provide the API method you need to call?
public void compressFileList(List<File> fileList, OutputStream outputStream)
throws IOException {
try (ZipOutputStream zipOutputStream =
new ZipOutputStream(new BufferedOutputStream(outputStream));
for (File file: fileList) {
try (FileInputStream fileInputStream = new FileInputStream(file)) {
ZipEntry zipEntry = new ZipEntry(file.getName());
zipOutputStream.putNextEntry(zipEntry);
byte[] tmp = new byte[4*1024];
int size = 0;
while((size = fileInputStream.read(tmp)) != -1){
zipOutputStream.write(tmp, 0, size);
}
zipOutputStream.flush();
} catch (FileNotFoundException e) { // Maybe skip not found files.
Logger.log(Level.INFO, "File not found {}", file.getPath());
}
}
}
}
Usage:
if (fileList.isEmpty()) {
...
return;
}
try {
compressFileList(fileList, servletRequest.getOutputStream())) {
} catch (FileNotFoundException e) {
...
} catch (IOException e) {
...
}
Related
So iam working on a project where i get to a point when i need to zip multiple folders (4folders to be specific) and one file in one output.zip file using java
So is there anyway for me to do it and by the way putting all the folders and file in one directory and then zipping it doesn't give the same result in other words the folders have to be in root level of the zip file
There are several solutions.
This one for example:
public static void zipDirectory(ZipOutputStream zos, File fileToZip, String parentDirectoryName) throws Exception
{
if (fileToZip == null || !fileToZip.exists())
{
return;
}
String zipEntryName = fileToZip.getName();
if (parentDirectoryName!=null && !parentDirectoryName.isEmpty())
{
zipEntryName = parentDirectoryName + "/" + fileToZip.getName();
}
// If we are dealing with a directory:
if (fileToZip.isDirectory())
{
System.out.println("+" + zipEntryName);
if(parentDirectoryName == null) // if parentDirectory is null, that means it's the first iteration of the recursion, so we do not include the first container folder
{
zipEntryName = "";
}
for (File file : fileToZip.listFiles()) // we iterate over all the folders/files and archive them by keeping the structure too.
{
zipDirectory(zos, file, zipEntryName);
}
} else // If we are dealing with a file, then we zip it directly
{
System.out.println(" " + zipEntryName);
byte[] buffer = new byte[1024];
FileInputStream fis = new FileInputStream(fileToZip);
zos.putNextEntry(new ZipEntry(zipEntryName));
int length;
while ((length = fis.read(buffer)) > 0)
{
zos.write(buffer, 0, length);
}
zos.closeEntry();
fis.close();
}
}
then you could use this function like this:
try
{
File directoryToBeZipped = new File("C:\\New\\test");
FileOutputStream fos = new FileOutputStream("C:\\New\\test\\archive.zip");
ZipOutputStream zos = new ZipOutputStream(fos);
zipDirectory(zos, directoryToBeZipped, null);
zos.flush();
fos.flush();
zos.close();
fos.close();
} catch (Exception e)
{
e.printStackTrace();
}
Or you could use the ZeroTurnAround ZIP Library. And do this in one line:
ZipUtil.pack(new File("D:\\sourceFolder\\"), new File("D:\\generatedZipFile.zip"));
Dead easy way (though you'll get warnings about proprietary classes)
final String[] ARGS = { "-cfM", "x.zip", "folder1", "folder2", "folder3", "folder4", "file.txt" };
sun.tools.jar.Main.main(ARGS);
It might be worth getting a similar thing that won't give you warnings
I take a multipartfile (i.e. SAMPLE.csv) in input.
I should zip it (i.e. SAMPLE.zip) and store it via FTP.
public void zipAndStore(MultipartFile file) {
try (ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
InputStream is = file.getInputStream()) {
ZipEntry zipEntry = new ZipEntry("SAMPLE.zip");
zos.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = is.read(bytes)) >= 0) {
zos.write(bytes, 0, length);
}
zos.closeEntry();
storeFtp("SAMPLE.zip", new ByteArrayInputStream(baos.toByteArray()));
} catch (Exception e) {
}
}
The storeFtp use the org.apache.commons.net.ftp.FTPClient.storeFile(String remote, InputStream local) method.
The problem is that the uploaded file is corrupted and i'm unable to manually decompress.
What's wrong?
A zipfile has a list of DirEntries and a endLocator at the end of the file (after all the ZipFileRecords, i.e. the ZipEntries in the code).
So you probably have to close the zipfile before calling storeFtp() to make sure the DirEntries and the endLocator are written to the file:
zos.closeEntry();
zos.close();
storeFtp("SAMPLE.zip", new ByteArrayInputStream(baos.toByteArray()));
(I don't know Java that well, so I can't check or test the full code)
Also check out this answer.
Is there a direct way to unpack a java.util.zip.ZipEntry to a File?
I want to specify a location (like "C:\temp\myfile.java") and unpack the Entry to that location.
There is some code with streams on the net, but I would prefer a tested library function.
Use ZipFile class
ZipFile zf = new ZipFile("zipfile");
Get entry
ZipEntry e = zf.getEntry("name");
Get inpustream
InputStream is = zf.getInputStream(e);
Save bytes
Files.copy(is, Paths.get("C:\\temp\\myfile.java"));
Use the below code to extract the "zip file" into File's then added in the list using ZipEntry. Hopefully, this will help you.
private List<File> unzip(Resource resource) {
List<File> files = new ArrayList<>();
try {
ZipInputStream zin = new ZipInputStream(resource.getInputStream());
ZipEntry entry = null;
while((entry = zin.getNextEntry()) != null) {
File file = new File(entry.getName());
FileOutputStream os = new FileOutputStream(file);
for (int c = zin.read(); c != -1; c = zin.read()) {
os.write(c);
}
os.close();
files.add(file);
}
} catch (IOException e) {
log.error("Error while extract the zip: "+e);
}
return files;
}
Use ZipInputStream to move to the desired ZipEntry by iterating using the getNextEntry() method. Then use the ZipInputStream.read(...) method to read the bytes for the current ZipEntry. Output those bytes to a FileOutputStream pointing to a file of your choice.
I have this little piece of code
public void doBuild() throws IOException {
ZipEntry sourceEntry=new ZipEntry(sourcePath);
ZipEntry assetEntry=new ZipEntry(assetPath);
ZipOutputStream out = new ZipOutputStream(new FileOutputStream("output/"+workOn.getName().replaceAll(".bld"," ")+buildNR+".zip"));
out.putNextEntry(sourceEntry);
out.putNextEntry(assetEntry);
out.close();
System.err.println("Build success!");
increaseBuild();
}
So, if I run it it runs trough it fine, creates the .zip and all, but the zip file is empty. sourceEntry and assetEntry are both directories. How could I get those directories to my .zip easily?
For those interested this is a MC mod build system and can be found at https://bitbucket.org/makerimages/makerbuild-system NOTE: the code above is not commited or pushed to there yet!!!!!!!!
Try something like this. The parameter useFullFileNames specifies
whether you want to preserve the full names of the paths to the
files which you're about to zip.
So if you have two files
/dir1/dir2/a.txt
and
/dir1/b.txt
the useFullFileNames specifies if you want to finally see in
the zip file those original paths to the two files or just
the two files with no paths like this
a.txt
and
b.txt
in the root of the zip file which you create.
Note that in my example, the files which are zipped
are actually read and then written to out.
I think you're missing that part.
public static boolean createZip(String fNameZip, boolean useFullFileNames, String... fNames) throws Exception {
try {
int cntBufferSize = 256 * 1024;
BufferedInputStream origin = null;
FileOutputStream dest = new FileOutputStream(fNameZip);
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(dest));
byte bBuffer[] = new byte[cntBufferSize];
File ftmp = null;
for (int i = 0; i < fNames.length; i++) {
if (fNames[i] != null) {
FileInputStream fi = new FileInputStream(fNames[i]);
origin = new BufferedInputStream(fi, cntBufferSize);
ftmp = new File(fNames[i]);
ZipEntry entry = new ZipEntry(useFullFileNames ? fNames[i] : ftmp.getName());
out.putNextEntry(entry);
int count;
while ((count = origin.read(bBuffer, 0, cntBufferSize)) != -1) {
out.write(bBuffer, 0, count);
}
origin.close();
}
}
out.close();
return true;
} catch (Exception e) {
return false;
}
}
I'm zipping up the contents of a directory, but running into an error when trying to open the zipped up files.
Can anyone tell what's going on with my code? Perhaps I'm not allocating enough bytes?
Look inside zipDirectory() and you will see that I'm zipping up folders which contain special extension files.
Not sure where the error's occurring, so maybe someone can help me out there!
Much appreciated
private void zipDirectory() {
File lazyDirectory = new File(defaultSaveLocation);
File[] files = lazyDirectory.listFiles();
for (File file : files) {
if (file.isDirectory()) {
System.out.println("Zipping up " + file);
zipContents(file);
}
}
}
public static void addToZip(String fileName, ZipOutputStream zos) throws FileNotFoundException, IOException {
System.out.println("Writing '" + fileName + "' to zip file");
File file = new File(fileName);
FileInputStream fis = new FileInputStream(file);
ZipEntry zipEntry = new ZipEntry(fileName);
zos.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while ((length = fis.read(bytes)) >= 0) {
zos.write(bytes, 0, length);
}
zos.closeEntry();
fis.close();
}
public static void zipContents(File dirToZip) {
List<File> fileList = new ArrayList<File>();
File[] filesToZip = dirToZip.listFiles();
for (File zipThis : filesToZip) {
String ext = "";
int i = zipThis.toString().lastIndexOf('.');
if (i > 0) {
ext = zipThis.toString().substring(i+1);
}
if(ext.matches("cpp|bem|gz|h|hpp|pl|pln|ppcout|vec|xml|csv")){
fileList.add(zipThis);
}
}
try {
FileOutputStream fos = new FileOutputStream(dirToZip.getName() + ".zip");
ZipOutputStream zos = new ZipOutputStream(fos);
for (File file : fileList) {
addToZip(file.toString(), zos);
}
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
Like most issues with IO streams in Java, your fault is almost certainly that you are not closing the streams properly. You need to add:
zos.finish(); // good practice
zos.close();
after the for loop.
For me the fix is that you need to do this for EVERY file entry
zos.finish()
zos.flush()
zos.closeEntry()
Then do the above things again to close the zos. Otherwise, the default windows can't open the zipped folder properly, but third-party application works.
This may also happen if you use a character-oriented writer (e.g. FileWriter) to create the unzipped files (that actually contain binary data)
I was unable to read files that I was extracting and switching to a binary output stream (FileOutputStream) fixed the issue