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Copy a file into a double archive in Java (a zip file which is in another zip file)

I have an ear archive, "archive.ear". This archive contains an war file, "archive.war". In this file, I want to replace a file, "/myFile.properties" with a new one, which exists on disk.
The content of the new file is saved in a java.io.File object, named "file". I saved the output stream from "/myFile.properties" from the archive in a java.io.OutputStream object. And after that, I tried to use
org.apache.commons.io.FileUtils.copy(File input, OutputStream output)
The current code is:
// Java method from extracting the output stream
public OutputStream getOutputStream(OutputStream out, String entry) throws IOException {
ZipOutputStream zos = new ZipOutputStream(out);
ZipEntry zipEntry = new ZipEntry(entry);
while (zipEntry != null) {
if (zipEntry.toString().equals(entry)) {
return zos;
}
}
throw new IllegalStateException("No entry '" + entry + "' found");
}
// copy the file content to output stream
// extract output stream "archive.war" from "archive.ear"
OutputStream warOs = zu.getOutputStream(new FileOutputStream("archive.ear"), "archive.war");
// extract output stream "<path>/myFile.properties" from "archive.war"
OutputStream myFileOutput = zu.getOutputStream(warOs, "<path>/myFile.properties" );
FileUtils.copyFile(file, myFileOutput);
I also tried to use, insted of copyFile():
myFileOutput.write(getBytesFromFile(file));
The method "getBytesFromFile()" returns an array of byte from a file object.
When I open the war archive, I expect "myFile.properties" to have the new content, which is in the java object "file". This file has the right content.
The result is a ZipException:
Exception in thread "main" java.util.zip.ZipException: no current ZIP entry
at java.util.zip.ZipOutputStream.write(Unknown Source)
at org.apache.commons.io.IOUtils.copyLarge(IOUtils.java:2315)
at org.apache.commons.io.IOUtils.copy(IOUtils.java:2270)
at org.apache.commons.io.IOUtils.copyLarge(IOUtils.java:2291)
at org.apache.commons.io.FileUtils.copyFile(FileUtils.java:1094)
at main.Main.main(Main.java:69)

You can't (easily) manipulate the existing zip file with Java.
You will have to do this in a roundabout way.
Open the current "archive.ear" as ZipInputStream.
Open a new "archive.ear.new" as ZipOutputStream.
Transfer all ZipEntries from 1. to 2.
When you come to your entry "archive.war"
Open a new ZipInputStream for it
Open a new ZipOutputStream for the entry for 2.
Transfer all of the ZipEntries except your "myFile.properties"
Transfer the content of "myFile.properties" for the entry
Flush your ZipOutputStream, close the entry
Rename the new file to the old file

Why is my file doesn't read to a zip archive

I have a code snippet, that in theory should read a path to an archive first, when write a file to the archive. (But that thing takes zip and for instance some txt file, and really move it to zip, but the file is empty) First, i thought this thing doesn't work since i didn't close streams, but now i use try-with, so the problem should be gone, but it is not.
public void createZip(Path source) throws Exception
{
try(ZipInputStream zipIn = new ZipInputStream(Files.newInputStream(source));
ZipOutputStream zipOut = new ZipOutputStream(Files.newOutputStream(zipFile)))
{
ZipEntry zipEntry = new ZipEntry(source.getFileName().toString());
zipOut.putNextEntry(zipEntry);
int data;
while((data = zipIn.read()) > 0)
{
zipOut.write(data);
}
}
}

To use correctly the ZipOutputStream after adding the entry you need to flush the zip.
Try adding the following at the endo of your method:
zipOut.flush();

Getting specific file from ZipInputStream

I can go through ZipInputStream, but before starting the iteration I want to get a specific file that I need during the iteration. How can I do that?
ZipInputStream zin = new ZipInputStream(myInputStream)
while ((entry = zin.getNextEntry()) != null)
{
println entry.getName()
}

If the myInputStream you're working with comes from a real file on disk then you can simply use java.util.zip.ZipFile instead, which is backed by a RandomAccessFile and provides direct access to the zip entries by name. But if all you have is an InputStream (e.g. if you're processing the stream directly on receipt from a network socket or similar) then you'll have to do your own buffering.
You could copy the stream to a temporary file, then open that file using ZipFile, or if you know the maximum size of the data in advance (e.g. for an HTTP request that declares its Content-Length up front) you could use a BufferedInputStream to buffer it in memory until you've found the required entry.
BufferedInputStream bufIn = new BufferedInputStream(myInputStream);
bufIn.mark(contentLength);
ZipInputStream zipIn = new ZipInputStream(bufIn);
boolean foundSpecial = false;
while ((entry = zin.getNextEntry()) != null) {
if("special.txt".equals(entry.getName())) {
// do whatever you need with the special entry
foundSpecial = true;
break;
}
}
if(foundSpecial) {
// rewind
bufIn.reset();
zipIn = new ZipInputStream(bufIn);
// ....
}
(I haven't tested this code myself, you may find it's necessary to use something like the commons-io CloseShieldInputStream in between the bufIn and the first zipIn, to allow the first zip stream to close without closing the underlying bufIn before you've rewound it).

use the getName() method on ZipEntry to get the file you want.
ZipInputStream zin = new ZipInputStream(myInputStream)
String myFile = "foo.txt";
while ((entry = zin.getNextEntry()) != null)
{
if (entry.getName().equals(myFileName)) {
// process your file
// stop looking for your file - you've already found it
break;
}
}
From Java 7 onwards, you are better off using ZipFile instead of ZipStream if you only want one file and you have a file to read from:
ZipFile zfile = new ZipFile(aFile);
String myFile = "foo.txt";
ZipEntry entry = zfile.getEntry(myFile);
if (entry) {
// process your file
}

Look at Finding a file in zip entry
ZipFile file = new ZipFile("file.zip");
ZipInputStream zis = searchImage("foo.png", file);
public searchImage(String name, ZipFile file)
{
for (ZipEntry e : file.entries){
if (e.getName().endsWith(name)){
return file.getInputStream(e);
}
}
return null;
}

I'm late to the party, but all above "answers" does not answer the question and accepted "answer" suggest create temp file which is inefficient.
Lets create sample zip file:
seq 10000 | sed "s/^.*$/a/"> /tmp/a
seq 10000 20000 | sed "s/^.*$/b/"> /tmp/b
seq 20000 30000 | sed "s/^.*$/c/"> /tmp/c
zip /tmp/out.zip /tmp/a /tmp/b /tmp/c
so now we have /tmp/out.zip file, which contains 3 files, each of them full of chars a, b or c.
Now lets read it:
public static void main(String[] args) throws IOException {
ZipInputStream zipStream = new ZipInputStream(new FileInputStream("/tmp/out.zip"));
ZipEntry zipEntry;
while ((zipEntry = zipStream.getNextEntry()) != null) {
String name = zipEntry.getName();
System.out.println("Entry: "+name);
if (name.equals("tmp/c")) {
byte[] bytes = zipStream.readAllBytes();
String s = new String(bytes);
System.out.println(s);
}
}
}
method readAllBytes seems weird, while we're in processing of stream, but it seems to work, I tested it also on some images, where there is higher chance of failure. So it's probably just unintuitive api, but it seems to work.

How does ZipInputStream.getNextEntry() work?

Say we have code like:
File file = new File("zip1.zip");
ZipInputStream zis = new ZipInputStream(new FileInputStream(file));
Let's assume you have a .zip file that contains the following:
zip1.zip
hello.c
world.java
folder1
foo.c
bar.java
foobar.c
How would zis.getNextEntry() iterate through that?
Would it return hello.c, world.java, folder1, foobar.c and completely ignore the files in folder1?
Or would it return hello.c, world.java, folder1, foo.c, bar.java, and then foobar.c?
Would it even return folder1 since it's technically a folder and not a file?
Thanks!

Well... Lets see:
ZipInputStream zis = new ZipInputStream(new FileInputStream("C:\\New Folder.zip"));
try
{
ZipEntry temp = null;
while ( (temp = zis.getNextEntry()) != null )
{
System.out.println( temp.getName());
}
}
Output:
New Folder/
New Folder/folder1/
New Folder/folder1/bar.java
New Folder/folder1/foo.c
New Folder/foobar.c
New Folder/hello.c
New Folder/world.java

Yes. It will print the folder name too, since it's also an entry within the zip. It will also print in the same order as it is displayed inside the zip. You can use below test to verify your output.
public class TestZipOrder {
#Test
public void testZipOrder() throws Exception {
File file = new File("/Project/test.zip");
ZipInputStream zis = new ZipInputStream(new FileInputStream(file));
ZipEntry entry = null;
while ( (entry = zis.getNextEntry()) != null ) {
System.out.println( entry.getName());
}
}
}

Excerpt from: https://blogs.oracle.com/CoreJavaTechTips/entry/creating_zip_and_jar_files
java.util.zip libraries offer some level of control for the added entries of the ZipOutputStream.
First, the order you add entries to the ZipOutputStream is the order they are physically located in the .zip file.
You can manipulate the enumeration of entries returned back by the entries() method of ZipFile to produce a list in alphabetical or size order, but the entries are still stored in the order they were written to the output stream.
So I would believe that you have to use the entries() method to see the order in which it will be iterated through.
ZipFile zf = new ZipFile("your file path with file name");
for (Enumeration<? extends ZipEntry> e = zf.entries();
e.hasMoreElements();) {
System.out.println(e.nextElement().getName());
}

The zip file internal directory is a "flat" list of all the files and directories in the zip. getNextEntry will iterate through the list and sequentially identify every file and directory in the zip file.
There is a variant of the zip file format that has no central directory, in which case (if it's handled at all) I suspect you'd iterate through all actual files in the zip, skipping directories (but not skipping files in directories).

java.util.zip.ZipException: too many entries in ZIP file

I am trying to write a Java class to extract a large zip file containing ~74000 XML files. I get the following exception when attempting to unzip it utilizing the java zip library:
java.util.zip.ZipException: too many entries in ZIP file
Unfortunately due to requirements of the project I can not get the zip broken down before it gets to me, and the unzipping process has to be automated (no manual steps). Is there any way to get around this limitation utilizing java.util.zip or with some 3rd party Java zip library?
Thanks.

Using ZipInputStream instead of ZipFile should probably do it.

Using apache IOUtils:
FileInputStream fin = new FileInputStream(zip);
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
FileOutputStream fout = new FileOutputStream(new File(
outputDirectory, ze.getName()));
IOUtils.copy(zin, fout);
IOUtils.closeQuietly(fout);
zin.closeEntry();
}
IOUtils.closeQuietly(zin);

The Zip standard supports a max of 65536 entries in a file.
Unless the Java library supports ZIP64 extensions, it won't work properly if you are trying to read or write an archive with 74,000 entries.

I reworked the method to deal with directory structures more convenient and to zip a whole bunch of targets at once.
Plain files will be added to the root of the zip file, if you pass a directory, the underlying structure will be preserved.
def zip (String zipFile, String [] filesToZip){
def result = new ZipOutputStream(new FileOutputStream(zipFile))
result.withStream { zipOutStream ->
filesToZip.each {fileToZip ->
ftz = new File(fileToZip)
if(ftz.isDirectory()){
pathlength = new File(ftz.absolutePath).parentFile.absolutePath.size()
ftz.eachFileRecurse {f ->
if(!f.isDirectory()) writeZipEntry(f, zipOutStream, f.absolutePath[pathlength..-1])
}
}
else writeZipEntry(ftz, zipOutStream, '')
}
}
}
def writeZipEntry(File plainFile, ZipOutputStream zipOutStream, String path) {
zipOutStream.putNextEntry(new ZipEntry(path+plainFile.name))
new FileInputStream(plainFile).withStream { inStream ->
def buffer = new byte[1024]
def count
while((count = inStream.read(buffer, 0, 1024)) != -1)
zipOutStream.write(buffer)
}
zipOutStream.closeEntry()
}