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Not able to read ZipInputStream returned by ZipFile.getInputStream(ZipEntry) method

I am trying to read extract a given file from zip file. Zip file contains directories & sub-directories as well. I tried Java7 nio file apis but since my zip has subdirectories as well, I need to provide complete path to extract the file, which is not suitable in my scenario. As I have to take filetobeextracted input from user. I have been trying below code for it but somehow read method of ZipInputStream not reading any contents to buffer. On debugging I found out that ZipEntry object value is null inside ZipInputStream due to its read method simply returns -1.But now I am stuck as I am not able to figure out how that value is being set for it.
try(OutputStream out=new FileOutputStream("filetoExtract");) {
zipFile = new ZipFile("zipFile");
Enumeration<? extends ZipEntry> e = zipFile.entries();
while (e.hasMoreElements()) {
ZipEntry entry = e.nextElement();
if (!entry.isDirectory()) {
String entryName = entry.getName();
String fileName = entryName.substring(entryName.lastIndexOf("/") + 1);
System.out.println(i++ + "." + entryName);
if (searchFile.equalsIgnoreCase(fileName)) {
System.out.println("File Found");
BufferedInputStream bufferedInputStream = new BufferedInputStream(zipFile.getInputStream(entry));
ZipInputStream zin = new ZipInputStream(bufferedInputStream);
byte[] buffer = new byte[9000];
int len;
while ((len = zin.read(buffer)) != -1) {
out.write(buffer, 0, len);
}
out.close();
break;
}
}
}
} catch (IOException ioe) {
System.out.println("Error opening zip file" + ioe);
}
Please advice what I am doing wrong here. Thanks
EDIT:
After debugging little more I found out that ZipFile class has inner class of similar name(ZipFileInputStream). So it was creating object of it rather than the outside ZipFileInputStream class. So I tried out below code and it worked out well. But I don't quite understand things here, what has happened. If someone could help me logic behind the scenes would be really great.
// BufferedInputStream bufferedInputStream = new
//BufferedInputStream(zipFile.getInputStream(entry));
//ZipInputStream zin = new ZipInputStream(bufferedInputStream);
InputStream zin= zipFile.getInputStream(entry);

The second line is unnecessary, as zipFile.getInputStream(entry) already returns an InputStream that represents the decompressed data. Therefore there's no need (or in fact it's wrong) to wrap that InputStream in yet another ZipInputStream:
BufferedInputStream bufferedInputStream = new BufferedInputStream(zipFile.getInputStream(entry));
ZipInputStream zin = new ZipInputStream(bufferedInputStream);

unzip a specific file in the zip file uploaded by sp using java and google app engine

I have trying to find out the solution for to unzip a particular file using java and Google app engine. I have tried using ZipInputStream but cant able to access the zip file that is uploaded in jsp. can any one help me to come out of this?
ServletFileUpload upload = new ServletFileUpload();
resp.setContentType("text/plain");
FileItemIterator iterator = upload.getItemIterator(req);
while (iterator.hasNext()) {
FileItemStream fileItemStream = iterator.next();
InputStream InputStream = fileItemStream.openStream();
if (!fileItemStream.isFormField()) {
ZipInputStream zis = new ZipInputStream(new BufferedInputStream(InputStream));
ZipEntry entry;
while ((entry = zis.getNextEntry()) != null) {
//code to access required file in the zip file
}
}
}

My guess would be that ZipInputStream requires stream that is seekable. The streams returned by servlets (and networking in general) aren't seekable.
Try reading the whole stream, then wrapping it in ByteArrayInputStream:
byte[] bytes = readBytes(fileItemStream.openStream());
InputStream bufferedStream = new ByteArrayInputStream(bytes);
public static byte[] readBytes(InputStream is) throws IOException {
ByteArrayOutputStream buffer = new ByteArrayOutputStream();
int len;
byte[] data = new byte[100000];
while ((len = is.read(data, 0, data.length)) != -1) {
buffer.write(data, 0, len);
}
buffer.flush();
return buffer.toByteArray();
}

Unable to zip multiple files using Java

Getting data onto inputStream object from web url
inputStream = AWSFileUtil.getInputStream(
AWSConnectionUtil.getS3Object(null),
"cdn.generalsentiment.com", filePath);
If they are mutliple files then i want to zip them and sent the filetype as "zip" to struts.xml which does the download.
actually am converting the inputstream into byteArrayInputStream
ByteArrayInputStream byteArrayInputStream = new
ByteArrayInputStream(inputStream.toString().getBytes());
while (byteArrayInputStream.read(inputStream.toString().getBytes()) > 0) {
zipOutputStream.write(inputStream.toString().getBytes());
}
and then
zipOutputStream.close();
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
fileInputStream = new FileInputStream(file);
while (fileInputStream.read(buffer) > 0) {
byteArrayOutputStream.write(buffer);
}
byteArrayOutputStream.close();
inputStream = new ByteArrayInputStream(byteArrayOutputStream.toByteArray());
reportName = "GS_MediaValue_Reports.zip";
fileType = "zip";
}
return fileType;
But the downloaded zip when extracted gives corrupt files.
Please suggest me a solution for this issue.

The short answer is that it's not how ZipOutputStream works. Since it was designed to store multiple files, along with their file names, directory structures and so on, you need to tell the stream about that explicitly.
Furthermore, converting a stream to a string is a bad idea in general, plus it's slow, especially when you're doing it in a loop.
So your solution will be something like:
ZipEntry entry = new ZipEntry( fileName ); // You have to give each entry a different filename
zipOutputStream.putNextEntry( entry );
byte buffer[] = new byte[ 1024 ]; // 1024 is the buffer size here, but it could be anything really
int count;
while( (count = inputStream.read( buffer, 0, 1024 ) ) != -1 ) {
zipOutputStream.write( buffer, 0, count );
}

Java servlet and IO: Create a file without saving to disk and sending it to the user

I`m hoping can help me out with a file creation/response question.
I know how to create and save a file. I know how to send that file back to the user via a ServletOutputStream.
But what I need is to create a file, without saving it on the disk, and then send that file via the ServletOutputStream.
The code above explains the parts that I have. Any help appreciated. Thanks in Advance.
// This Creates a file
//
String text = "These days run away like horses over the hill";
File file = new File("MyFile.txt");
Writer writer = new BufferedWriter(new FileWriter(file));
writer.write(text);
writer.close();
// Missing link goes here
//
// This sends file to browser
//
InputStream inputStream = null;
inputStream = new FileInputStream("C:\\MyFile.txt");
byte[] buffer = new byte[8192];
ByteArrayOutputStream baos = new ByteArrayOutputStream();
int bytesRead;
while ( (bytesRead = inputStream.read(buffer)) != -1)
baos.write(buffer, 0, bytesRead);
response.setContentType("text/html");
response.addHeader("Content-Disposition", "attachment; filename=Invoice.txt");
byte[] outBuf = baos.toByteArray();
stream = response.getOutputStream();
stream.write(outBuf);

You don't need to save off a file, just use a ByteArray stream, try something like this:
inputStream = new ByteArrayInputStream(text.getBytes());
Or, even simpler, just do:
stream.write(text.getBytes());
As cHao suggests, use text.getBytes("UTF-8") or something similar to specify a charset other than the system default. The list of available charsets is available in the API docs for Charset.

How do I get the InputStream of decompressed data from an InputStream of GZIPed data?

I call a service which returns a gzipped file. I have the data as an InputStream (courtesy of javax.activation.DataHandler.getInputStream();) from the response.
What I would like to do is, without writing anything to disk, get an InputStream of the decompressed data in the file that is in the archive. The compressed file in this case is an xml document that I am trying to unmarshal using javax.xml.bind.Unmarshaller which takes an InputStream.
I'm currently trying to write the InputStream to an OutputStream (decompressing the data) and then I'll need to write it back to an InputStream. It's not working yet so I thought I would see if there was a better (I would hope so) approach.
I can write the initial InputStream to disk and get a gz file, and then read that file, get the compressed file out of it and go from there but I'd rather keep it all in memory is possible.
Update 1: Here is my current (not working - get a "Not in GZIP format" exception):
ByteArrayInputStream xmlInput = null;
try {
InputStream in = dh.getInputStream(); //dh is a javax.activation.DataHandler
BufferedInputStream bis = new BufferedInputStream(in);
ByteArrayOutputStream bo = new ByteArrayOutputStream();
int bytes_read = 0;
byte[] dataBuf = new byte[4096];
while ((bytes_read = bis.read(dataBuf)) != -1) {
bo.write(dataBuf, 0, bytes_read);
}
ByteArrayInputStream bin = new ByteArrayInputStream(bo.toByteArray());
GZIPInputStream gzipInput = new GZIPInputStream(bin);
ByteArrayOutputStream out = new ByteArrayOutputStream();
dataBuf = new byte[4096];;
bytes_read = 0;
while ((bytes_read = gzipInput.read(dataBuf)) > 0) {
out.write(dataBuf, 0, bytes_read);
}
xmlInput = new ByteArrayInputStream(out.toByteArray());
If instead of writing to a ByteArrayOutputStream I write to a FileOutputStream the first time around I get a compressed file (which I can manually open to get the xml file within) and the service (eBay) says it should be a gzip file so I'm not sure why I get a "Not in GZIP format" error.
Update 2: I tried something a little different - same error ("Not in GZIP format"). Wow, I just tried to end that parenthesis with a semi-colon. Anyways, here is my second attempt, which still does not work:
ByteArrayInputStream xmlInput = null;
try {
GZIPInputStream gzipInput = new GZIPInputStream(dh.getInputStream());
ByteArrayOutputStream bo = new ByteArrayOutputStream();
int bytes_read = 0;
byte[] dataBuf = new byte[4096];
while ((bytes_read = gzipInput.read(dataBuf)) != -1) {
bo.write(dataBuf, 0, bytes_read);
}
xmlInput = new ByteArrayInputStream(bo.toByteArray());

Decorate the input stream with a GZIPInputStream.
InputStream decompressed = new GZIPInputStream(compressed);

The following code should work. Keep in mind you'll have to handle exceptions properly.
OutputStream out = null;
InputStream in = null;
try {
out = /* some output stream */;
in = new java.util.GZIPInputStream(/*some stream*/);
byte[] buffer = new byte[4096];
int c = 0;
while (( c = in.read(buffer, 0, 4096)) > 0) {
out.write(buffer, 0, c);
}
} finally {
if (in != null) {
in.close();
}
if (out != null) {
out.close();
}
}

Take a look at GZIPInputStream. Here's an example; the class handles this very transparently, it's almost no work to use.