I'm creating a xlsx using poi and saving it on fileSystem. I need to download the file on a servlet call and due to memory constraints I did not create a xssf workbook object and used the following code instead :
byte[] buf = new byte[1024];
ServletOutputStream sOut = response.getOutputStream();
FileInputStream input = null;
try {
long length = fileToRead.length();
input = new FileInputStream(fileToRead);
while ((input != null) && ((length = input.read(buf)) != -1)) {
sOut.write(buf, 0, (int) length);
}
Where fileToRead is the file present at the file system.
How can I integrate this with How to create a zip file in Java
You could use
ZipOutputStream zos = new ZipOutputStream(response.getOutputStream());
....
ze = new ZipEntry("xlsData");
zos.putEntry (ze);
// loop
zos.write(buf, 0, (int) length);
// finally
zos.close();
Related
FileOutputStream fos = new FileOutputStream(FILE_NAME);
ZipOutputStream zipOut = new ZipOutputStream(fos);
for (File file : files) {
FileInputStream fis = new FileInputStream(file);
ZipEntry zipEntry = new ZipEntry(file.getName());
zipOut.putNextEntry(zipEntry);
byte[] bytes = new byte[1024];
int length;
while((length = fis.read(bytes)) >= 0) {
zipOut.write(bytes, 0, length);
}
fis.close();
}
zipOut.close();
fos.close();
I have a zip file, I want to download it when I call the api.With the zip file, I don't know how to download. Hope to get help from you.
If you are using Spring/Spring Boot. You can follow this stackoverflow thread Download File in Spring
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();
}
I´m trying to implement some "over the air" update mechanism for OSGi bundles. For that, I need to be able to create a jar file from a String (basically the content of the jar file read by JarInputStream). The following example code should illustrate my needs:
//read bundle to be copied!
File originalFile = new File(
"/Users/stefan/Documents/Projects/OSGi/SimpleBundle_1.0.0.201404.jar");
JarInputStream fis = new JarInputStream(new FileInputStream(originalFile));
StringBuilder stringBuilder = new StringBuilder();
int ch;
while ((ch = fis.read()) != -1) {
stringBuilder.append((char) ch);
}
fis.close();
//Create content string
String content = stringBuilder.toString();
if (logger.isInfoEnabled()) {
logger.info(content);
}
//Init new jar input stream
JarInputStream jarInputStream = new JarInputStream(
new ByteArrayInputStream(content.getBytes()));
if (logger.isInfoEnabled()) {
logger.info("Save content to disc!");
}
File newFile = new File(
"/Users/stefan/Documents/Projects/OSGi/equinox/SimpleBundle_1.0.0.201404.jar");
//Init new jar output stream
JarOutputStream fos = new JarOutputStream(
new FileOutputStream(newFile));
if (!newFile.exists()) {
newFile.createNewFile();
}
int BUFFER_SIZE = 10240;
byte buffer[] = new byte[BUFFER_SIZE];
while (true) {
int nRead = jarInputStream.read(buffer, 0,
buffer.length);
if (nRead <= 0)
break;
fos.write(buffer, 0, nRead);
}
//Write content to new jar file.
fos.flush();
fos.close();
jarInputStream.close();
Unfortunately, the created jar file is empty and throws an "Invalid input file" error if I try to open it with JD-GUI. Is it possible to create a jar file from the String "content"?
Best regards and thank you very much
Stefan
Your jar is empty because you do not read anything from the JarInputStream. If you want to read JarInputStream, you should iterate its entries. If you want to change the Manifest, the first entry should be skipped, use the getManifest() of the jarInputStream and the constructor of the JarOutputStream, where Manifest can be specified. Based on your code (no manifest change but plain jar copy):
ZipEntry zipEntry = jarInputStream.getNextEntry();
while (zipEntry != null) {
fos.putNextEntry(zipEntry);
// Simple stream copy comes here
int BUFFER_SIZE = 10240;
byte buffer[] = new byte[BUFFER_SIZE];
int l = jarInputStream.read(buffer);
while(l >= 0) {
fos.write(buffer, 0, l);
l = jarInputStream.read(buffer);
}
zipEntry = jarInputStream.getNextEntry();
}
You only need this if you want to change the content (Manifest or entries) of the JAR file during the copy. Otherwise, simple InputStream and FileOutputStream will do the work (as Tim said).
I have a list of files from different locations. I create a zip file using the following the code which works without error. But when I try to unzip the file in Windows using Extract All it fails seeing unable to find any bytes, yet if I double click into the zip file itself with Windows Explorer I can see the files and individual ones can be opened and contains the correct data
ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(zipFile));
for (File next : files)
{
ZipEntry zipEntry = new ZipEntry(next.getName());
zos.putNextEntry(zipEntry);
FileInputStream in = new FileInputStream(next);
byte[] buf = new byte[1024];
int len;
while ((len = in.read(buf)) > 0)
{
zos.write(buf, 0, len);
}
zos.closeEntry();
in.close();
}
zos.close();
This may or may not be related but I've found using fixed byte length can lead to a loss of new line characters.
This may help:
final byte[] newLine = System.getProperty(
"line.separator").getBytes("UTF-8");
while ((line = in.readLine()) != null)
final byte[] buffer = line.getBytes("UTF-8");
out.write(buffer, 0, buffer.length);
out.write(newLine, 0, newLine.length);
}
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.