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Creating a download for zip file containing text/csv from string variables

I currently need to create a zip file for downloading. This should contain two (2) csv files that are to be created from string variables. I'm at a loss on how I should do this. My draft is below.
public #ResponseBody Object getFileV1(HttpServletRequest request, HttpServletResponse response) {
try {
response.setContentType("application/zip");
response.setHeader("Content-Disposition", "attachment; filename=Reassigned Tickets Report " + new Date().toString() + ".zip");
String stringValue1 = "This is a test value for csv1";
String stringValue2 = "This is a test value for csv2";
InputStream is1 = new ByteArrayInputStream(stringValue1.getBytes("UTF-8"));
InputStream is2 = new ByteArrayInputStream(stringValue2.getBytes("UTF-8"));
ZipInputStream zin;
ZipEntry entry;
ZipOutputStream zout= new ZipOutputStream(response.getOutputStream());
zin = new ZipInputStream(is1);
entry = zin.getNextEntry();
zout.putNextEntry(entry);
zin = new ZipInputStream(is2);
entry = zin.getNextEntry();
zout.putNextEntry(entry);
zout.closeEntry();
zin.close();
zout.close();
response.flushBuffer();
return null;
} catch (Exception e) {
e.printStackTrace();
return e;
}
}
Obviously this is not working. Probably because I'm still a novice at this. Please bear with me.
I get a "java.lang.NullPointerException" at the line where "zout.putNextEntry" is called. Would appreciate your advice. Thank you in advance.

I solved my problem after a day of looking around. This works for me. But I'm not sure if this is the most efficient way.
public #ResponseBody Object getFileV1(HttpServletRequest request, HttpServletResponse response) {
try {
response.setContentType("application/zip");
response.setHeader("Content-Disposition", "attachment; filename=Test Report " + new Date().toString() + ".zip");
String stringValue1 = "This is a test value for csv1";
String stringValue2 = "This is a test value for csv2";
PrintWriter writer1 = new PrintWriter(new OutputStreamWriter(new FileOutputStream("stringValue1.csv"), "UTF-8"));
writer1.print(stringValue1);
writer1.close();
PrintWriter writer2 = new PrintWriter(new OutputStreamWriter(new FileOutputStream("stringValue2.csv"), "UTF-8"));
writer2.print(stringValue2);
writer2.close();
File file1 = new File("stringValue1.csv");
File file2 = new File("stringValue2.csv");
filesToZip(response, file1, file2);
file1.delete();
file2.delete();
response.flushBuffer();
return null;
} catch (Exception e) {
e.printStackTrace();
return e;
}
}
This is the method I got from another thread with a few edits.
public static void filesToZip(HttpServletResponse response, File... files) throws IOException {
// Create a buffer for reading the files
byte[] buf = new byte[1024];
// create the ZIP file
ZipOutputStream out = new ZipOutputStream(response.getOutputStream());
// compress the files
for(int i=0; i<files.length; i++) {
FileInputStream in = new FileInputStream(files[i].getName());
// add ZIP entry to output stream
out.putNextEntry(new ZipEntry(files[i].getName()));
// transfer bytes from the file to the ZIP file
int len;
while((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// complete the entry
out.closeEntry();
in.close();
}
// complete the ZIP file
out.close();
}
The only thing I don't love is that I had to create temporary files and delete them after processing.

Download dynamically created zip files in play framework

Hi i am trying to write a play framework service where i can download multiple files. I create zip of multiple file on the fly but i am not sure how to send it as a response in Play Framework i will show what i have done so far.
public Result download() {
String[] items = request().queryString().get("items[]");
String toFilename = request().getQueryString("toFilename");
ByteArrayOutputStream baos = new ByteArrayOutputStream();
try (ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(baos))) {
for (String item : items) {
Path path = Paths.get(REPOSITORY_BASE_PATH, item);
if (Files.exists(path)) {
ZipEntry zipEntry = new ZipEntry(path.getFileName().toString());
zos.putNextEntry(zipEntry);
byte buffer[] = new byte[2048];
try (BufferedInputStream bis = new BufferedInputStream(Files.newInputStream(path))) {
int bytesRead = 0;
while ((bytesRead = bis.read(buffer)) != -1) {
zos.write(buffer, 0, bytesRead);
}
} finally {
zos.closeEntry();
}
}
}
response().setHeader("Content-Type", "application/zip");
response().setHeader("Content-Disposition", "inline; filename=\"" + MimeUtility.encodeWord(toFilename) + "\"");
//I am confused here how to output the response of zip file i have created
//I tried with the `baos` and with `zos` streams but not working
return ok(baos.toByteArray());
} catch (IOException e) {
LOG.error("copy:" + e.getMessage(), e);
return ok(error(e.getMessage()).toJSONString());
}
return null;
}
i tried sending response with return ok(baos.toByteArray()); i was able to download file but when i open the downloaded file it give me error An error occurred while loading the archive.

You need to close the zip file. After adding all entries, do: zos.close()
On a side note, I would recommend writing the zip file to disk rather than keeping it in a memory buffer. You could then use return ok(File content, String filename) to send its content to the client.

I am adding this answer if someone wants to know what was the final code:
String[] items = request().queryString().get("items[]");
String toFilename = request().getQueryString("toFilename");
ByteArrayOutputStream baos = new ByteArrayOutputStream();
try (ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(baos))) {
for (String item : items) {
Path path = Paths.get(REPOSITORY_BASE_PATH, item);
if (Files.exists(path)) {
ZipEntry zipEntry = new ZipEntry(path.getFileName().toString());
zos.putNextEntry(zipEntry);
byte buffer[] = new byte[2048];
try (BufferedInputStream bis = new BufferedInputStream(Files.newInputStream(path))) {
int bytesRead = 0;
while ((bytesRead = bis.read(buffer)) != -1) {
zos.write(buffer, 0, bytesRead);
}
} finally {
zos.closeEntry();
}
}
}
zos.close(); //closing the Zip
response().setHeader("Content-Type", "application/zip");
response().setHeader("Content-Disposition", "attachment; filename=\"" + MimeUtility.encodeWord(toFilename) + "\"");
return ok(baos.toByteArray());
} catch (IOException e) {
LOG.error("copy:" + e.getMessage(), e);
return ok(error(e.getMessage()).toJSONString());
}

Thank you for the help! There are two extra changes I made so it works for me in scala playframework 2.5.x
instead of return ok(baos.toByteArray()) ,
use Ok.chunked(StreamConverters.fromInputStream(fileByteData))
Instead of reading byte to byte the file,FileUtils.readFileToByteArray(file) can be very helpful here.
Attached is the complete version of my code.
import java.io.{BufferedOutputStream, ByteArrayInputStream, ByteArrayOutputStream}
import java.util.zip.{ZipEntry, ZipOutputStream}
import akka.stream.scaladsl.{StreamConverters}
import org.apache.commons.io.FileUtils
import play.api.mvc.{Action, Controller}
class HomeController extends Controller {
def single() = Action {
Ok.sendFile(
content = new java.io.File("C:\\Users\\a.csv"),
fileName = _ => "a.csv"
)
}
def zip() = Action {
Ok.chunked(StreamConverters.fromInputStream(fileByteData)).withHeaders(
CONTENT_TYPE -> "application/zip",
CONTENT_DISPOSITION -> s"attachment; filename = test.zip"
)
}
def fileByteData(): ByteArrayInputStream = {
val fileList = List(
new java.io.File("C:\\Users\\a.csv"),
new java.io.File("C:\\Users\\b.csv")
)
val baos = new ByteArrayOutputStream()
val zos = new ZipOutputStream(new BufferedOutputStream(baos))
try {
fileList.map(file => {
zos.putNextEntry(new ZipEntry(file.toPath.getFileName.toString))
zos.write(FileUtils.readFileToByteArray(file))
zos.closeEntry()
})
} finally {
zos.close()
}
new ByteArrayInputStream(baos.toByteArray)
}
}

Android unzip function not working

The following unzip finction doesn't work for all zip files.
My zip file pattern is as follows-
The Zip file contains one xml file and one folder(name- "images").
The name of the xml file is same as the zip file name.
The folder("images") may or may not contain any files.
I have validated the xml file before putting it into the zip file.
It throws exception at this line for some zip files-
FileOutputStream fout = new ileOutputStream(path.substring(0,path.length()-4)+"/"+filename);
The function is:
public boolean unZip(String path)
{
InputStream is;
ZipInputStream zis;
try
{
String filename;
is = new FileInputStream(path);
zis = new ZipInputStream(new BufferedInputStream(is));
ZipEntry ze;
byte[] buffer = new byte[1024];
int count;
while ((ze = zis.getNextEntry()) != null)
{
filename = ze.getName();
if (ze.isDirectory()) {
File fmd = new File(path.substring(0,path.length()-4)+"/"+filename);
fmd.mkdirs();
continue;
}
FileOutputStream fout = new FileOutputStream(path.substring(0,path.length()-4)+"/"+filename);
while ((count = zis.read(buffer)) != -1)
{
fout.write(buffer, 0, count);
}
fout.close();
zis.closeEntry();
}
zis.close();
}
catch(IOException e)
{
e.printStackTrace();
return false;
}
return true;
}

This method works fine. It was a permission issue while creating the zips in Linux platform. But function starts working properly when I changed the file permission.

Writing ZipOutputStream to OutPutStream blocking the connection

Hi I am working on an android file explorer application. When a large folder is requested for download I am zipping the folder and writing it to the outputstream using the code below.But any further requests are blocked until the file is downloaded.What could be the problem ?Should I consider moving this to a thread?I have tried many variations already and none seem to work.Any pointers in the right direction would be greatly appreciated.
if(f.isDirectory()&& action.equalsIgnoreCase("download")){
File zip=new File(file);
entity = new EntityTemplate(new ContentProducer() {
public void writeTo(final OutputStream outstream) throws IOException {
action=null;
OutputStreamWriter writer=fileop.createZipFile(file, outstream);
writer.flush();
}
});response.setHeader("Content-Type", "application/zip");
response.setHeader("Content-Disposition","attachment; filename=\"" +zip.getName()+".zip\"");
response.setEntity(entity);
}
createZipFileMethod :
OutputStreamWriter createZipFile(String sourceDir,OutputStream os) {
// TODO Auto-generated method stub
OutputStreamWriter osw=null;
try
{
//wrap object of OutputStream
osw=new OutputStreamWriter(os,"UTF-8");
//create object of ZipOutputStream from OutputStream
ZipOutputStream zout = new ZipOutputStream(os);
//create File object from source directory
File fileSource = new File(sourceDir);
addDirectory(zout, fileSource);
//close the ZipOutputStream
zout.close();
System.out.println("Zip file has been created!");
}
catch(IOException ioe)
{
System.out.println("IOException :" + ioe);
}
return osw;
}
private static void addDirectory(ZipOutputStream zout, File fileSource) {
//get sub-folder/files list
File[] files = fileSource.listFiles();
System.out.println("Adding directory " + fileSource.getName());
for(int i=0; i < files.length; i++)
{
//if the file is directory, call the function recursively
if(files[i].isDirectory())
{
addDirectory(zout, files[i]);
continue;
}
/*
* we are here means, its file and not directory, so
* add it to the zip file
*/
try
{
System.out.println("Adding file " + files[i].getName());
//create byte buffer
byte[] buffer = new byte[1024];
//create object of FileInputStream
FileInputStream fin = new FileInputStream(files[i]);
zout.putNextEntry(new ZipEntry(files[i].getName()));
/*
* After creating entry in the zip file, actually
* write the file.
*/
int length;
while((length = fin.read(buffer)) > 0)
{
zout.write(buffer, 0, length);
}
/*
* After writing the file to ZipOutputStream, use
*
* void closeEntry() method of ZipOutputStream class to
* close the current entry and position the stream to
* write the next entry.
*/
zout.closeEntry();
//close the InputStream
fin.close();
}
catch(IOException ioe)
{
System.out.println("IOException :" + ioe);
}
}
}

Overwriting ZipEntrys

Simple question,
I'm writing a series of text files into a zip, just wrapping a fileoutputstream in a zipoutputstream and then in a printwriter.
public static int saveData(File outfile, DataStructure input) {
//variables
ArrayList<String> out = null;
FileOutputStream fileout = null;
ZipOutputStream zipout = null;
PrintWriter printer = null;
//parameter tests
try {
fileout = new FileOutputStream(outfile);
zipout = new ZipOutputStream(fileout);
printer = new PrintWriter(zipout);
} catch (Exception e) {
e.printStackTrace();
return util.FILE_INVALID;
}
for(DataItem data : input){
//process the data into a list of strings
try {
zipout.putNextEntry(new ZipEntry( dataFileName ));
for(String s : out) {
printer.println(s);
}
zipout.closeEntry();
} catch (Exception e) {
try {
fileout.close();
} catch (Exception x) {
x.printStackTrace();
return util.CRITICAL_ERROR;
}
e.printStackTrace();
return util.CRITICAL_ERROR;
}
}
try {
fileout.close();
} catch (Exception e) {
e.printStackTrace();
return util.CRITICAL_ERROR;
}
return util.SUCCESS;
}
Previously in the app i've been developing I've just been saving to the current directory for testing and I know in the case of a file already existing that the file will be overwritten (and have been exploiting this). What I dont know is the behaviour for zips. Will it overwrite entries of the same name? Or will it simply overwrite the whole zip file (which would be convenient for my purposes.
K.Barad

As Joel said, If you try to add a duplicate ZipEntry you will get an exception. If you want to replace the current entry you need to delete it and re-insert it.
You might want to do something like here below to achieve it:
private ZipFile addFileToExistingZip(File zipFile, File versionFile) throws IOException{
// get a temp file
File tempFile = File.createTempFile(zipFile.getName(), null);
// delete it, otherwise you cannot rename your existing zip to it.
tempFile.delete();
boolean renameOk=zipFile.renameTo(tempFile);
if (!renameOk)
{
throw new RuntimeException("could not rename the file "+zipFile.getAbsolutePath()+" to "+tempFile.getAbsolutePath());
}
byte[] buf = new byte[4096 * 1024];
ZipInputStream zin = new ZipInputStream(new FileInputStream(tempFile));
ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFile));
ZipEntry entry = zin.getNextEntry();
while (entry != null) {
String name = entry.getName();
boolean toBeDeleted = false;
if (versionFile.getName().indexOf(name) != -1) {
toBeDeleted = true;
}
if(!toBeDeleted){
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(name));
// Transfer bytes from the ZIP file to the output file
int len;
while ((len = zin.read(buf)) > 0) {
out.write(buf, 0, len);
}
}
entry = zin.getNextEntry();
}
// Close the streams
zin.close();
// Compress the files
InputStream in = new FileInputStream(versionFile);
String fName = versionFile.getName();
// Add ZIP entry to output stream.
out.putNextEntry(new ZipEntry(fName));
// Transfer bytes from the file to the ZIP file
int len;
while ((len = in.read(buf)) > 0) {
out.write(buf, 0, len);
}
// Complete the entry
out.closeEntry();
in.close();
// Complete the ZIP file
out.close();
tempFile.delete();
return new ZipFile(zipFile);
}
The above code worked for me where the need was to add a new zip entry to an existing zip file. If the entry is already present inside the zip, then overwrite it.
Comments/improvements in the code are welcome!
Thanks!

If you try to add a duplicate ZipEntry you will get an exception. If you want to replace the current entry you need to delete it and re-insert it. I suspect the exception you get is much the same as this one.