I want to compress the dynamic created content and write to ServletOutputStream directly, without saving it as a file on the server before compressing.
For example, I created an Excel Workbook and a StringBuffer that includes strings with the SQL template. I don't want to save the dynamic content to .xlsx and .sql file on the server before zipping the files and writing to ServletOutputStream for downloading.
Sample Code:
ServletOutputStream out = response.getOutputStream();
workbook.write(byteArrayOutputStream);
zipIt(byteArrayOutputStream,out);
public static boolean zipIt(ByteArrayOutputStream input, ServletOutputStream output) {
try {
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(output));
ZipEntry zipEntry = new ZipEntry("test.xlsx");
zos.putNextEntry(zipEntry);
if (input != null) {
zipEntry.setSize(input.size());
zos.write(input.toByteArray());
zos.closeEntry();
}
} catch (IOException e) {
logger.error("error {}", e);
return false;
}
return true;
}
Create an HttpServlet and in the doGet() or doPost() method create a ZipOutputStream initialized with the ServletOutputStream and write directly to it:
resp.setContentType("application/zip");
// Indicate that a file is being sent back:
resp.setHeader("Content-Disposition", "attachment;filename=test.zip");
// workbook.write() closes the stream, so we first have to
// write it to a "buffer", a ByteArrayOutputStream
ByteArrayOutputStream baos = new ByteArrayOutputStream();
workbook.write(baos);
byte[] data = baos.toByteArray();
try (ZipOutputStream out = new ZipOutputStream(resp.getOutputStream())) {
// Here you can add your content to the zip
ZipEntry e = new ZipEntry("test.xlsx");
// Configure the zip entry, the properties of the file
e.setSize(data.length);
e.setTime(System.currentTimeMillis());
// etc.
out.putNextEntry(e);
// And the content of the XLSX:
out.write(data);
out.closeEntry();
// You may add other files here if you want to
out.finish();
} catch (Exception e) {
// Handle the exception
}
}
Related
I`m trying to compress a list of Xml converted on Strings, save them in only one zip file and returning as a body of a POST on restful. But everytime I save the file I get the error "The archive is either in unknown format or damaged".
protected ByteArrayOutputStream zip(Map<String, String> mapConvertedXml) {
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
try {
for(Map.Entry<String, String> current : mapConvertedXml.entrySet()){
ZipEntry entry = new ZipEntry(current.getKey() + ".xml");
entry.setSize(current.getValue().length());
zos.putNextEntry(entry);
zos.write(current.getValue().getBytes());
zos.flush();
}
zos.close();
} catch (IOException ioe) {
ioe.printStackTrace();
}
return baos;
}
Can anyone help me with that?
To test your zip file please save it temporarily in your filesystem and open it manualy.
FileOutputStream fos = new FileOutputStream(pathToZipFile);
ZipOutputStream zos = new ZipOutputStream(fos);
Then, you can use something like that to construct your Rest Server and return your file.
public static Response getFileLast(#Context UriInfo ui, #Context HttpHeaders hh) {
Response response = null;
byte[] sucess = null;
ByteArrayOutputStream baos = getYourFile();
sucess = baos.toByteArray();
if(sucess!=null) {
response = Response.ok(sucess, MediaType.APPLICATION_OCTET_STREAM).header("content-disposition","attachment; filename = YourFileName").build();
} else {
response = Response.status(Status.NOT_FOUND).type(MediaType.APPLICATION_JSON).entity(new Error("Error downloading file.")).build();
}
return response;
}
You can test this Advanced Rest Client for example.
I want to create an Excel file from a method in Java and download it in a browser.
I have found an example on this post where you create the Excel file, but I want to create the .xls file and download it from a web browser.
How can I do that?
I finally found a solution for my problem...!!
This is working for me:
#RequestMapping("/downloadFile")
public void downloadFile(HttpServletRequest request, HttpServletResponse response) {
try {
String fileName = "C:/excelFile.xls";
HSSFWorkbook workbook = new HSSFWorkbook();
HSSFSheet sheet = workbook.createSheet("firstSheet");
HSSFRow rowhead = sheet.createRow((short) 0);
rowhead.createCell(0).setCellValue("No.");
rowhead.createCell(1).setCellValue("Name");
rowhead.createCell(2).setCellValue("Address");
rowhead.createCell(3).setCellValue("Email");
HSSFRow row = sheet.createRow((short) 1);
row.createCell(0).setCellValue("1");
row.createCell(1).setCellValue("Carlos");
row.createCell(2).setCellValue("Costa Rica");
row.createCell(3).setCellValue("myNameh#gmail.com");
FileOutputStream fileOut = new FileOutputStream(fileName);
workbook.write(fileOut);
fileOut.close();
System.out.println("Your excel file has been generated!");
//Code to download
File fileToDownload = new File(fileName);
InputStream in = new FileInputStream(fileToDownload);
// Gets MIME type of the file
String mimeType = new MimetypesFileTypeMap().getContentType(fileName);
if (mimeType == null) {
// Set to binary type if MIME mapping not found
mimeType = "application/octet-stream";
}
System.out.println("MIME type: " + mimeType);
// Modifies response
response.setContentType(mimeType);
response.setContentLength((int) fileToDownload.length());
// Forces download
String headerKey = "Content-Disposition";
String headerValue = String.format("attachment; filename=\"%s\"", fileToDownload.getName());
response.setHeader(headerKey, headerValue);
// obtains response's output stream
OutputStream outStream = response.getOutputStream();
byte[] buffer = new byte[4096];
int bytesRead = -1;
while ((bytesRead = in.read(buffer)) != -1) {
outStream.write(buffer, 0, bytesRead);
}
in.close();
outStream.close();
System.out.println("File downloaded at client successfully");
} catch (Exception ex) {
System.out.println(ex);
}
}
If you want to trigger a Java process from a Web Browser (HTTP request), then you need an application server (like Tomcat) to accept your HTTP request and execute some server-side code (Servlet). A main method like in the example cannot be launched by a HTTP request. Look here if you never wrote a Servlet before.
This has nothing to do with Excel, but this post shows how to download a file from a Spring MVC controler.
#RequestMapping(value = "/files/{file_name}", method = RequestMethod.GET)
public void getFile(#PathVariable("file_name") String fileName, HttpServletResponse response) {
try {
// get your file as InputStream
InputStream is = ...;
// copy it to response's OutputStream
org.apache.commons.io.IOUtils.copy(is, response.getOutputStream());
response.flushBuffer();
} catch (IOException ex) {
log.info("Error writing file to output stream. Filename was '{}'", fileName, ex);
throw new RuntimeException("IOError writing file to output stream");
}
}
I want to create a ZIP file that contains my archived files that I received from the backend, and then send this file to a user. For 2 days I have been looking for the answer and can't find proper solution, maybe you can help me :)
For now, the code is like this (I know I shouldn't do it all in the Spring controller, but don't care about that, it is just for testing purposes, to find the way to make it works):
#RequestMapping(value = "/zip")
public byte[] zipFiles(HttpServletResponse response) throws IOException {
// Setting HTTP headers
response.setContentType("application/zip");
response.setStatus(HttpServletResponse.SC_OK);
response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");
// Creating byteArray stream, make it bufferable and passing this buffer to ZipOutputStream
ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream();
BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(byteArrayOutputStream);
ZipOutputStream zipOutputStream = new ZipOutputStream(bufferedOutputStream);
// Simple file list, just for tests
ArrayList<File> files = new ArrayList<>(2);
files.add(new File("README.md"));
// Packing files
for (File file : files) {
// New zip entry and copying InputStream with file to ZipOutputStream, after all closing streams
zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
FileInputStream fileInputStream = new FileInputStream(file);
IOUtils.copy(fileInputStream, zipOutputStream);
fileInputStream.close();
zipOutputStream.closeEntry();
}
if (zipOutputStream != null) {
zipOutputStream.finish();
zipOutputStream.flush();
IOUtils.closeQuietly(zipOutputStream);
}
IOUtils.closeQuietly(bufferedOutputStream);
IOUtils.closeQuietly(byteArrayOutputStream);
return byteArrayOutputStream.toByteArray();
}
But the problem is, that using the code, when I enter the URL localhost:8080/zip, I get a file test.zip.html instead of .zip file.
When I remove .html extension and leave just test.zip it opens correctly. So my questions are:
How to avoid returning this .html extension?
Why is it added?
I have no idea what else can I do. I was also trying replace ByteArrayOuputStream with something like:
OutputStream outputStream = response.getOutputStream();
and set the method to be void so it returns nothing, but It created .zip file which was damaged?
On my MacBook after unpacking the test.zip I was getting test.zip.cpgz which was again giving me test.zip file and so on.
On Windows the .zip file was damaged as I said and couldn't even open it.
I also suppose, that removing .html extension automatically will be the best option, but how?
Hope it is no as hard as It seems to be :)
Thanks
The problem is solved.
I replaced:
response.setContentType("application/zip");
with:
#RequestMapping(value = "/zip", produces="application/zip")
And now I get a clear, beautiful .zip file.
If any of you have either better or faster proposition, or just want to give some advice, then go ahead, I am curious.
#RequestMapping(value="/zip", produces="application/zip")
public void zipFiles(HttpServletResponse response) throws IOException {
//setting headers
response.setStatus(HttpServletResponse.SC_OK);
response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");
ZipOutputStream zipOutputStream = new ZipOutputStream(response.getOutputStream());
// create a list to add files to be zipped
ArrayList<File> files = new ArrayList<>(2);
files.add(new File("README.md"));
// package files
for (File file : files) {
//new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
FileInputStream fileInputStream = new FileInputStream(file);
IOUtils.copy(fileInputStream, zipOutputStream);
fileInputStream.close();
zipOutputStream.closeEntry();
}
zipOutputStream.close();
}
#RequestMapping(value="/zip", produces="application/zip")
public ResponseEntity<StreamingResponseBody> zipFiles() {
return ResponseEntity
.ok()
.header("Content-Disposition", "attachment; filename=\"test.zip\"")
.body(out -> {
var zipOutputStream = new ZipOutputStream(out);
// create a list to add files to be zipped
ArrayList<File> files = new ArrayList<>(2);
files.add(new File("README.md"));
// package files
for (File file : files) {
//new zip entry and copying inputstream with file to zipOutputStream, after all closing streams
zipOutputStream.putNextEntry(new ZipEntry(file.getName()));
FileInputStream fileInputStream = new FileInputStream(file);
IOUtils.copy(fileInputStream, zipOutputStream);
fileInputStream.close();
zipOutputStream.closeEntry();
}
zipOutputStream.close();
});
}
I am using REST Web Service of Spring Boot and I have designed the endpoints to always return ResponseEntity whether it is JSON or PDF or ZIP and I came up with the following solution which is partially inspired by denov's answer in this question as well as another question where I learned how to convert ZipOutputStream into byte[] in order to feed it to ResponseEntity as output of the endpoint.
Anyway, I created a simple utility class with two methods for pdf and zip file download
#Component
public class FileUtil {
public BinaryOutputWrapper prepDownloadAsPDF(String filename) throws IOException {
Path fileLocation = Paths.get(filename);
byte[] data = Files.readAllBytes(fileLocation);
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.parseMediaType("application/pdf"));
String outputFilename = "output.pdf";
headers.setContentDispositionFormData(outputFilename, outputFilename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");
return new BinaryOutputWrapper(data, headers);
}
public BinaryOutputWrapper prepDownloadAsZIP(List<String> filenames) throws IOException {
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.parseMediaType("application/zip"));
String outputFilename = "output.zip";
headers.setContentDispositionFormData(outputFilename, outputFilename);
headers.setCacheControl("must-revalidate, post-check=0, pre-check=0");
ByteArrayOutputStream byteOutputStream = new ByteArrayOutputStream();
ZipOutputStream zipOutputStream = new ZipOutputStream(byteOutputStream);
for(String filename: filenames) {
File file = new File(filename);
zipOutputStream.putNextEntry(new ZipEntry(filename));
FileInputStream fileInputStream = new FileInputStream(file);
IOUtils.copy(fileInputStream, zipOutputStream);
fileInputStream.close();
zipOutputStream.closeEntry();
}
zipOutputStream.close();
return new BinaryOutputWrapper(byteOutputStream.toByteArray(), headers);
}
}
And now the endpoint can easily return ResponseEntity<?> as shown below using the byte[] data and custom headers that is specifically tailored for pdf or zip.
#GetMapping("/somepath/pdf")
public ResponseEntity<?> generatePDF() {
BinaryOutputWrapper output = new BinaryOutputWrapper();
try {
String inputFile = "sample.pdf";
output = fileUtil.prepDownloadAsPDF(inputFile);
//or invoke prepDownloadAsZIP(...) with a list of filenames
} catch (IOException e) {
e.printStackTrace();
//Do something when exception is thrown
}
return new ResponseEntity<>(output.getData(), output.getHeaders(), HttpStatus.OK);
}
The BinaryOutputWrapper is a simple immutable POJO class I created with private byte[] data; and org.springframework.http.HttpHeaders headers; as fields in order to return both data and headers from utility method.
I am trying to add a function to my web app which lets the users download an excel file.
I'm trying to achieve this with the following code:
#Override
public void doPost(HttpServletRequest request, HttpServletResponse response) {
File file = new File("d:/test/test.xls");
response.setContentType("application/xls");
response.addHeader("Content-Disposition", "attachment; filename=test.xls");
response.setContentLength((int) file.length());
try {
FileInputStream fileInputStream = new FileInputStream(file);
OutputStream responseOutputStream = response.getOutputStream();
int bytes;
while ((bytes = fileInputStream.read()) != -1) {
responseOutputStream.write(bytes);
}
fileInputStream.close();
responseOutputStream.close();
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
I'm able to download the excel file with the above code, however the file is corrupted. If I open it with microsoft excel, I get a popup with the message:
"the file format and extension of don't match. the file could be corrupted or unsafe".
And the excel file is empty.
After running the code, the original file(d:/test/test.xls) gets also corrupted.
What am I doing wrong?
The official MIME type for Excel file .xls is application/vnd.ms-excel and for .xlsx is application/vnd.openxmlformats-officedocument.spreadsheetml.sheet.
Also, I would suggest doing response.reset() prior to writing to the output stream and responseOutputStream.flush() (important) prior to closing the response.
Try below code :
File file = null;
InputStream in = null;
OutputStream outstream = null;
try {
response.reset();
in = new FileInputStream(file);
response.setContentType("application/vnd.ms-excel");
response.addHeader("content-disposition", "attachment; filename=data.xls");
outstream = response.getOutputStream();
IOUtils.copyLarge(in, outstream);
}
catch (Exception e) {
out.write("Unable to download file");
}finally {
IOUtils.closeQuietly(outstream);
IOUtils.closeQuietly(in);
IOUtils.closeQuietly(out);
if (file != null)
file.delete();
}
dont forgot to add apache commons-io-2.4 in your dependency
I have the below code got from mkyong, to zip files on local. But, my requirement is to zip files on server and need to download that. Could any one help.
code wrote to zipFiles:
public void zipFiles(File contentFile, File navFile)
{
byte[] buffer = new byte[1024];
try{
// i dont have idea on what to give here in fileoutputstream
FileOutputStream fos = new FileOutputStream("C:\\MyFile.zip");
ZipOutputStream zos = new ZipOutputStream(fos);
ZipEntry ze= new ZipEntry(contentFile.toString());
zos.putNextEntry(ze);
FileInputStream in = new FileInputStream(contentFile.toString());
int len;
while ((len = in.read(buffer)) > 0) {
zos.write(buffer, 0, len);
}
in.close();
zos.closeEntry();
//remember close it
zos.close();
System.out.println("Done");
}catch(IOException ex){
ex.printStackTrace();
}
}
what could i provide in fileoutputstream here? contentfile and navigationfile are files i created from code.
If your server is a servlet container, just write an HttpServlet which does the zipping and serving the file.
You can pass the output stream of the servlet response to the constructor of ZipOutputStream and the zip file will be sent as the servlet response:
ZipOutputStream zos = new ZipOutputStream(response.getOutputStream());
Don't forget to set the response mime type before zipping, e.g.:
response.setContentType("application/zip");
The whole picture:
public class DownloadServlet extends HttpServlet {
#Override
public void doGet( HttpServletRequest request, HttpServletResponse response ) throws ServletException, IOException {
response.setContentType("application/zip");
response.setHeader("Content-Disposition", "attachment; filename=data.zip");
// You might also wanna disable caching the response
// here by setting other headers...
try ( ZipOutputStream zos = new ZipOutputStream(response.getOutputStream()) ) {
// Add zip entries you want to include in the zip file
}
}
}
Try this:
#RequestMapping(value="download", method=RequestMethod.GET)
public void getDownload(HttpServletResponse response) {
// Get your file stream from wherever.
InputStream myStream = someClass.returnFile();
// Set the content type and attachment header.
response.addHeader("Content-disposition", "attachment;filename=myfilename.txt");
response.setContentType("txt/plain");
// Copy the stream to the response's output stream.
IOUtils.copy(myStream, response.getOutputStream());
response.flushBuffer();
}
Reference