I have a webapp that allows users to select images and then download them. For a single image, I use HTML5's anchor download and it works beautifully. Now I need to allow them to select multiple images, and download them as a .zip file. I'm using an api to get each image as an InputStream and returning a Jersey Response.
I'm new to zipping and I'm a bit confused with how zipping with InputStream should work.
For single images, it works like so:
try {
InputStream imageInputStream = ImageStore.getImage(imageId);
if (imageInputStream == null) {
XLog.warnf("Unable to find image [%s].", imageId);
return Response.status(HttpURLConnection.HTTP_GONE).build();
}
Response.ResponseBuilder response = Response.ok(imageInputStream);
response.header("Content-Type", imageType.mimeType());
response.header("Content-Disposition", "filename=image.jpg");
return response.build();
}
It's not much, but here's the java I have so far for multiple images
public Response zipAndDownload(List<UUID> imageIds) {
try {
// TODO: instantiate zip file?
for (UUID imageId : imageIds) {
InputStream imageInputStream = ImageStore.getImage(imageId);
// TODO: add image to zip file (ZipEntry?)
}
// TODO: return zip file
}
...
}
I just don't know how to deal with multiple InputStreams, and it seems that I shouldn't have multiple, right?
An InputStream per image is ok. To zip the files you need to create a .zip file for them to live in and get a ZipOutputStream to write to it:
File zipFile = new File("/path/to/your/zipFile.zip");
ZipOutputStream zos = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));
For each image, create a new ZipEntry, add it to the ZipOutputSteam, then copy the bytes from your image's InputStream to the ZipOutputStream:
ZipEntry ze = new ZipEntry("PrettyPicture1.jpg");
zos.putNextEntry(ze);
byte[] bytes = new byte[1024];
int count = imageInputStream.read(bytes);
while (count > -1)
{
zos.write(bytes, 0, count);
count = imageInputStream.read(bytes);
}
imageInputStream.close();
zos.closeEntry();
After you add all the entries, close the ZipOutputStream:
zos.close();
Now your zipFile points to a zip file full of pictures you can do whatever you want with. You can return it like you do with a single image:
BufferedInputStream zipFileInputStream = new BufferedInputStream(new FileInputStream(zipFile));
Response.ResponseBuilder response = Response.ok(zipFileInputStream);
But the content type and disposition are different:
response.header("Content-Type", MediaType.APPLICATION_OCTET_STREAM_TYPE);
response.header("Content-Disposition", "attachment; filename=zipFile.zip");
Note: You can use the copy method from Guava's ByteStreams helper to copy the streams instead of copying the bytes manually. Simply replace the while loop and the 2 lines before it with this line:
ByteStreams.copy(imageInputStream, zos);
Related
I'm trying to create a zip file to be able to send multiple files over http.
My issue is that the Zip file that is generated is "corrupted" before and after the file has been send. The issue is i'm not able to find what i did wrong as i'm getting no errors inside the console.
So does someone has an idea file my generated zip file is corrupted ?
This is my code :
OutputStream responseBody = t.getResponseBody();
ByteArrayOutputStream baos = new ByteArrayOutputStream();
ZipOutputStream zos = new ZipOutputStream(baos);
int counter = 1;
for (PDDocument doc : documents)
{
ZipEntry zipEntry = new ZipEntry("document" + counter);
zos.putNextEntry(zipEntry);
ByteArrayOutputStream docOs = new ByteArrayOutputStream();
doc.save(docOs);
docOs.close();
zos.write(docOs.toByteArray());
zos.closeEntry();
zos.finish();
zos.flush();
counter++;
}
zos.close();
baos.close();
responseBody.write(baos.toByteArray());
responseBody.flush();
Thank you for your help !
You need to remove zos.finish() from inside the loop as it terminates the ZIP entries, as it is handled by zos.close() at end of the stream.
With very large streams you will be better off sending ZIP directly to responseBody bypassing ByteArrayOutputStream memory buffer.
If you are still having problems check the content type of the output is set. It might be easier to debug by temporarily writing the byte[] to file to check the ZIP format you are sending with:
Files.write(Path.of("temp.zip"), baos.toByteArray());
This outline below shows sending a simple ZIP over http (from a servlet, adjust the first 2 lines to appropriate calls for "t"). This may help you check which step of your code causes the corruption if you work back to adding your own document objects inside the loop:
// MUST set response content type:
// resp.setContentType("application/zip");
OutputStream out = resp.getOutputStream(); // or t.getResponseBody();
try(ZipOutputStream zos = new ZipOutputStream(out))
{
while (counter-- > 0)
{
ZipEntry zipEntry = new ZipEntry("document" + counter+".txt");
zos.putNextEntry(zipEntry);
zos.write(("This is ZipEntry: "+zipEntry.getName()+"\r\n").getBytes());
}
}
I am in a situation where on calling an API i want to download a excel file along with a zip file.
I have cracked the code for downloading them separately, but when put together only one file gets downloaded and the other one just doesnt gets downloaded.
I guss the problem is I cannot use response.getOutPutStream().flush() or response.flushBuffer() simultaneously.
String absolutePath = context.getRealPath("resources/ZipFolders");
String inputFile = Paths.get(absolutePath + "/Attachments.zip").toAbsolutePath().toString();
File finalFile = new File(inputFile);
ZipOutputStream zipOut = new ZipOutputStream(new FileOutputStream(finalFile));
String absolutePath2 = context.getRealPath("resources/Spreadsheets");
String inputFile2 = Paths.get(absolutePath2 + "/Validation_Spreadsheet.xlsx").toAbsolutePath().toString();
File file = new File(inputFile2);
byte[] bytes = IOUtils.toByteArray(new FileInputStream(file));
ZipEntry zipEntry = new ZipEntry("Validation_Spreadsheet.xlsx");
zipOut.putNextEntry(zipEntry);
zipOut.write(bytes);
zipOut.closeEntry();
zipOut.close();
response.setContentType("application/zip");
response.setHeader("Content-disposition", "attachment;filename=attachment_trial.zip");
response.getOutputStream().write(IOUtils.toByteArray(new FileInputStream(finalFile)));
System.err.println("above flush>>>>>>>>>>>>>>");
response.getOutputStream().flush();
responsetrial.setContentType("application/vnd.openxmlformats-officedocument.spreadsheetml.sheet");
responsetrial.setHeader("Content-disposition", "attachment; filename=TransactionErrors.xlsx");
responsetrial.getOutputStream().write(IOUtils.toByteArray(new FileInputStream(file)));
System.err.println("above flush2>>>>>>>>>>>>>>");
responsetrial.getOutputStream().flush();
It's not possible to download 2 files over a single HTTP request.
You will need to make 2 separately requests for this task.
If you need to download many files in a single "HTML button", you need to write some javascript logic two make this.
Trying to read a zip file stored in a database as a byte array.
.zip is getting downloaded using the following code but the size of files included in the zip is none. No data is there.
I already went through lots of answers but not sure what's wrong with the following code.
Please assist.
#RequestMapping(value = ApplicationConstants.ServiceURLS.TRANSLATIONS + "/{resourceId}/attachments", produces = "application/zip")
public void attachments(HttpServletResponse response, #PathVariable("resourceId") Long resourceId) throws IOException {
TtTranslationCollection tr = translationManagementDAO.getTranslationCollection(resourceId);
byte[] fileData = tr.getFile();
// setting headers
response.setStatus(HttpServletResponse.SC_OK);
response.addHeader("Content-Disposition", "attachment; filename=\"attachements.zip\"");
ZipOutputStream zipOutputStream = new ZipOutputStream(response.getOutputStream());
ZipInputStream zipStream = new ZipInputStream(new ByteArrayInputStream(fileData));
ZipEntry ent = null;
while ((ent = zipStream.getNextEntry()) != null) {
zipOutputStream.putNextEntry(ent);
}
zipStream.close();
zipOutputStream.close();
}
You have to copy the byte data (content) of the zip file to the output as well...
This should work (untested):
while ((ent = zipStream.getNextEntry()) != null) {
zipOutputStream.putNextEntry(ent);
// copy byte stream
org.apache.commons.io.IOUtils.copy(zis.getInputStream(ent), zipOutputStream);
}
BTW: why you do not just simply forward the original zip byte content?
try (InputStream is = new ByteArrayInputStream(fileData));) {
IOUtils.copy(is, response.getOutputStream());
}
or even better (thanks to comment by #M. Deinum)
IOUtils.copy(fileData, response.getOutputStream());
I am seeking for most efficient way (in terms of speed) to retrieve some file out of the middle of a ZIP file.
e.g. I have ZIP file, which includes 700 folders (tagged 1 to 700). Each folder equals picture and mp3 file. There is special folder called Info, which contains XML file. Problem is, I need to iterate through this ZIP file to find XML file and then I am displaying images from desired folders. I am using ZipFile approach (thus I am iterating through whole ZIP file, even if I want folder 666, I need to go through 665 items in ZIP file) -> selecting from ZIP file is extremely slow.
I would like to ask you, If you have faced similar issue, how have you solved this? Is there any approach in Java, which turns my ZIP file into virtual folder to browse it much more quicker? Is there any external library, which is the most efficient in terms of time?
Source Code snippet:
try {
FileInputStream fin = new FileInputStream(
"sdcard/external_sd/mtp_data/poi_data/data.zip");
ZipInputStream zin = new ZipInputStream(fin);
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
// Log.d("ZE", ze.getName());
if (ze.getName().startsWith("body/665/")) {
// Log.d("FILE F", "soubor: "+ze.getName());
if (ze.getName().endsWith(".jpg")
|| ze.getName().endsWith(".JPG")) {
Log.d("OBR", "picture: " + ze.getName());
ByteArrayOutputStream baos = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
int count;
while ((count = zin.read(buffer)) != -1) {
baos.write(buffer, 0, count);
}
byte[] bytes = baos.toByteArray();
bmp = BitmapFactory.decodeByteArray(bytes, 0,
bytes.length);
photoField.add(bmp);
i++;
}
}
}
}
The ZipFile.getEntry() and ZipFile.getInputStream() methods can be used to access a specific file in a ZIP archive. For example:
ZipFile file = ...
ZipEntry entry = file.getEntry("folder1/picture.jpg");
InputStream in = file.getInputStream(entry);
I need to create a File object out of a file path to an image that is contained in a jar file after creating a jar file. If tried using:
URL url = getClass().getResource("/resources/images/image.jpg");
File imageFile = new File(url.toURI());
but it doesn't work. Does anyone know of another way to do it?
To create a file on Android from a resource or raw file I do this:
try{
InputStream inputStream = getResources().openRawResource(R.raw.some_file);
File tempFile = File.createTempFile("pre", "suf");
copyFile(inputStream, new FileOutputStream(tempFile));
// Now some_file is tempFile .. do what you like
} catch (IOException e) {
throw new RuntimeException("Can't create temp file ", e);
}
private void copyFile(InputStream in, OutputStream out) throws IOException {
byte[] buffer = new byte[1024];
int read;
while((read = in.read(buffer)) != -1){
out.write(buffer, 0, read);
}
}
Don't forget to close your streams etc
This should work.
String imgName = "/resources/images/image.jpg";
InputStream in = getClass().getResourceAsStream(imgName);
ImageIcon img = new ImageIcon(ImageIO.read(in));
Usually, you can't directly get a java.io.File object, since there is no physical file for an entry within a compressed archive. Either you live with a stream (which is best most in the cases, since every good API can work with streams) or you can create a temporary file:
URL imageResource = getClass().getResource("image.gif");
File imageFile = File.createTempFile(
FilenameUtils.getBaseName(imageResource.getFile()),
FilenameUtils.getExtension(imageResource.getFile()));
IOUtils.copy(imageResource.openStream(),
FileUtils.openOutputStream(imageFile));
You cannot create a File object to a reference inside an archive. If you absolutely need a File object, you will need to extract the file to a temporary location first. On the other hand, most good API's will also take an input stream instead, which you can get for a file in an archive.