问题
I want to create .zip file that contains my zipped files that I recieve from backend, and then send this file to the user. For 2 days I have been looking for the answer and cant find proper solution, maybe you can help me :)
For now, the code is like this: (I know I shouldnt do it all in the spring controller, but dont 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 headers
response.setContentType("application/zip");
response.setStatus(HttpServletResponse.SC_OK);
response.addHeader("Content-Disposition", "attachment; filename=\"test.zip\"");
//creating byteArray stream, make it bufforable and passing this buffor 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 URL: localhost:8080/zip I get file: test.zip.html instead of .zip file
When I remove .html extension and leave just test.zip it opens correctly 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
回答1:
seems to be solved. I replaced:
response.setContentType("application/zip");
with:
@RequestMapping(value = "/zip", produces="application/zip")
And now I get 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.
回答2:
@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();
}
回答3:
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.
回答4:
@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();
});
}
来源：https://stackoverflow.com/questions/27952949/spring-rest-create-zip-file-and-send-it-to-the-client