问题
I have a java/spring boot application where I want to build an API endpoint that creates and returns a downloadable excel file. Here is my controller endpoint:
@RestController
@RequestMapping("/Foo")
public class FooController {
private final FooService fooService;
@GetMapping("/export")
public ResponseEntity export() {
Resource responseFile = fooService.export();
return ResponseEntity.ok()
.header(HttpHeaders.CONTENT_DISPOSITION, "attachment; filename="+responseFile.getFilename())
.contentType(MediaType.MULTIPART_FORM_DATA)
.body(responseFile);
}
}
Then the service class
public class FooService {
public Resource export() throws IOException {
StringBuilder filename = new StringBuilder("Foo Export").append(" - ")
.append("Test 1.xlsx");
return export(filename);
}
private ByteArrayResource export(String filename) throws IOException {
byte[] bytes = new byte[1024];
try (Workbook workbook = generateExcel()) {
FileOutputStream fos = write(workbook, filename);
fos.write(bytes);
fos.flush();
fos.close();
}
return new ByteArrayResource(bytes);
}
private Workbook generateExcel() {
Workbook workbook = new XSSFWorkbook();
Sheet sheet = workbook.createSheet();
//create columns and rows
return workbook;
}
private FileOutputStream write(final Workbook workbook, final String filename) throws IOException {
FileOutputStream fos = new FileOutputStream(filename);
workbook.write(fos);
fos.close();
return fos;
}
}
This code successfully creates the proper excel file using the Apache POI library. But this won't return it out of the controller properly because ByteArrayResource::getFilename always returns null:
/**
* This implementation always returns {@code null},
* assuming that this resource type does not have a filename.
*/
@Override
public String getFilename() {
return null;
}
What type of resource can I use to return the generated excel file?
回答1:
Since you are using ByteArrayResource, you can use the below controller code assuming that the FooService is autowired in the controller class.
@RequestMapping(path = "/download_excel", method = RequestMethod.GET)
public ResponseEntity<Resource> download(String fileName) throws IOException {
ByteArrayResource resource = fooService.export(fileName);
return ResponseEntity.ok()
.headers(headers) // add headers if any
.contentLength(resource.contentLength())
.contentType(MediaType.parseMediaType("application/vnd.ms-excel"))
.body(resource);
}
回答2:
Basically , there are few points that you first need to understand & then decide what you want to do ,
1.Is excel creation on disk needed or can you stream it from memory?
If its a download pop up, user might keep it open for long time & memory be occupied during that period ( disadvantage of in memory approach ) .
Secondly, if generated file has to be new for each request ( i.e. data to be exported is different ) then there is no point in keeping it at disk ( disadvantage of in disk approach ) .
Thirdly, it will be hard for an API code to do disk clean up because you never know in advance as when user will finish up his down load ( disadvantage of in disk approach ) .
Answer by Fizik26 is this In - Memory approach where you don't create a file on disk. . Only thing from that answer is that you need to keep track of length of array out.toByteArray() & that can easily be done via a wrapper class.
2.While downloading a file , your code needs to stream a file chunk by chunk - thats what Java streams are for. Code like below does that.
return ResponseEntity.ok().contentLength(inputStreamWrapper.getByteCount())
.contentType(MediaType.parseMediaType("application/vnd.ms-excel"))
.cacheControl(CacheControl.noCache())
.header("Content-Disposition", "attachment; filename=" + "SYSTEM_GENERATED_FILE_NM")
.body(new InputStreamResource(inputStreamWrapper.getByteArrayInputStream()));
and inputStreamWrapper is like ,
public class ByteArrayInputStreamWrapper {
private ByteArrayInputStream byteArrayInputStream;
private int byteCount;
public ByteArrayInputStream getByteArrayInputStream() {
return byteArrayInputStream;
}
public void setByteArrayInputStream(ByteArrayInputStream byteArrayInputStream) {
this.byteArrayInputStream = byteArrayInputStream;
}
public int getByteCount() {
return byteCount;
}
public void setByteCount(int byteCount) {
this.byteCount = byteCount;
}
}
Regarding file name, if file name is not an input to end point - that means ..its system generated ( a combination of constant string plus a variable part per user ). I am not sure why you need to get that from resource.
You won't need this wrapper if use - org.springframework.core.io.ByteArrayResource
回答3:
You have to set the file name to the response header using Content-disposition. Try this
@GetMapping("/export")
public ResponseEntity export(HttpServletResponse response) {
fooService.export(response);
}
Change your service method like this
public Resource export(HttpServletResponse response) throws IOException {
StringBuilder filename = new StringBuilder("Foo Export").append(" - ")
.append("Test 1.xlsx");
return export(filename, response);
}
private void export(String filename, HttpServletResponse response) throws IOException {
try (Workbook workbook = generateExcel()) {
FileOutputStream fos = write(workbook, filename);
IOUtils.copy(new FileInputStream(fos.getFD()),
servletResponse.getOutputStream());//IOUtils is from apache commons io
response.setContentType("application/vnd.ms-excel");
response.setHeader("Content-disposition", "attachment; filename=" + filename);
}catch(Exception e) {
//catch if any checked exception
}finally{
//Close all the streams
}
}
回答4:
Letting controller know is always better what it is going to write using ReponseEntity. At service level just create and play around the objects. @RestController or @Controller doesn't matter here.
What you are looking forward for in your controller is somewhat like this (sample) -
@GetMapping(value = "/alluserreportExcel")
public ResponseEntity<InputStreamResource> excelCustomersReport() throws IOException {
List<AppUser> users = (List<AppUser>) userService.findAllUsers();
ByteArrayInputStream in = GenerateExcelReport.usersToExcel(users);
// return IO ByteArray(in);
HttpHeaders headers = new HttpHeaders();
// set filename in header
headers.add("Content-Disposition", "attachment; filename=users.xlsx");
return ResponseEntity.ok().headers(headers).body(new InputStreamResource(in));
}
Generate Excel Class -
public class GenerateExcelReport {
public static ByteArrayInputStream usersToExcel(List<AppUser> users) throws IOException {
...
...
//your list here
int rowIdx = 1;
for (AppUser user : users) {
Row row = sheet.createRow(rowIdx++);
row.createCell(0).setCellValue(user.getId().toString());
...
}
workbook.write(out);
return new ByteArrayInputStream(out.toByteArray());
and finally, somewhere, in your view -
<a href="<c:url value='/alluserreportExcel' />"
target="_blank">Export all users to MS-Excel</a>
For full example, take a peek - here, here and here.
回答5:
You can use this :
headers.add("Content-Disposition", "attachment; filename=NAMEOFYOURFILE.xlsx");
ByteArrayInputStream in = fooService.export();
return ResponseEntity
.ok()
.headers(headers)
.body(new InputStreamResource(in));
It will download the Excel file when you call this endpoint.
In your export method in your service, you have to return something like that :
ByteArrayOutputStream out = new ByteArrayOutputStream();
try {
workbook.write(out);
} catch (IOException e) {
e.printStackTrace();
}
return new ByteArrayInputStream(out.toByteArray());
来源：https://stackoverflow.com/questions/52078128/spring-boot-controller-export-an-excel