Can't display the image in the html file. Where did I go wrong?
#RequestMapping(value = "/image/{usr.id}", headers = "Accept=image/jpeg, image/jpg, image/png, image/gif", method = RequestMethod.GET)
public #ResponseBody BufferedImage getImage(#PathVariable("usr.id") Long id) {
Attachment att = attSvc.getPicById(id);
try {
InputStream in = new ByteArrayInputStream(att.getAttachmentFile());
return ImageIO.read(in);
} catch (IOException e) {
System.out.println("ERROR:" + e);
throw new RuntimeException(e);
}
}
Please tell me if you need more references! Thank you so much.
This should solve your issue.
...
public #ResponseBody byte[] getImage(#PathVariable("usr.id") Long id) {
...
InputStream in = new ByteArrayInputStream(att.getAttachmentFile());
BufferedImage img = ImageIO.read(in);
ByteArrayOutputStream bao = new ByteArrayOutputStream();
ImageIO.write(img, "jpg", bao);
return bao.toByteArray();
...
Related
Shall I remove this from application.properties
spring.http.multipart.enabled=true
What should be my approach towards this file upload without using multipart?
This way, I'm able to uploading file using where I'm using multipart.
#RequestMapping(value = "/dog/create/{name}", method = RequestMethod.POST)
public JsonNode dogCreation(HttpServletRequest httpRequest, #RequestParam(value = "picture", required = false) MultipartFile multipartFile,
#PathVariable("name") String name) throws IOException, InterruptedException {
JSONObject response = new JSONObject();
Dog dog = new Dog();
String DOG_IMAGES_BASE_LOCATION = "resource\\images\\dogImages";
try {
File file = new File(DOG_IMAGES_BASE_LOCATION);
if (!file.exists()) {
file.mkdirs();
}
} catch (Exception e) {
e.printStackTrace();
}
dog = dogService.getDogByName(name);
if (dog == null) {
if (!multipartFile.isEmpty()) {
String multipartFileName = multipartFile.getOriginalFilename();
String format = multipartFileName.substring(multipartFileName.lastIndexOf("."));
try {
Path path = Paths.get(DOG_IMAGES_BASE_LOCATION + "/" + name + format);
byte[] bytes = multipartFile.getBytes();
File file = new File(path.toString());
file.createNewFile();
Files.write(path, bytes);
if (file.length() == 0) {
response = utility.createResponse(500, Keyword.ERROR, "Image upload failed");
} else {
String dbPath = path.toString().replace('\\', '/');
dog = new Dog();
dog.setName(name);
dog.setPicture(dbPath);
dog = dogService.dogCreation(dog);
response = utility.createResponse(200, Keyword.SUCCESS, "Image upload successful");
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
return objectMapper.readTree(response.toString());
}
I want to do it without using multipart, what would you suggest?
This is what I've done till now to solve this
#RequestMapping(value = "/dog/create/{name}", method = RequestMethod.POST)
public JsonNode dogCreation(HttpServletRequest httpRequest, #RequestParam("picture") String picture,
#PathVariable("name") String name) throws IOException, InterruptedException {
JSONObject response = new JSONObject();
Dog dog = new Dog();
String DOG_IMAGES_BASE_LOCATION = "resource\\images\\dogImages";
try {
File file = new File(DOG_IMAGES_BASE_LOCATION);
if (!file.exists()) {
file.mkdirs();
}
} catch (Exception e) {
e.printStackTrace();
}
dog = dogService.getDogByName(name);
if (dog == null) {
if (!picture.isEmpty()) {
String dogPicture = picture;
byte[] encodedDogPicture = Base64.encodeBase64(dogPicture.getBytes());
String format = dogPicture.substring(picture.lastIndexOf("."));
try {
} catch (Exception e) {
e.printStackTrace();
}
}
}
return objectMapper.readTree(response.toString());
}
I just have to say that this should probably only be used as a workaround.
On your frontend, convert the file to base64 in js:
const reader = new FileReader();
reader.readAsDataURL(file);
reader.onload = function(evt) {
console.log(evt.target.result);
//do POST here - something like this:
$.ajax("/upload64", {
method: "POST",
contentType: "application/text"
data: evt.target.result
}
};
On the server with an example of a decoder - more decoding options here Decode Base64 data in Java
import sun.misc.BASE64Decoder;
#PostMapping("/upload64")
public String uploadBase64(#RequestBody String payload){
BASE64Decoder decoder = new BASE64Decoder();
byte[] decodedBytes = decoder.decodeBuffer(encodedBytes);
//use your bytes
}
I want to an image while I hit an API like localhost:8080:/getImage/app/path={imagePath}
While I hit this API it will return me an Image.
Is this possible?
Actually, I have tried this but it is giving me an ERROR.
Here is my code,
#GET
#Path("/app")
public BufferedImage getFullImage(#Context UriInfo info) throws MalformedURLException, IOException {
String objectKey = info.getQueryParameters().getFirst("path");
return resizeImage(300, 300, objectKey);
}
public static BufferedImage resizeImage(int width, int height, String imagePath)
throws MalformedURLException, IOException {
BufferedImage bufferedImage = ImageIO.read(new URL(imagePath));
final Graphics2D graphics2D = bufferedImage.createGraphics();
graphics2D.setComposite(AlphaComposite.Src);
// below three lines are for RenderingHints for better image quality at cost of
// higher processing time
graphics2D.setRenderingHint(RenderingHints.KEY_INTERPOLATION, RenderingHints.VALUE_INTERPOLATION_BILINEAR);
graphics2D.setRenderingHint(RenderingHints.KEY_RENDERING, RenderingHints.VALUE_RENDER_QUALITY);
graphics2D.setRenderingHint(RenderingHints.KEY_ANTIALIASING, RenderingHints.VALUE_ANTIALIAS_ON);
graphics2D.drawImage(bufferedImage, 0, 0, width, height, null);
graphics2D.dispose();
System.out.println(bufferedImage.getWidth());
return bufferedImage;
}
My ERROR,
java.io.IOException: The image-based media type image/webp is not supported for writing
Is there any way to return an Image while hitting any URL in java?
You can use IOUtils. Here is code sample.
#RequestMapping(path = "/getImage/app/path/{filePath}", method = RequestMethod.GET)
public void getImage(HttpServletResponse response, #PathVariable String filePath) throws IOException {
File file = new File(filePath);
if(file.exists()) {
String contentType = "application/octet-stream";
response.setContentType(contentType);
OutputStream out = response.getOutputStream();
FileInputStream in = new FileInputStream(file);
// copy from in to out
IOUtils.copy(in, out);
out.close();
in.close();
}else {
throw new FileNotFoundException();
}
}
i didn't test it due to i don't have the environment in this machine, but logically it should work like the following, read it as input stream and let your method returns #ResponseBody byte[]
#GET
#Path("/app")
public #ResponseBody byte[] getFullImage(#Context UriInfo info) throws MalformedURLException, IOException {
String objectKey = info.getQueryParameters().getFirst("path");
BufferedImage image = resizeImage(300, 300, objectKey);
ByteArrayOutputStream os = new ByteArrayOutputStream();
ImageIO.write(image, "jpg", os);
InputStream is = new ByteArrayInputStream(os.toByteArray());
return IOUtils.toByteArray(is);
}
UPDATE
depending on #Habooltak Ana suggestion there is no need to create an input stream, the code should be look like the following
#GET
#Path("/app")
public #ResponseBody byte[] getFullImage(#Context UriInfo info) throws
MalformedURLException, IOException {
String objectKey = info.getQueryParameters().getFirst("path");
BufferedImage image = resizeImage(300, 300, objectKey);
ByteArrayOutputStream os = new ByteArrayOutputStream();
ImageIO.write(image, "jpg", os);
return os.toByteArray();
}
Just return a file object with correct HTTP-Headers (Content-Type and Content-Disposition) will work in most cases/environments.
Pseudocode
File result = createSomeJPEG();
/*
e.g.
RenderedImage rendImage = bufferedImage;
File file = new File("filename.jpg");
ImageIO.write(rendImage, "jpg", file);
*/
response().setHeader("Content-Disposition", "attachment;filename=filename.jpg;");
response().setHeader("Content-Type", "image/jpeg");
return ok(result);
See also:
file downloading in restful web services
what's the correct way to send a file from REST web service to client?
I have to create an application that allows the user upload a specific Excel file. Sumarizing, the system needs to receive the file, write something in a specific cell and return the same file modificated to the view, then, the user can download this new file.
I'm using Apache POI for modify the excel file, but I don't know how to return this file to the view.
Upload view:
<form method="POST" enctype="multipart/form-data" th:action="#{/pessoa/lote}">
<table>
<tr><td>File to upload:</td><td><input type="file" name="file" /></td></tr>
<tr><td></td><td><input type="submit" value="Upload" /></td></tr>
</table>
</form>
Controller: Here I don't know what I need to do, I call a method named "consultaLote" from a Service who is named "ConsultaPessoaService".
#RequestMapping(value = "/lote", method = RequestMethod.POST)
public String handleFileUpload(#RequestParam("file") MultipartFile file,
RedirectAttributes redirectAttributes) {
try {
cpService.consultaLote(file);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
redirectAttributes.addFlashAttribute("message",
"You successfully uploaded " + file.getOriginalFilename() + "!");
return "redirect:/";
}
In the service, I can read and edit the file, then its generated a FileOutputStream. At this time I don't know how to continue, I need to return the file to the view using the controller, but I don't know how:
#Service
public class ConsultaPessoaService {
public FileOutputStream consultaLote(MultipartFile file) throws IOException {
DataFormatter formatter = new DataFormatter();
File convFile = new File(file.getOriginalFilename());
convFile.createNewFile();
FileOutputStream fos = new FileOutputStream(convFile);
fos.write(file.getBytes());
fos.close();
FileInputStream inputStream = new FileInputStream(convFile);
Workbook workbook = new XSSFWorkbook(inputStream);
Sheet firstSheet = workbook.getSheetAt(0);
Iterator<Row> iterator = firstSheet.iterator();
while (iterator.hasNext()) {
Row nextRow = iterator.next();
System.out.println(formatter.formatCellValue(nextRow.getCell(3)));
nextRow.getCell(3).setCellValue("test");
}
FileOutputStream outputStream = new FileOutputStream("arquivot.xlsx");
workbook.write(outputStream);
workbook.close();
outputStream.flush();
outputStream.close();
return outputStream;
}
}
You should return a byte array from consultaLote method. Method return type will be byte[] because you need this byte array to write the httpservlet outputstream.
public byte[] consultaLote(MultipartFile file) throws IOException {
//whatever changes you want, do it here. I am going to converting part
ByteArrayOutputStream baos = new ByteArrayOutputStream();
workbook.write(baos);
// before returning you can close your your ByteArrayOutputStream
//baos.close();
return baos.toByteArray();
}
Now comes to your controller. We write the byte array to OutputStream from HttpservletResponse.
#RequestMapping(value = "/lote", method = RequestMethod.POST)
public String handleFileUpload(#RequestParam("file") MultipartFile file,
RedirectAttributes redirectAttributes,HttpServletResponse response) {
byte[] fileBytes = null;
String reportName="yourReportName";
try {
fileBytes = cpService.consultaLote(file);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
if(fileBytes !=null){
response.setHeader("Content-Disposition", "attachment; filename=" + reportName + ".xls");
response.setContentType("application/xls");
response.getOutputStream().write(fileBytes);
response.getOutputStream().flush();
}
redirectAttributes.addFlashAttribute("message",
"You successfully uploaded " + file.getOriginalFilename() + "!");
return "redirect:/";
}
Hope this will help.
I am using GridFS to store images. Now I want to write stored image to spring view page directly I have tried a lot, but not succeded. I can write image to my local system by using
gfs.writeTo("my location of local directory");
but how can I write same image to JSP view in spring? Any help is appreciated.
This is worked for me
Serve method:
#Override
public void serveImage(String imageId , HttpServletResponse response ) {
InputStream is = null;
ApplicationContext ctx = new AnnotationConfigApplicationContext(MongoDBConfiguration.class);
GridFsOperations gridOperations = (GridFsOperations) ctx.getBean("yourBeanName");
List<GridFSDBFile> result = gridOperations.find(new Query().addCriteria(Criteria.where("_id").is(imageId)));
for (GridFSDBFile file : result) {
try {
response.setHeader("Content-Disposition", "inline; filename=image.jpg");
response.setContentType("image/jpg");
response.setContentLengthLong(file.getLength());
is = file.getInputStream();
IOUtils.copy(is, response.getOutputStream());
response.flushBuffer();
} catch (java.nio.file.NoSuchFileException e) {
response.setStatus(HttpStatus.NOT_FOUND.value());
} catch (Exception e) {
response.setStatus(HttpStatus.INTERNAL_SERVER_ERROR.value());
}
}
}
Html code:
<image>
<source th:src="#{/get/${imageId}}" type="image/jpg">
</image>
Controller class:
#RequestMapping(value = "/get/{imageId}", method = RequestMethod.GET)
public void handleFileDownload(#PathVariable String imageId, HttpServletResponse response) {
try {
vaskService.serveImage(imageId, response);
} catch (Exception e) {
response.setStatus(HttpStatus.INTERNAL_SERVER_ERROR.value());
}
}
I have my controller like this, i am using POSTMAN REST client for upload pdf file. setting content-type: multipart/form-data;boundary=randomBoundaryNotInAnyOfParts
It is creating the file but not writing anything in it. Am i missing something here?
#RequestMapping(value = "/uploadfile", method = RequestMethod.POST)
public #ResponseBody String upload(HttpServletRequest request) {
InputStream is = null;
OutputStream out = null;
try {
is = request.getInputStream();
byte[] b = new byte[1024];
out = new FileOutputStream(new File("C:\\tmp\\upload.txt"));
out.write(b);
} catch (IOException e) {
e.printStackTrace();
} finally{
try{
if(out != null){
out.close();
}
if(is != null){
is.close();
}
}catch(IOException io){
io.printStackTrace();
}
}
return null;
}
Thanks,
Vinay
Your byte[] is empty hence it is writing nothing in the file. Get content of the request into the byte[] and then write it into file.
Or you can use BufferedWriter also it is more fast.
Thanks,
Brijesh