i am using GWTUpload to upload a file.
i am getting the server info, file name, content type etc.. in onFinishHandler, but there's no option to get the file content from server to client.
Note : am trying to upload XML File and EXCEL File
i am using GWT 2.4, GXT 3.0.1, GWTUpload 0.6.6
here's the sample code
Client Code - OnFinishHandler
u.addOnFinishUploadHandler(new OnFinishUploaderHandler() {
#Override
public void onFinish(IUploader uploader) {
if (uploader.getStatus() == Status.SUCCESS) {
System.err.println(uploader.getServerResponse());
UploadedInfo info = uploader.getServerInfo();
System.out.println("File name " + info.name);
System.out.println("File content-type " + info.ctype);
System.out.println("File size " + info.size);
System.out.println("Server message " + info.message);
}
}
});
Servlet Code
public class GWTFileUploadServlet extends UploadAction {
private static final long serialVersionUID = -6742854283091447922L;
String fileContent;
File uploadedFile;
#Override
public String executeAction(HttpServletRequest request,
List<FileItem> sessionFiles) throws UploadActionException {
String response = "";
int cont = 0;
for (FileItem item : sessionFiles) {
if (false == item.isFormField()) {
cont++;
try {
File file = File.createTempFile("upload-", ".bin");
item.write(file);
uploadedFile = file;
fileContent = item.getContentType();
response += "File saved as " + file.getAbsolutePath();
} catch (Exception e) {
throw new UploadActionException(e.getMessage());
}
}
}
removeSessionFileItems(request);
return response;
}
#Override
public void getUploadedFile(HttpServletRequest request,
HttpServletResponse response) throws IOException {
if (fileContent != null && !fileContent.isEmpty()) {
response.setContentType(fileContent);
FileInputStream is = new FileInputStream(uploadedFile);
copyFromInputStreamToOutputStream(is, response.getOutputStream());
} else {
renderXmlResponse(request, response, XML_ERROR_ITEM_NOT_FOUND);
}
}
}
please help me....
You can read the file you have created in the filesystem when you called item.write(...), but it is better if you get an InputStream from the FileItem you received and write its content anywhere. For instance if the content is a String you can use a StringWritter to get it:
InputStream inputStream = item.getInputStream();
StringWriter writer = new StringWriter();
IOUtils.copy(inputStream, writer);
String theContentString = writer.toString();
[EDITED]
To get the content of the file in client side, so you have to retrieve it from the server using any method:
As as a customized message in your gwtupload servlet if the content of the file is ascii: use return String of executeAction.
Do a RequestBuilder call to the servlet to get the file using the uploader url value.
Use any GWT ajax mechanism like RPC.
In modern browsers you can get the content of a file in client side without sending it to server side. Take a look to lib-gwt-file
In your code You can just use
byte[] file ;
file = item.get();
And You will get all the file content in an encoded format in the "file" variable .
Related
I am able to download a single file but how I can download a zip file which contain multiple files.
Below is the code to download a single file but I have multiples files to download. Any help would greatly appreciated as I am stuck on this for last 2 days.
#GET
#Path("/download/{fname}/{ext}")
#Produces(MediaType.APPLICATION_OCTET_STREAM)
public Response downloadFile(#PathParam("fname") String fileName,#PathParam("ext") String fileExt){
File file = new File("C:/temp/"+fileName+"."+fileExt);
ResponseBuilder rb = Response.ok(file);
rb.header("Content-Disposition", "attachment; filename=" + file.getName());
Response response = rb.build();
return response;
}
Here is my working code I have used response.getOuptStream()
#RestController
public class DownloadFileController {
#Autowired
DownloadService service;
#GetMapping("/downloadZip")
public void downloadFile(HttpServletResponse response) {
response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition", "attachment;filename=download.zip");
response.setStatus(HttpServletResponse.SC_OK);
List<String> fileNames = service.getFileName();
System.out.println("############# file size ###########" + fileNames.size());
try (ZipOutputStream zippedOut = new ZipOutputStream(response.getOutputStream())) {
for (String file : fileNames) {
FileSystemResource resource = new FileSystemResource(file);
ZipEntry e = new ZipEntry(resource.getFilename());
// Configure the zip entry, the properties of the file
e.setSize(resource.contentLength());
e.setTime(System.currentTimeMillis());
// etc.
zippedOut.putNextEntry(e);
// And the content of the resource:
StreamUtils.copy(resource.getInputStream(), zippedOut);
zippedOut.closeEntry();
}
zippedOut.finish();
} catch (Exception e) {
// Exception handling goes here
}
}
}
Service Class:-
public class DownloadServiceImpl implements DownloadService {
#Autowired
DownloadServiceDao repo;
#Override
public List<String> getFileName() {
String[] fileName = { "C:\\neon\\FileTest\\File1.xlsx", "C:\\neon\\FileTest\\File2.xlsx", "C:\\neon\\FileTest\\File3.xlsx" };
List<String> fileList = new ArrayList<>(Arrays.asList(fileName));
return fileList;
}
}
Use these Spring MVC provided abstractions to avoid loading of whole file in memory.
org.springframework.core.io.Resource & org.springframework.core.io.InputStreamSource
This way, your underlying implementation can change without changing controller interface & also your downloads would be streamed byte by byte.
See accepted answer here which is basically using org.springframework.core.io.FileSystemResource to create a Resource and there is a logic to create zip file on the fly too.
That above answer has return type as void, while you should directly return a Resource or ResponseEntity<Resource> .
As demonstrated in this answer, loop around your actual files and put in zip stream. Have a look at produces and content-type headers.
Combine these two answers to get what you are trying to achieve.
public void downloadSupportBundle(HttpServletResponse response){
File file = new File("supportbundle.tar.gz");
Path path = Paths.get(file.getAbsolutePath());
logger.debug("__path {} - absolute Path{}", path.getFileName(),
path.getRoot().toAbsolutePath());
response.setContentType("application/octet-stream");
response.setHeader("Content-Disposition", "attachment;filename=supportbundle.tar.gz");
response.setStatus(HttpServletResponse.SC_OK);
System.out.println("############# file name ###########" + file.getName());
try (ZipOutputStream zippedOut = new ZipOutputStream(response.getOutputStream())) {
FileSystemResource resource = new FileSystemResource(file);
ZipEntry e = new ZipEntry(resource.getFilename());
e.setSize(resource.contentLength());
e.setTime(System.currentTimeMillis());
zippedOut.putNextEntry(e);
StreamUtils.copy(resource.getInputStream(), zippedOut);
zippedOut.closeEntry();
zippedOut.finish();
} catch (Exception e) {
}
}
I want to stream video file instead of Downloading it. I tried with the below code. But it is downloading.
#GET
#Path("video")
#Produces(MediaType.APPLICATION_OCTET_STREAM)
public Response video() {
File file = new File("/home/lvaddi/Downloads/demoVideoFile.flv");
ResponseBuilder response = Response.ok(file, MediaType.APPLICATION_OCTET_STREAM);
response.header("Content-Disposition", "filename=videofile.flv");
return response.build();
}
You should use StreamingOutput to do that not return the file as body (which will send the whole file in bulk), something like:
new StreamingOutput() {
#Override
public void write(OutputStream output) throws IOException, WebApplicationException {
try {
InputStream input = <get input stream from your file>
IOUtils.copy(input, output);
output.flush();
} finally {
EntityUtils.consume(response.getEntity());
IOUtils.closeQuietly(response);
}
}
};
There's also an example here
I have some JavaScript code that upload file to server using ajax and form data and server side java code that accept it. I can upload English file name. But when I uploaded other Unicode file name, the file name I got in server side is unreadable. The following is code snippet.
JavaScript
var f = new FormData();
f.append("file", file);
xhr.send(f);
Java
public void upload(MultipartFormDataInput input) {
Map<String, List<InputPart>> uploadForm = input.getFormDataMap();
List<InputPart> inputParts = uploadForm.get("user_file[]");
IFileInfo file = null;
for (InputPart inputPart : inputParts) {
try {
MultivaluedMap<String, String> header = inputPart.getHeaders();
fileName = getFileName(header);
System.out.println("File name is " + fileName);
} catch (IOException e) {
e.printStackTrace();
}
}
}
private String getFileName(MultivaluedMap<String, String> header) {
System.out.println("Headers is " + header.getFirst("Content-Disposition"));
String[] contentDisposition = header.getFirst("Content-Disposition")
.split(";");
for (String filename : contentDisposition) {
if ((filename.trim().startsWith("filename"))) {
String[] name = filename.split("=");
String finalFileName = name[1].trim().replaceAll("\"", "");
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return finalFileName;
}
}
return "unknown";
}
When I upload "大家好.jpg" , I got server side log showing the following.
Headers is form-data; name="user_file[]"; filename="���������.jpg"
File name is ���������.jpg
I think browser encode file name before uploading it.But I don't know which encoding did it used or how to decode it back. Any help is much appreciated.
I am using GWT.
I have to download a file file from server to client.
Document is in the external repository.
Client sends the id of the document through a Servlet.
On server side: Using this ID document is retrieved:
Document document = (Document)session.getObject(docId);
ContentStream contentStream = document.getContentStream();
ByteArrayInputStream inputStream = (ByteArrayInputStream) contentStream.getStream();
int c;
while ((c = inputStream.read()) != -1) {
System.out.print((char) c);
}
String mime = contentStream.getMimeType();
String name = contentStream.getFileName();
InputStream strm = contentStream.getStream();
Here I can read the document.
I want to send this to the client.
How do I make this a file and send it back to the client?
In Your Servlet:
Document document =(Document)session.getObject(docId);
ContentStream contentStream = document.getContentStream();
String name = contentStream.getFileName();
response.setHeader("Content-Type", "application/octet-stream;");
response.setHeader("Content-Disposition", "attachment;filename=\"" + name + "\"");
OutputStream os = response.getOutputStream();
InputStream is =
(ByteArrayInputStream) contentStream.getStream();
BufferedInputStream buf = new BufferedInputStream(is);
int readBytes=0;
while((readBytes=buf.read())!=-1) {
os.write(readBytes);
}
os.flush();
os.close();// *important*
return;
You can create a standard servlet (which extends HttpServlet and not RemoteServiceServlet) on server side and opportunity to submit the id as servlet parameter on client side.
Now you need after getting request create the excel file and send it to the client. Browser shows automatically popup with download dialog box.
But you should make sure that you set the right content-type response headers. This header will instruct the browser which type of file is it.
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
String fileId = reguest.getParameter("fileId"); // value of file id from request
File file = CreatorExel.getFile(fileId); // your method to create file from helper class
// setting response headers
response.setHeader("Content-Type", getServletContext().getMimeType(file.getName()));
response.setHeader("Content-Length", file.length());
response.setHeader("Content-Disposition", "inline; filename=\"" + file.getName() + "\"");
BufferedInputStream input = null;
BufferedOutputStream output = null;
try {
InputStream inputStream = new FileInputStream(file);
ServletOutputStream outputStream = response.getOutputStream();
input = new BufferedInputStream(fileInput);
output = new BufferedOutputStream(outputStream);
int count;
byte[] buffer = new byte[8192]; // buffer size is 512*16
while ((count = input.read(buffer)) > 0) {
output.write(buffer, 0, count);
}
} finally {
if (output != null) {
try {
output.close();
} catch (IOException ex) {
}
}
if (input != null) {
try {
input.close();
} catch (IOException ex) {
}
}
}
I need to read form TXT file on the server side and send its contents to the client side to print in a label or any thing, i need to know where i place the TXT file is it on the server package or in WAR and how to code it?? thanks
It doens't really matter where you place the file. It must be on the server and a php script must be able to open the file. You can ready the text file the following way with php.
Then make a http request with GWT to that file.
Read the file:
<?php
// get contents of a file into a string
$filename = "/usr/local/something.txt";
$handle = fopen($filename, "r");
$contents = fread($handle, filesize($filename));
fclose($handle);
echo $contents;
?>
Make http request:
public class GetExample implements EntryPoint {
public static final int STATUS_CODE_OK = 200;
public static void doGet(String url) {
RequestBuilder builder = new RequestBuilder(RequestBuilder.GET, url);
try {
Request response = builder.sendRequest(null, new RequestCallback() {
public void onError(Request request, Throwable exception) {
// Code omitted for clarity
}
public void onResponseReceived(Request request, Response response) {
String content = response.getText();
}
});
} catch (RequestException e) {
// Code omitted for clarity
}
}
public void onModuleLoad() {
doGet("/");
}
}