I am using org.apache.commons.fileupload to upload.
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
boolean isMultipart = ServletFileUpload.isMultipartContent(request);
GWT.log("is multipart? " + Boolean.toString(isMultipart), null);
ServletFileUpload upload = new ServletFileUpload();
try{
FileItemIterator iter = upload.getItemIterator(request);
while (iter.hasNext()) {
FileItemStream item = iter.next();
String name = item.getFieldName();
InputStream stream = item.openStream() ;
/**
* Save th uploaded file
*/
}
}
catch(Exception e){
e.printStackTrace();
}
}
How can I save the uploaded file?
This looks like server-side code, which (aside from GWT-RPC servlets) is not specific to GWT at all. That GWT.log() is unnecessary -- replace it with a regular logging call, and handle the upload as you normally would in non-GWT code, since that's what it is.
Here is a helpful example of using apache's fileupload to get you started.
I think this can help you.
if (!item.isFormField()) {
String fieldName = item.getFieldName();
String fileName = item.getName();
String contentType = item.getContentType();
boolean isInMemory = item.isInMemory();
long sizeInBytes = item.getSize();
File saveTo = new File("/file_uploads/" + fileName);
try {
item.write(saveTo);
...
}
catch (Exception e){
...
}
Keep in mind that uploaded file may already be automatically saved by org.apache.commons.fileupload
You can set size threshold for file to be saved on disk or loaded in memory using
// Create a factory for disk-based file items
DiskFileItemFactory factory = new DiskFileItemFactory();
// Set factory constraints
factory.setSizeThreshold(yourMaxMemorySize);
factory.setRepository(yourTempDirectory);
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
// Set overall request size constraint
upload.setSizeMax(yourMaxRequestSize);
// Parse the request
List /* FileItem */ items = upload.parseRequest(request);
Everything you need to know about org.apache.commons.fileupload is here: Using FileUpload
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've seen this question here before, but none of the solutions work for me.
I have a SmartGWT app with Spring MVC. This all works great, and I have working RESTful web-services.
I have a form to upload not only the file, but also some meta data as well.
There is an associated DataSource with this form:
private final String DEFAULT_FILE_UPLOAD_SERVICE_PATH = "upload";
private final String TARGET = "uploadTarget";
public FileUploadForm()
{
setEncoding(Encoding.MULTIPART);
setMethod(FormMethod.POST);
setAutoFetchData(false);
setDataSource(fileUploadDS);
setTitleOrientation(TitleOrientation.TOP);
setNumCols(1);
setColWidths("*");
uploadFileIdItem.setRequired(true);
uploadFileIdItem.setDefaultValue(0);
uploadFileIdItem.setVisible(false);
uploadFileIdItem.setShowTitle(false);
// ==========================================================================
fileUploadTypeSelectItem.setShowTitle(false);
fileUploadTypeSelectItem.setName(Constants.FILE_UPLOAD_UPLOADTYPE);
fileUploadTypeSelectItem.setPickListWidth(TEXT_SIZE);
fileUploadTypeSelectItem.setTitle(Constants.TITLE_FILE_UPLOAD_UPLOADTYPE);
fileUploadTypeSelectItem.setOptionDataSource(fileUploadTypeDS);
fileUploadTypeSelectItem.setRequired(true);
fileUploadTypeSelectItem.setDisplayField(Constants.FILE_UPLOAD_UPLOADTYPE_NAME);
fileUploadTypeSelectItem.setValueField(Constants.FILE_UPLOAD_UPLOADTYPE_ID);
fileUploadTypeSelectItem.setDataPath("fileUploadType/fileUploadTypeId");
// ==========================================================================
setAction(GWT.getHostPageBaseURL() + "rest/" + DEFAULT_FILE_UPLOAD_SERVICE_PATH);
ButtonItem uploadButton = new ButtonItem("Upload");
uploadButton.addClickHandler(new com.smartgwt.client.widgets.form.fields.events.ClickHandler()
{
#Override
public void onClick(com.smartgwt.client.widgets.form.fields.events.ClickEvent event)
{
submitForm();
}
});
FileItem uploadItem = new FileItem(Constants.FILENAME);
uploadItem.setTitle(Constants.FILENAME);
setFields(uploadFileIdItem, fileUploadTypeSelectItem, uploadItem, uploadButton);
}
So, I don't know if I need to use:
setAction(GWT.getHostPageBaseURL() + "rest/" + DEFAULT_FILE_UPLOAD_SERVICE_PATH);
or
setAction(GWT.getHostPageBaseURL() + DEFAULT_FILE_UPLOAD_SERVICE_PATH);
or
setAction(GWT.getHostPageBaseURL() + DEFAULT_FILE_UPLOAD_SERVICE_PATH);
None of these seem to work, I submit my data to upload the filename, and I constantly get the HTTP 404 error.
I did not define anything extra special in the web.xml file for servlets.
Instead, the springmvc-servlet contains:
<context:component-scan base-package="com.myself.products.app.server.controller" />
And the servlet is actually defined like:
#SuppressWarnings("serial")
#Controller
#RequestMapping("/upload")
public class FileUploadServlet extends HttpServlet
{
private final Logger logger = LoggerFactory.getLogger(FileUploadServlet.class);
#Override
public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
this.process(request, response);
}
#Override
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
this.process(request, response);
}
private void process(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
// check that we have a file upload request
if (ServletFileUpload.isMultipartContent(request))
{
processFiles(request, response);
}
}
private File tmpDir;
private static final String DESTINATION_DIR_PATH = "/files/upload";
private File destinationDir;
public void init(ServletConfig config) throws ServletException
{
super.init(config);
tmpDir = new File(((File) getServletContext().getAttribute("javax.servlet.context.tempdir")).toString());
if (!tmpDir.isDirectory())
{
throw new ServletException(tmpDir.toString() + " is not a directory");
}
logger.debug("tmpDir: " + tmpDir.toString());
String realPath = getServletContext().getRealPath(DESTINATION_DIR_PATH);
destinationDir = new File(realPath);
if (!destinationDir.isDirectory())
{
throw new ServletException(DESTINATION_DIR_PATH + " is not a directory");
}
}
private void processFiles(HttpServletRequest request, HttpServletResponse response) throws ServletException,
IOException
{
// create a factory for disk-based file items
DiskFileItemFactory factory = new DiskFileItemFactory();
// set the size threshold, above which content will be stored on disk
factory.setSizeThreshold(1 * 1024 * 1024); // 1 MB
// set the temporary directory (this is where files that exceed the threshold will be stored)
factory.setRepository(tmpDir);
// create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
try
{
// parse the request
List<?> items = upload.parseRequest(request);
// process the uploaded items
Iterator<?> itr = items.iterator();
while (itr.hasNext())
{
FileItem item = (FileItem) itr.next();
// write the uploaded file to the application's file staging area
File file = new File(destinationDir, item.getName());
item.write(file);
}
}
catch (FileUploadException e)
{
logger.error("Error encountered while parsing the request", e);
}
catch (Exception e)
{
logger.error("Error encountered while uploading file", e);
}
}
You've seen this code before along this web-site, and several others.
I'd like to submit the file, AND data if possible, but if not, then how can I submit the form, and then metadata for it?
Any help would be much appreciated.
Simple File Upload GWT Example:
Available here:
http://www.gwtproject.org/javadoc/latest/com/google/gwt/user/client/ui/FileUpload.html
For sending Metadata along with request, need to set the hidden field to panel:
import com.google.gwt.user.client.ui.Hidden;
Hidden hidden = new Hidden();
hidden.setName("json");
hidden.setVisible(false);
hidden.setValue("simpleMetadata:testData");
panel.add(hidden);
I will suggest you to seperate saving metadata from uploding a file and have 2 forms. This is what I'm doing and it is working for me:
uploadForm.setAction(GWT.getHostPageBaseURL() + "importServiceName");
uploadForm.setEncoding(Encoding.MULTIPART);
uploadForm.setTarget(TARGET);
uploadForm.setMethod(FormMethod.POST);
fileItem = new UploadItem("file");
fileItem.setTitle("File");
fileItem.setWidth(300);
NamedFrame frame = new NamedFrame(TARGET);
frame.setWidth("1");
frame.setHeight("1");
frame.setVisible(false);
uploadForm.setItems(fileItem);
I'm using NamedFrame to be able to fetch servlet response in gwt code, but this is different story. I'm defining servler manually in web.xml
I have a rest service written to receive a file and save it.
The problem is that when I receive more than 2 requests, the files are not written only the last request is taken into consideration and written.
Here is my code:
#POST
#RequestMapping(value = "/media/{mediaName}/{mediaType}")
#Produces(MediaType.APPLICATION_OCTET_STREAM)
#ResponseBody
public String updateResourceLocally(#FormDataParam("rawData") InputStream rawData, #PathVariable("mediaName") String mediaName, #PathVariable("mediaType") String mediaType) {
logger.info("Entering updateResourceLocally for " + jobId + "; for media type: " + mediaType);
final String storeDir = "/tmp/test/" + mediaName + ("/");
final String finalExtension = mediaType;
final InputStream finalRawData = rawData;
// new Thread(new Runnable() {
// public void run() {
// writeToFile(finalRawData, storeDir, finalExtension);
// }
// }).start();
writeToFile(finalRawData, storeDir, finalExtension);
// int poolSize = 100;
// ExecutorService executor = Executors.newFixedThreadPool(poolSize);
// executor.execute(new Runnable() {
// #Override
// public void run() {
// writeToFile(rawData, storeDir, finalExtension);
// }
// });
logger.info("File uploaded to : " + storeDir);
return "Success 200";
}
I tried to put the writeToFile into threads, but still no success. Here is what writeToFile does
public synchronized void writeToFile(InputStream rawData,
String uploadedFileLocation, String extension) {
StringBuilder finalFileName = null;
String currentIncrement = "";
String fileName = "raw";
try {
File file = new File(uploadedFileLocation);
if (!file.exists()) {
file.mkdirs();
}
while (true) {
finalFileName = new StringBuilder(fileName);
if (!currentIncrement.equals("")) {
finalFileName.append("_").append(currentIncrement).append(extension);
}
File f = new File(uploadedFileLocation + finalFileName);
if (f.exists()) {
if (currentIncrement.equals("")) {
currentIncrement = "1";
} else {
currentIncrement = (Integer.parseInt(currentIncrement) + 1) + "";
}
} else {
break;
}
}
int read = 0;
byte[] bytes = new byte[1024];
OutputStream out = new FileOutputStream(new File(uploadedFileLocation + finalFileName));
while ((read = rawData.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
} catch (IOException e) {
throw new RuntimeException(e.getMessage());
}
}
The writeToFile creates a folder and writes a file, if the file already exists, it appends 1 and then increments the 1 accordingly and writes the file, so I would get raw.zip, raw-1.zip, etc.
I think the inputstream bytes are being lost, am I correct in my assumption?
NOTE: I do not have a UI client, I am using Poster a Firefox extension.
Update: What I am trying to achieve here is very simple
I receive number of requests with files attached
I need to save them. If the mediaName and mediaType are the same, then I need to append something to the filename and save it in the same location
If they are different I do not have a problem
The problem I am facing with the current code is that, when I post multiple time to the same URL, I have file-names created according to what I want, but the file content is not right, they vary depending on when the request came in and only the last POST's request is written properly.
Eg. I have a zip file of size 250MB, when I post 5 time, the 1st four will have random sizes and the 5th will have the complete 250MB, but the previous four should also have the same content.
You must separate the stream copy from the free filename assignation. The stream copy must be done within the calling thread (jersey service). Only the file naming operation must be common to all threads/requests.
Here is your code with a little refactoring :
getNextFilename
This file naming operation must be synchronized to guarantee each call gives a free name. This functions creates an empty file to guarantee the next call to work, because the function relies on file.exists().
public synchronized File getNextFilename(String uploadedFileLocation, String extension)
throws IOException
{
// This function MUST be synchronized to guarantee unicity of files names
// Synchronized functions must be the shortest possible to avoid threads waiting each other.
// No long job such as copying streams here !
String fileName = "raw";
//Create directories (if not already existing)
File dir = new File(uploadedFileLocation);
if (!dir.exists())
dir.mkdirs();
//Search for next free filename (raw.<extension>, else raw_<increment>.<extension>)
int currentIncrement = 0;
String finalFileName = fileName + "." + extension;
File f = new File(uploadedFileLocation + finalFileName);
while (f.exists())
{
currentIncrement++;
finalFileName = fileName + "_" + currentIncrement + "." + extension;
f = new File(uploadedFileLocation + finalFileName);
}
//Creates the file with size 0 in order to physically reserve the file "raw_<n>.extension",
//so the next call to getNextFilename will find it (f.exists) and will return "raw_<n+1>.extension"
f.createNewFile();
//The file exists, let the caller fill it...
return f;
}
writeToFile
Must not be synchronized !
public void writeToFile(InputStream rawData, String uploadedFileLocation, String extension)
throws IOException
{
//(1) Gets next available filename (creates the file with 0 size)
File file = getNextFilename(uploadedFileLocation, extension);
//(2) Copies data from inputStream to file
int read = 0;
byte[] bytes = new byte[1024];
OutputStream out = new FileOutputStream(file);
while ((read = rawData.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
}
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 .
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) {
}
}
}