"multipart config was not present on Servlet" when trying to upload a file with JAX-RS class - file-upload

I have the following JAX-RS class to upload a file from a browser (implemented in Wildfly 14). Problem is I get the error multipart config was not present on Servlet. Since I annotated the class with #Consumes({ MediaType.MULTIPART_FORM_DATA }) I'm not sure what is missing. How to fix this problem?
#Produces({ MediaType.APPLICATION_JSON })
#Consumes({ MediaType.MULTIPART_FORM_DATA })
public class FileUploadService {
#Context
private HttpServletRequest request;
#POST
#Path("/upload")
public Response processUpload() throws IOException, ServletException {
String path = "/mypath";
for (Part part : request.getParts()) {
String fileName = getFileName(part);
String fullPath = path + File.separator + fileName;
// delete file if exists
java.nio.file.Path path2 = FileSystems.getDefault().getPath(fullPath);
Files.deleteIfExists(path2);
// get file input stream
InputStream fileContent = part.getInputStream();
byte[] buffer = new byte[fileContent.available()];
fileContent.read(buffer);
File targetFile = new File(fullPath);
// write output file
OutputStream outStream = new FileOutputStream(targetFile);
outStream.write(buffer);
outStream.close();
}
return Response.ok("OK").build();
}
private String getFileName(Part part) {
for (String content : part.getHeader("content-disposition").split(";")) {
if (content.trim().startsWith("filename"))
return content.substring(content.indexOf("=") + 2, content.length() - 1);
}
return "";
}
}

Related

Files uploaded but not appearing on server

I use the code stated here to upload files through a webapi http://bartwullems.blogspot.pe/2013/03/web-api-file-upload-set-filename.html. I also made the following api to list all the files I have :
[HttpPost]
[Route("sharepoint/imageBrowser/listFiles")]
[SharePointContextFilter]
public async Task<HttpResponseMessage> Read()
{
string pathImages = HttpContext.Current.Server.MapPath("~/Content/images");
DirectoryInfo d = new DirectoryInfo(pathImages);//Assuming Test is your Folder
FileInfo[] Files = d.GetFiles(); //Getting Text files
List<object> lst = new List<object>();
foreach (FileInfo f in Files)
{
lst.Add(new
{
name = f.Name,
type = "f",
size = f.Length
});
}
return Request.CreateResponse(HttpStatusCode.OK, lst);
}
When calling this api, all the files uploaded are listed. But when I go to azure I dont see any of them (Content.png is a file I manually uploaded to azure)
Why are the files listed if they dont appear on azure.
According to your description, I suggest you could firstly use azure kudu console to locate the right folder in the azure web portal to see the image file.
Open kudu console:
In the kudu click the debug console and locate the site\wwwroot\yourfilefolder
If you find your file is still doesn't upload successfully, I guess there maybe something wrong with your upload codes. I suggest you could try below codes.
Notice: You need add image folder in the wwwort folder.
{
public class UploadingController : ApiController
{
public async Task<HttpResponseMessage> PostFile()
{
// Check if the request contains multipart/form-data.
if (!Request.Content.IsMimeMultipartContent())
{
throw new HttpResponseException(HttpStatusCode.UnsupportedMediaType);
}
string root = Environment.GetEnvironmentVariable("HOME").ToString() + "\\site\\wwwroot\\images";
//string root = HttpContext.Current.Server.MapPath("~/images");
var provider = new FilenameMultipartFormDataStreamProvider(root);
try
{
StringBuilder sb = new StringBuilder(); // Holds the response body
// Read the form data and return an async task.
await Request.Content.ReadAsMultipartAsync(provider);
// This illustrates how to get the form data.
foreach (var key in provider.FormData.AllKeys)
{
foreach (var val in provider.FormData.GetValues(key))
{
sb.Append(string.Format("{0}: {1}\n", key, val));
}
}
// This illustrates how to get the file names for uploaded files.
foreach (var file in provider.FileData)
{
FileInfo fileInfo = new FileInfo(file.LocalFileName);
sb.Append(string.Format("Uploaded file: {0} ({1} bytes)\n", fileInfo.Name, fileInfo.Length));
}
return new HttpResponseMessage()
{
Content = new StringContent(sb.ToString())
};
}
catch (System.Exception e)
{
return Request.CreateErrorResponse(HttpStatusCode.InternalServerError, e);
}
}
}
public class FilenameMultipartFormDataStreamProvider : MultipartFormDataStreamProvider
{
public FilenameMultipartFormDataStreamProvider(string path) : base(path)
{
}
public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
{
var name = !string.IsNullOrWhiteSpace(headers.ContentDisposition.FileName) ? headers.ContentDisposition.FileName : Guid.NewGuid().ToString();
return name.Replace("\"", string.Empty);
}
}
}
Result:

ASP.NET WebApi file upload using guid and file extension

I currently am able to save a file being uploaded to a WebAPI controller, but I'd like to be able to save the file as a guid with the correct file name extension so it can be viewed correctly.
Code:
[ValidationFilter]
public HttpResponseMessage UploadFile([FromUri]string AdditionalInformation)
{
var task = this.Request.Content.ReadAsStreamAsync();
task.Wait();
using (var requestStream = task.Result)
{
try
{
// how can I get the file extension of the content and append this to the file path below?
using (var fileStream = File.Create(HttpContext.Current.Server.MapPath("~/" + Guid.NewGuid().ToString())))
{
requestStream.CopyTo(fileStream);
}
}
catch (IOException)
{
throw new HttpResponseException(HttpStatusCode.InternalServerError);
}
}
HttpResponseMessage response = new HttpResponseMessage();
response.StatusCode = HttpStatusCode.Created;
return response;
}
I can't seem to get a handle on the actual filename of the content. I thought headers.ContentDisposition.FileName might be a candidate but that doesn't seem to get populated.
Thanks for the comments above which pointed me in the right direction.
To clarify the final solution, I used a MultipartFormDataStreamProvider which streams the file automatically. The code is in another question I posted to a different problem here:
MultipartFormDataStreamProvider and preserving current HttpContext
My full provider code is listed below. The key to generating the guid file name is to override the GetLocalFileName function and use the headers.ContentDisposition property. The provider handles the streaming of the content to file.
public class MyFormDataStreamProvider : MultipartFormDataStreamProvider
{
public MyFormDataStreamProvider (string path)
: base(path)
{ }
public override Stream GetStream(HttpContent parent, HttpContentHeaders headers)
{
// restrict what images can be selected
var extensions = new[] { "png", "gif", "jpg" };
var filename = headers.ContentDisposition.FileName.Replace("\"", string.Empty);
if (filename.IndexOf('.') < 0)
return Stream.Null;
var extension = filename.Split('.').Last();
return extensions.Any(i => i.Equals(extension, StringComparison.InvariantCultureIgnoreCase))
? base.GetStream(parent, headers)
: Stream.Null;
}
public override string GetLocalFileName(System.Net.Http.Headers.HttpContentHeaders headers)
{
// override the filename which is stored by the provider (by default is bodypart_x)
string oldfileName = headers.ContentDisposition.FileName.Replace("\"", string.Empty);
string newFileName = Guid.NewGuid().ToString() + Path.GetExtension(oldfileName);
return newFileName;
}
}

com.sun.jersey.api.client.UniformInterfaceException (returned a response status of 400)

I am trying to set up file upload example using JAX RS. I could set up the project and successfully upload file in a server location. But i get the following error when file size is more than 10KB (weird!!)
com.sun.jersey.api.client.UniformInterfaceException: POST http://localhost:9090/DOAFileUploader/rest/file/upload returned a response status of 400
at com.sun.jersey.api.client.WebResource.handle(WebResource.java:607)
at com.sun.jersey.api.client.WebResource.access$200(WebResource.java:74)
at com.sun.jersey.api.client.WebResource$Builder.post(WebResource.java:507)
at com.sony.doa.rest.client.DOAClient.upload(DOAClient.java:75)
at com.sony.doa.rest.client.DOAMain.main(DOAMain.java:34)
I am new to JAX RS and i'm not sure what exactly the issue is. Do i need to set some parameters client side or server side (like size, timeout etc)?
This is the client side code calling webservice:
public void upload() {
File file = new File(inputFilePath);
FormDataMultiPart part = new FormDataMultiPart();
part.bodyPart(new FileDataBodyPart("file", file, MediaType.APPLICATION_OCTET_STREAM_TYPE));
WebResource resource = Client.create().resource(url);
String response = resource.type(MediaType.MULTIPART_FORM_DATA_TYPE).post(String.class, part);
System.out.println(response);
}
This is the server side code:
#Path("/file")
public class UploadFileService {
#POST
#Path("/upload")
#Consumes(MediaType.MULTIPART_FORM_DATA)
public Response uploadFile(
#FormDataParam("file") InputStream uploadedInputStream,
#FormDataParam("file") FormDataContentDisposition fileDetail) {
String uploadedFileLocation = "e://uploaded/"
+ fileDetail.getFileName();
writeToFile(uploadedInputStream, uploadedFileLocation);
String output = "File uploaded to : " + uploadedFileLocation;
return Response.status(200).entity(output).build();
}
private void writeToFile(InputStream uploadedInputStream,
String uploadedFileLocation) {
try {
OutputStream out = new FileOutputStream(new File(
uploadedFileLocation));
int read = 0;
byte[] bytes = new byte[16000];
out = new FileOutputStream(new File(uploadedFileLocation));
while ((read = uploadedInputStream.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
} catch (IOException e) {
e.printStackTrace();
} } }
Please let me know what settings i have to change for file sizes greater than 10KB?
Thanks!
I use org.apache.commons.fileupload.servlet.ServletFileUpload in a Jersey context, and it works fine., and yes, it set the max file size, sorry I missed this before.
here is a snipet of code I use (this is a multipart form, so there are other fields along with the file)
private LibraryUpload parseLibraryUpload(HttpServletRequest request) {
LibraryUpload libraryUpload;
File libraryZip = null;
String name = null;
String version = null;
ServletFileUpload upload = new ServletFileUpload();
upload.setFileSizeMax(MAX_FILE_SIZE);
FileItemIterator iter;
try {
iter = upload.getItemIterator(request);
while (iter.hasNext()) {
....
if (item.isFormField()) {
....
}else{
BufferedInputStream buffer = new BufferedInputStream(stream);
buffer.mark(MAX_FILE_SIZE);
libraryZip = File.createTempFile("fromUpload", null);
IOUtils.copy(buffer, new FileOutputStream(libraryZip));
...
}
I have encountered the same problem with Jersey. I have activated jersey trace but nothing help me.
I have changed the library by an apache Library and I see than the problem with linked to a repository for temporary files for tomcat. The repository was not exist. For files under 10k, the repository was not used.
So, after the repository creation, I used jersey library and all works fine.

Writing PDF content on response stream in openrasta

I want to render a pdf in Iframe. So if I do a GET request to http://localhost/pdf/2, it should return PDF content in the response stream. The other way can be redirecting user to full URL of the PDF file which I don't want to do.
Thanks in advance
you can use the InMemoryFile and InMemoryDownloadableFile classes.
An example:
private class AttachmentFile : InMemoryDownloadableFile
{
public AttachmentFile(byte[] file, string filename, string contenttype)
{
OpenStream().Write(file, 0, file.Length);
this.FileName = filename;
this.ContentType = new MediaType(contenttype);
}
}
private class InlineFile : InMemoryFile
{
public InlineFile(byte[] file, string filename, string contenttype)
{
OpenStream().Write(file, 0, file.Length);
this.FileName = filename;
this.ContentType = new MediaType(contenttype);
}
}
[HttpOperation(HttpMethod.GET)]
public object Get(string filename)
{
bool inline = false; //server attachments inline or as download
try
{
inline = Convert.ToBoolean(Params["INLINE"]);
}
catch { }
string contenttype = set contentttype...
byte[] attachment = read file....
if (inline)
return new InlineFile(attachment, filename, contenttype);
else return new AttachmentFile(attachment, filename, contenttype);
}
else return new OperationResult.Forbidden();
}

JavaFX: File upload to REST service / servlet fails because of missing boundary

I'm trying to upload a file using JavaFX using the HttpRequest. For this purpose I have written the following function.
function uploadFile(inputFile : File) : Void {
// check file
if (inputFile == null or not(inputFile.exists()) or inputFile.isDirectory()) {
return;
}
def httpRequest : HttpRequest = HttpRequest {
location: urlConverter.encodeURL("{serverUrl}");
source: new FileInputStream(inputFile)
method: HttpRequest.POST
headers: [
HttpHeader {
name: HttpHeader.CONTENT_TYPE
value: "multipart/form-data"
}
]
}
httpRequest.start();
}
On the server side, I am trying to handle the incoming data using the Apache Commons FileUpload API using a Jersey REST service. The code used to do this is a simple copy of the FileUpload tutorial on the Apache homepage.
#Path("Upload")
public class UploadService {
public static final String RC_OK = "OK";
public static final String RC_ERROR = "ERROR";
#POST
#Produces("text/plain")
public String handleFileUpload(#Context HttpServletRequest request) {
if (!ServletFileUpload.isMultipartContent(request)) {
return RC_ERROR;
}
FileItemFactory factory = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(factory);
List<FileItem> items = null;
try {
items = upload.parseRequest(request);
}
catch (FileUploadException e) {
e.printStackTrace();
return RC_ERROR;
}
...
}
}
However, I get a exception at items = upload.parseRequest(request);:
org.apache.commons.fileupload.FileUploadException: the request was rejected because no multipart boundary was found
I guess I have to add a manual boundary info to the InputStream. Is there any easy solution to do this? Or are there even other solutions?
Have you tried just using the InputStream from HttpServletRequest like so
InputStream is = httpRequest.getInputStream();
BufferedInputStream in = new BufferedInputStream(is);
//Write out bytes
out.close();
is.close();