In my Spring Boot application, I accept an audiofile as MultipartFile with #RequestParam. I know, I can convert the file into some InputStream. I am also able to convert it into some byte array:
#PostMapping("/microsoft")
void transcribe(#RequestParam("file") MultipartFile file) {
InputStream inputStream = new BufferedInputStream(file.getInputStream());
byte[] byteArr = file.getBytes();
AudioConfig audioConfig = AudioConfig.???; //here the correct method with my file needs to be called
}
For transcription using Microsoft API, I need to create some Audioconfig object. This is the link to the class.
I used in the past fromWavFileInput(...) when I loaded a local audio file. But now, I need to be able to use the MultipartFile. I have no idea which method of the AudioConfig class I can use and how to convert the file correctly.
The idea is, to create a temp file and transfer the MultipartFile into it:
File tempFile = File.createTempFile("prefix-", "-suffix");
file.transferTo(tempFile);
The use fromWavFileOutput with the path of the temporary file:
AudioConfig audioConfig = AudioConfig.fromWavFileOutput(tempFile.getPath());
For me, the temp file exceeds its maximum permitted size. To get this solved, make sure to add spring.servlet.multipart.max-file-size=-1 in your application.properties file to set limit to infinity.
Related
I have a file on my local disk, I want to convert that file to multipart file as I need to upload this file to another server. I am converting file to multipart file using MockMultipartFile, but it is not working as this package is only available in the test environment
I am trying to use CommonsMultipartFile instead of MockMultipartFile, but CommonsMultipartFile constructor needs an object of file item. I am not able to utilize that constructor
Using MockMultipartFile::
FileInputStream fileInputStream = new FileInputStream(unzippedFile);
return new MockMultipartFile(FILE_STRING, unzippedFile.getName(), MULTIPART_FORM_DATA_VALUE, IOUtils.toByteArray(fileInputStream))
Using CommonsMultipartFile::
DiskFileItem diskFileItem = new DiskFileItem(FILE_STRING, MULTIPART_FORM_DATA_VALUE, false, unzippedFile.getName(), (int) unzippedFile.length(), unzippedFile.getParentFile());
diskFileItem.getOutputStream();
MultipartFile multipartFile = new CommonsMultipartFile(diskFileItem);
I expect CommonsMultipartFile constructor to take diskFileItem, but it is giving an error
I have a database where the user doesn't has access to.
Still I can go to the database and "read" the documents with for example
var db:NotesDatabase = sessionAsSigner.getDatabase("","somedir/some.nsf");
In this database there's a pdf file I would like to open or download. I have the filename and the unid . If the user had acces to the database I could do it with
http(s)://[yourserver]/[application.nsf] /xsp/.ibmmodres/domino/OpenAttachment/ [application.nsf]/[UNID|/$File/[AttachmentName]?Open
How can I do it with sessionAsSigner without putting a $PublicAccess=1 field on the form ?
edit:
the pdf file is stored as attachment in a richtextfield
second edit
I'm trying to use the XSnippet from Naveen and made some changes
The error message I get is : 'OutStream' not found
The code I tried is :
response.reset();
response.setContentType("application/pdf");
response.setHeader("Content-Disposition", "inline; filename=" + zipFileName);
var embeddedObj:NotesEmbeddedObject = null;
var bufferInStream:java.io.BufferedInputStream = null;
var outStream:java.io.OutputStream = response.getOutputStream();
embeddedObj = downloadDocument.getAttachment(fileName);
if (embeddedObj != null) {
bufferInStream = new java.io.BufferedInputStream(embeddedObj.getInputStream());
var bufferLength = bufferInStream.available();
var data = new byte[bufferLength];
bufferInStream.read(data, 0, bufferLength); // Read the attachment data
ON THE NEXT LINE IS THE PROBLEM
OutStream.write(data); // Write attachment into pdf
bufferInStream.close();
embeddedObj.recycle();
}
downloadDocument.recycle();
outStream.flush();
outStream.close();
facesContext.responseComplete();
Create an XAgent (= XPage without rendering) which takes datebase + documentid + filename as URL parameters and delivers the file as response OutputStream.
The URL would be
http(s)://[yourserver]/download.nsf/download.xsp?db=[application.nsf]&unid=[UNID]&attname=[AttachmentName]
for an XAgent download.xsp in a database download.nsf.
The code behind the XAgent runs as sessionAsSigner and is able to read the file even the user itself has no right to access file's database.
Use Eric's blog (+ Java code) as a starting point. Replace "application/json" with "application/pdf" and stream pdf file instead of json data.
As an alternative you can adapt this XSnippet code from Thomas Adrian. Use download() together with grabFile() to write your pdf-File to OutputStream.
Instead of extracting attachment file to path and reading it from there you can stream the attachment right from document to response's OutputStream. Here is an XSnippet from Naveen Maurya as a good example.
If you can get the PDF file as a stream, you should be able to use the OutputStream of the external context's response.
Stephan Wissel has a blog posting about writing out an ODF file so you should be able to cut that up as a starting point.
http://www.wissel.net/blog/d6plinks/SHWL-8248MT
You already have the db so, you will just need to know the UNID of the document.
var doc = db.getDocumentByUNID(unid) 'unid is a supplied param
var itm:RichTextItem = doc.getFirstItem("Body") 'assuming file is in body field
Once you have the itm, you can loop round all of the embeddedObjects and get the pdf file. At this point, I don't know if you can stream it directly or if you have to detach it, but assuming you detach it, you will then use something like this.
File file = new File("path to file");
FileInputStream fileIn = new FileInputStream(file);
Don't forget to clean up the temporarily detached file
I have a file called data.dat that I would like to load into a byte[] in Vaadin. It is a data file that I need to load and then manipulate based on the user's input.
I tried:
String basepath = VaadinService.getCurrent().getBaseDirectory().getAbsolutePath();
FileResource resource = new FileResource(new File(basepath + MY_DATA_FILE));
The problem is that I don't know how to manipulate the resource to extract the byte[]. I see lots of information for how to put images and so on to the UI components, how to stream a user generated PDF, and so on, but I can't seem to find any examples on how to load your own data file that you then manipulate within code...
From the docs of vaadin you have:
java.io.File getSourceFile()
Gets the source file.
or
DownloadStream getStream()
Gets resource as stream.
And in DownloadStream:
java.io.InputStream getStream()
Gets downloadable stream.
So you can easily operate on File or InputStream to read the contents of a FileResource
For example in Java 8:
byte[] bytes = Files.readAllBytes(Paths.get(resource.getSourceFile().getAbsolutePath()));
I have a Java Spring MVC web application. From client, through AngularJS, I am uploading a file and posting it to Controller as webservice.
In my Controller, I am gettinfg it as MultipartFile and I can copy it to local machine.
But I want to upload the file to Amazone S3 bucket. So I have to convert it to java.io.File. Right now what I am doing is, I am copying it to local machine and then uploading to S3 using jets3t.
Here is my way of converting in controller
MultipartHttpServletRequest mRequest=(MultipartHttpServletRequest)request;
Iterator<String> itr=mRequest.getFileNames();
while(itr.hasNext()){
MultipartFile mFile=mRequest.getFile(itr.next());
String fileName=mFile.getOriginalFilename();
fileLoc="/home/mydocs/my-uploads/"+date+"_"+fileName; //date is String form of current date.
Then I am using FIleCopyUtils of SpringFramework
File newFile = new File(fileLoc);
// if the directory does not exist, create it
if (!newFile.getParentFile().exists()) {
newFile.getParentFile().mkdirs();
}
FileCopyUtils.copy(mFile.getBytes(), newFile);
So it will create a new file in the local machine. That file I am uplaoding in S3
S3Object fileObject = new S3Object(newFile);
s3Service.putObject("myBucket", fileObject);
It creates file in my local system. I don't want to create.
Without creating a file in local system, how to convert a MultipartFIle to java.io.File?
MultipartFile, by default, is already saved on your server as a file when user uploaded it.
From that point - you can do anything you want with this file.
There is a method that moves that temp file to any destination you want.
http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartFile.html#transferTo(java.io.File)
But MultipartFile is just API, you can implement any other MultipartResolver
http://docs.spring.io/spring/docs/3.0.x/api/org/springframework/web/multipart/MultipartResolver.html
This API accepts input stream and you can do anything you want with it. Default implementation (usually commons-multipart) saves it to temp dir as a file.
But other problem stays here - if S3 API accepts a file as a parameter - you cannot do anything with this - you need a real file. If you want to avoid creating files at all - create you own S3 API.
The question is already more than one year old, so I'm not sure if the jets35 link provided by the OP had the following snippet at that time.
If your data isn't a File or String you can use any input stream as a data source, but you must manually set the Content-Length.
// Create an object containing a greeting string as input stream data.
String greeting = "Hello World!";
S3Object helloWorldObject = new S3Object("HelloWorld2.txt");
ByteArrayInputStream greetingIS = new ByteArrayInputStream(greeting.getBytes());
helloWorldObject.setDataInputStream(greetingIS);
helloWorldObject.setContentLength(
greeting.getBytes(Constants.DEFAULT_ENCODING).length);
helloWorldObject.setContentType("text/plain");
s3Service.putObject(testBucket, helloWorldObject);
It turns out you don't have to create a local file first. As #Boris suggests you can feed the S3Object with the Data Input Stream, Content Type and Content Length you'll get from MultipartFile.getInputStream(), MultipartFile.getContentType() and MultipartFile.getSize() respectively.
Instead of copying it to your local machine, you can just do this and replace the file name with this:
File newFile = new File(multipartFile.getOriginalName());
This way, you don't have to have a local destination create your file
if you are try to use in httpentity check my answer here
https://stackoverflow.com/a/68022695/7532946
This question already has answers here:
How to read file from ZIP using InputStream?
(7 answers)
Closed 1 year ago.
How can I create new File (from java.io) in memory, not on the hard disk?
I am using the Java language. I don't want to save the file on the hard drive.
I'm faced with a bad API (java.util.jar.JarFile). It's expecting File file of String filename. I have no file (only byte[] content) and can create temporary file, but it's not beautiful solution. I need to validate the digest of a signed jar.
byte[] content = getContent();
File tempFile = File.createTempFile("tmp", ".tmp");
FileOutputStream fos = new FileOutputStream(tempFile);
fos.write(archiveContent);
JarFile jarFile = new JarFile(tempFile);
Manifest manifest = jarFile.getManifest();
Any examples of how to achieve getting manifest without creating a temporary file would be appreciated.
How can I create new File (from java.io) in memory , not in the hard disk?
Maybe you are confusing File and Stream:
A File is an abstract representation of file and directory pathnames. Using a File object, you can access the file metadata in a file system, and perform some operations on files on this filesystem, like delete or create the file. But the File class does not provide methods to read and write the file contents.
To read and write from a file, you are using a Stream object, like FileInputStream or FileOutputStream. These streams can be created from a File object and then be used to read from and write to the file.
You can create a stream based on a byte buffer which resides in memory, by using a ByteArrayInputStream and a ByteArrayOutputStream to read from and write to a byte buffer in a similar way you read and write from a file. The byte array contains the "File's" content. You do not need a File object then.
Both the File... and the ByteArray... streams inherit from java.io.OutputStream and java.io.InputStream, respectively, so that you can use the common superclass to hide whether you are reading from a file or from a byte array.
It is not possible to create a java.io.File that holds its content in (Java heap) memory *.
Instead, normally you would use a stream. To write to a stream, in memory, use:
OutputStream out = new ByteArrayOutputStream();
out.write(...);
But unfortunately, a stream can't be used as input for java.util.jar.JarFile, which as you mention can only use a File or a String containing the path to a valid JAR file. I believe using a temporary file like you currently do is the only option, unless you want to use a different API.
If you are okay using a different API, there is conveniently a class in the same package, named JarInputStream you can use. Simply wrap your archiveContent array in a ByteArrayInputStream, to read the contents of the JAR and extract the manifest:
try (JarInputStream stream = new JarInputStream(new ByteArrayInputStream(archiveContent))) {
Manifest manifest = stream.getManifest();
}
*) It's obviously possible to create a full file-system that resides in memory, like a RAM-disk, but that would still be "on disk" (and not in Java heap memory) as far as the Java process is concerned.
You could use an in-memory filesystem, such as Jimfs
Here's a usage example from their readme:
FileSystem fs = Jimfs.newFileSystem(Configuration.unix());
Path foo = fs.getPath("/foo");
Files.createDirectory(foo);
Path hello = foo.resolve("hello.txt"); // /foo/hello.txt
Files.write(hello, ImmutableList.of("hello world"), StandardCharsets.UTF_8);
I think temporary file can be another solution for that.
File tempFile = File.createTempFile(prefix, suffix, null);
FileOutputStream fos = new FileOutputStream(tempFile);
fos.write(byteArray);
There is a an answer about that here.