Java multipartfile to path

Interface MultipartFile

The file contents are either stored in memory or temporarily on disk. In either case, the user is responsible for copying file contents to a session-level or persistent store as and if desired. The temporary storage will be cleared at the end of request processing.

Method Summary

Return whether the uploaded file is empty, that is, either no file has been chosen in the multipart form or the chosen file has no content.

Method Details

getName

getOriginalFilename

Return the original filename in the client’s filesystem. This may contain path information depending on the browser used, but it typically will not with any other than Opera. Note: Please keep in mind this filename is supplied by the client and should not be used blindly. In addition to not using the directory portion, the file name could also contain characters such as «..» and others that can be used maliciously. It is recommended to not use this filename directly. Preferably generate a unique one and save this one somewhere for reference, if necessary.

getContentType

isEmpty

Return whether the uploaded file is empty, that is, either no file has been chosen in the multipart form or the chosen file has no content.

getSize

getBytes

getInputStream

Return an InputStream to read the contents of the file from. The user is responsible for closing the returned stream.

getResource

Return a Resource representation of this MultipartFile. This can be used as input to the RestTemplate or the WebClient to expose content length and the filename along with the InputStream.

transferTo

Transfer the received file to the given destination file. This may either move the file in the filesystem, copy the file in the filesystem, or save memory-held contents to the destination file. If the destination file already exists, it will be deleted first. If the target file has been moved in the filesystem, this operation cannot be invoked again afterwards. Therefore, call this method just once in order to work with any storage mechanism. NOTE: Depending on the underlying provider, temporary storage may be container-dependent, including the base directory for relative destinations specified here (e.g. with Servlet multipart handling). For absolute destinations, the target file may get renamed/moved from its temporary location or newly copied, even if a temporary copy already exists.

transferTo

Transfer the received file to the given destination file. The default implementation simply copies the file input stream.

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How to convert a multipart file to file in Java?

When working with file uploads in Java, it is common to receive the uploaded file as a MultipartFile object from a framework such as Spring. However, sometimes it is necessary to convert this MultipartFile to a standard java.io.File object for further processing or storage. In this article, we will explore different methods to convert a MultipartFile to a File in Java.

Method 1: Using FileOutputStream and File

Converting a Multipart File to File in Java using FileOutputStream and File

Here are the steps to convert a multipart file to a File in Java using FileOutputStream and File:

File convertedFile = new File("path/to/converted/file");

1. Create a FileOutputStream object to write the contents of the multipart file to the converted file.

FileOutputStream fos = new FileOutputStream(convertedFile);

1. Use the getBytes() method of the MultipartFile object to retrieve the contents of the file as a byte array.

byte[] fileContent = multipartFile.getBytes();

Here is the complete code:

import org.springframework.web.multipart.MultipartFile; import java.io.File; import java.io.FileOutputStream; import java.io.IOException; public class MultipartFileToFileConverter  public static File convert(MultipartFile multipartFile) throws IOException  File convertedFile = new File("path/to/converted/file"); FileOutputStream fos = new FileOutputStream(convertedFile); byte[] fileContent = multipartFile.getBytes(); fos.write(fileContent); fos.close(); return convertedFile; > >

That’s it! You can now use this method to convert a MultipartFile to a File in Java using FileOutputStream and File.

Method 2: Using transferTo() method of MultipartFile

To convert a MultipartFile to a File in Java, you can use the transferTo() method of MultipartFile. Here are the steps:

1. Get the input stream from the MultipartFile object.
2. Create a new File object with the desired file name and location.
3. Use the transferTo() method to write the input stream to the File object.

import org.springframework.web.multipart.MultipartFile; import java.io.File; import java.io.IOException; public class MultipartFileToFileConverter  public static void convert(MultipartFile multipartFile, String filePath) throws IOException  File file = new File(filePath); multipartFile.transferTo(file); > >

In the above code, multipartFile is the MultipartFile object that you want to convert to a File, and filePath is the path and name of the target File object.

Note that the transferTo() method throws an IOException, so you need to handle it appropriately in your code.

That’s it! With these simple steps, you can easily convert a MultipartFile to a File using the transferTo() method.

Method 3: Using InputStream and FileOutputStream

To convert a multipart file to a File object in Java, you can use the InputStream and FileOutputStream classes. Here’s how you can do it:

InputStream inputStream = multipartFile.getInputStream();

File file = new File(multipartFile.getOriginalFilename());

FileOutputStream outputStream = new FileOutputStream(file);

byte[] buffer = new byte[1024]; int bytesRead; while ((bytesRead = inputStream.read(buffer)) != -1)  outputStream.write(buffer, 0, bytesRead); >

inputStream.close(); outputStream.close();

InputStream inputStream = multipartFile.getInputStream(); File file = new File(multipartFile.getOriginalFilename()); FileOutputStream outputStream = new FileOutputStream(file); byte[] buffer = new byte[1024]; int bytesRead; while ((bytesRead = inputStream.read(buffer)) != -1)  outputStream.write(buffer, 0, bytesRead); > inputStream.close(); outputStream.close();

This code will convert a MultipartFile object to a File object using InputStream and FileOutputStream .

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Converting MultipartFiles to Files in Java for Managing Large Files (Rephrased MSDTHOT)

The method I am using to convert MultipartFile to a File functions well for most files. However, for large files, an exception occurs despite increasing the heap size. If you require an object, you may utilize the provided method that returns a File object. Then, you can use the class to relocate the file to your desired directory after writing it out. My query is regarding how to extract the file path to upload it to Google Drive.

Java Convert MultipartFile to File for large files

The method I’m employing to transform MultipartFile to a File is as follows:

public File convert(MultipartFile file) throws IOException

The functionality is satisfactory, however, when dealing with sizable documents, an exception is thrown.

java.lang.OutOfMemoryError: Java heap space 

Despite increasing the heap size, the error persists.

Is there a possible solution to resolve this issue programmatically? Perhaps, dividing the multipart file into smaller segments while performing the conversion would be helpful, although I am uncertain about the coding process.

We would highly appreciate any assistance or recommendations that you may have to offer.

Is MultipartFile included in the Spring package org.springframework.web.multipart ? In that case, you have the option to perform.

 public File convert(MultipartFile file) throws IOException < File convFile = new File(file.getOriginalFilename()); convFile.createNewFile(); try(InputStream is = file.getInputStream()) < Files.copy(is, convFile.toPath()); >return convFile; > 

Utilize the provided code to split the file into segments and subsequently combine them.

try < file = new File(filePath.toString()); messageDigest = MessageDigest.getInstance("MD5"); checksum = getFileCheckSum(messageDigest, file); fileInputStream = new FileInputStream(file); int fileSize = (int) file.length(); fileChannel = fileInputStream.getChannel(); int numberOfChunks = (int) Math.ceil(fileChannel.size() / (double) chunkSize); int totalpacket = numberOfChunks; int totalFileSize = fileSize; int read = 0; while (numberOfChunks >0) < System.out.println("file is :" + file + "checksum is:" + checksum); fileSize -= read; if (numberOfChunks >1) < ByteBuffer bytebuffer = ByteBuffer.allocate(chunkSize); read = fileChannel.read(bytebuffer); bytebuffer.flip(); String content = new String(); if (bytebuffer.hasRemaining()) < content = new String(bytebuffer.array(), Charset.forName("UTF-8")); >bytebuffer.clear(); > else < String chunkData = new String(); ByteBuffer byteBuffer = ByteBuffer.allocate(fileSize); fileChannel.read(byteBuffer); chunkData = new String(byteBuffer.array(), Charset.forName("UTF-8")); byteBuffer.clear(); >numberOfChunks--; > > catch (IOException e) < log.info(e); >catch (NoSuchAlgorithmException nsae) < log.info(nsae); >finally < try < fileInputStream.close(); fileChannel.close(); >catch (IOException ioe) < log.info(ioe); >> 

public File convert(MultipartFile file) throws IOException

MultipartFile issue, unable to convert to File, 1 Answer. Sorted by: 3. getBytes () tries to load the whole byte array into memory which is causing your OOM what you need to do is stream the file and write it out. Try the following: private static Path convert (MultipartFile file) throws IOException < Path newFile = Paths.get (file.getOriginalFilename ()); try (InputStream is = file

MultipartFile issue, unable to convert to File

I am attempting to upload a file that exceeds 1 GB in size, and I am utilizing Spring Boot for this purpose.

I attempted using the code below, however, an error of «Out of Memory» occurred.

public void uploadFile(MultipartFile file) throws IOException < try < HttpHeaders headers = new HttpHeaders(); headers.setContentType(MediaType.APPLICATION_OCTET_STREAM); SimpleClientHttpRequestFactory requestFactory = new SimpleClientHttpRequestFactory(); requestFactory.setBufferRequestBody(false); restTemplate.setRequestFactory(requestFactory); String uploadFile= restTemplate.exchange(url, HttpMethod.POST, new HttpEntity<>(new FileSystemResource(convert(file)), headers), String.class).getBody(); > catch (Exception e) < throw new RuntimeException("Exception Occured", e); >> private static File convert(MultipartFile file) throws IOException

My primary challenge is the incapability to transform MultipartFile into a java.io.File.

Despite substituting FileSystemResource with ByteArrayResource , the OOM error persists even after my attempts.

I have also attempted to utilize the code below:

private static File convert(MultipartFile file) throws IOException

However, the above snippet is causing the following exception.

It is not possible to cast CommonsMultipartFile from org.springframework.web.multipart.commons to MultipartFile from org.springframework.web.multipart.

1. Could anyone please tell me on how to convert MultipartFile to java.io.File?
2. And also is there any other approach better than FileSystemResource bcoz I will have to create new file in server everytime before uploading. If file is more than 1GB, another 1 GB new file has to be created on server side, and has to manually delete that file again, which I personally didn’t like this approach.

To prevent your OOM , avoid loading the entire byte array into memory using getBytes() . Instead, stream the file and write it out.

private static Path convert(MultipartFile file) throws IOException < Path newFile = Paths.get(file.getOriginalFilename()); try(InputStream is = file.getInputStream(); OutputStream os = Files.newOutputStream(newFile))) < byte[] buffer = new byte[4096]; int read = 0; while((read = is.read(buffer)) >0) < os.write(buffer,0,read); >> return newFile; > 

I modified your approach to provide a Path output instead of File . This output is associated with the java.nio bundle, which is the recommended choice due to its enhanced optimization compared to java.io .

In case you require an item with the hashtag File , feel free to contact the identifier newFile.toFile() .

After obtaining the Path object, you can utilize the java.nio.file.Files class to relocate the file to a desired directory once it has been written out.

private static void relocateFile(Path original, Path newLoc) throws IOException < if(Files.isDirectory(newLoc)) < newLoc = newLoc.resolve(original.getFileName()); >Files.move(original, newLoc); > 

Java — How to convert a multipart file to File?, small correction on @PetrosTsialiamanis post , new File( multipart.getOriginalFilename()) this will create file in server location where sometime you will face write permission issues for the user, its not always possible to give write permission to every user who perform … Code sample) throws IOException Feedback

How can I extract file path from MultipartFile

How can I obtain the path from MultipartFile in order to extract file path and save it to Google Drive?

 //logger.info(""+uploadFile.getClass().getResource("").getFile().getPath()); //System.out.println(uploadFile.getClass().getResource("").getPath()); 

Attempted to change MultipartFile to File, but encountered compatibility issues between «import java.io.File» and «import com.google.api.services.drive.model.File;» due to extensive usage of the latter.

@Override public int uploadFile(HttpSession session, MultipartFile uploadFile) throws Exception < // TODO Auto-generated method stub File fileMetadata = new File(); Drive driveService = getCredential(session); fileMetadata.setName(uploadFile.getOriginalFilename()); logger.info(uploadFile.getResource().getFile().getAbsoluteFile().getPath()); //logger.info(""+uploadFile.getClass().getResource("").getFile().getPath()); //System.out.println(uploadFile.getClass().getResource("").getPath()); java.io.File filePath = new java.io.File("need to input file path here"); FileContent mediaContent = new FileContent("image/jpeg", filePath); File file = driveService.files().create(fileMetadata, mediaContent).setFields("id").execute(); System.out.println("File ID: " + file.getId()); return 0; >

An exception was thrown for the servlet [appServlet] in the context with an empty path while trying to access the [uploadFile] resource. java.io.FileNotFoundException: MultipartFile cannot be resolved to absolute file path

While it’s possible to obtain the name and size of the MultipartFile from the parameter, the question remains: how does one retrieve the path from it?

To upload a file, you do not require a file path. Simply utilize either multipartFile.getInputStream() or multipartFile.getBytes() . Generally, you cannot determine the path to a file when receiving it because this functionality is dependent on the browser.

Although it wasn’t possible to directly obtain the file path from MultipartFile, the getInputStream() property was accessible.

Utilize the provided code with the assumption that the input stream is derived from the variable MultPartFile and its corresponding .getInputStream() method.

 StringBuilder resultStringBuilder = new StringBuilder(); try (BufferedReader br = new BufferedReader(new InputStreamReader(**inputStream**))) < String line; while ((line = br.readLine()) != null) < resultStringBuilder.append(line).append("\n"); >> 

Once the value you desire is obtained, you can access it by reading the resultStringBuilder variable.

The approach I used was effective for me and I hope it can be of assistance to you.

Java — How to convert MultipartFile to UTF-8 always, What you did new InputStreamReader (csvFile.getInputStream (), StandardCharsets.UTF_8) tells the CSV parser that the content of the inputstream is UTF-8 encoded. Since UTF-8 is (usally) the standard encoding, this is actually the same as using new InputStreamReader (csvFile.getInputStream ()). If I get your …

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