Create file in memory java

write file in memory with java.nio?

With nio it is possible to map an existing file in memory. But is it possible to create it only in memory without file on the hard drive ? I want to mimic the CreateFileMapping windows functions which allow you to write in memory. Is there an equivalent system in Java ? The goal is to write in memory in order for another program ( c ) to read it.

You can tell Windows to create a memory mapping of the pagefile rather than a named file. Is that what you’re looking for?

2 Answers 2

Have a look at the following. A file is created but this might be as close as your going to get.

MappedByteBuffer MappedByteBuffer.load() FileChannel FileChannel.map() 

Here is a snippet to try and get you started.

 filePipe = new File(tempDirectory, namedPipe.getName() + ".pipe"); try < int pipeSize = 4096; randomAccessFile = new RandomAccessFile(filePipe, "rw"); fileChannel = randomAccessFile.getChannel(); mappedByteBuffer = fileChannel.map(FileChannel.MapMode.READ_WRITE, 0, pipeSize); mappedByteBuffer.load(); >catch (Exception e) < . 

This work and indeed, everything is in RAM (i check with some utilities, no hdd i/o) But, do you know how to retrieve the mapped file with c code ? using window function ?

Actually, after research, you can't write in RAM with java, you need to do it through the system api. This thing for exemple: msdn.microsoft.com/en-us/library/windows/desktop/aa366537 Java just let you to put a file in RAM in order to modify it faster, that's all.

Most libraries in Java deal with input and output streams as opposed to java.io.File objects.
Examples: image reading, XML, audio, zip

Where possible, when dealing with I/O, use streams.

This may not be what you want, however, if you need random access to the data.

When using memory mapped files, and you get a MappedByteBuffer from a FileChannel using FileChannel.map(), if you don't need a file just use a ByteBuffer instead, which exists totally in memory. Create one of these using ByteBuffer.allocate() or ByteBuffer.allocateDirect().

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How to create a new java.io.file in memory?

Java's java.io.File class provides a representation of a file or directory in the file system. Sometimes, it may be desirable to create a new File instance in memory instead of on the file system. This can be useful for cases where the file needs to be processed temporarily without being written to disk, or for testing purposes.

Method 1: Using a ByteArrayOutputStream

Here's a step-by-step guide on how to create a new java.io.File in memory using a ByteArrayOutputStream :

ByteArrayOutputStream outputStream = new ByteArrayOutputStream();

outputStream.write("This is the content of the file".getBytes());

byte[] byteArray = outputStream.toByteArray();

File file = new File("filename.txt"); FileOutputStream fileOutputStream = new FileOutputStream(file); fileOutputStream.write(byteArray); fileOutputStream.close();

This will create a new file named "filename.txt" in the current directory with the contents "This is the content of the file".

Note that the FileOutputStream is used to write the byte array to the file. You can also use other output streams such as BufferedOutputStream or DataOutputStream to write the byte array to the file.

Method 2: Using a temporary file with a custom prefix and suffix

To create a new java.io.File in memory using a temporary file with a custom prefix and suffix, you can use the File.createTempFile() method. This method creates a new temporary file in the default temporary-file directory with a generated name. You can specify a prefix and suffix for the file name using this method.

Here's an example code snippet:

try  // create a temporary file with a custom prefix and suffix File tempFile = File.createTempFile("myPrefix", ".txt"); // write some data to the file FileWriter writer = new FileWriter(tempFile); writer.write("Hello, World!"); writer.close(); // read the data from the file FileReader reader = new FileReader(tempFile); char[] buffer = new char[1024]; int length = reader.read(buffer); System.out.println(new String(buffer, 0, length)); reader.close(); // delete the temporary file tempFile.delete(); > catch (IOException e)  e.printStackTrace(); >

In this example, we create a temporary file with a prefix of "myPrefix" and a suffix of ".txt". We then write the string "Hello, World!" to the file using a FileWriter . We read the data back from the file using a FileReader and print it to the console. Finally, we delete the temporary file using the delete() method.

Note that the createTempFile() method throws an IOException if it is unable to create the temporary file. It is important to handle this exception appropriately in your code.

Method 3: Using a virtual file system (VFS)

Creating a new java.io.File in memory can be achieved using a virtual file system (VFS). Here are the steps to create a new file in memory using VFS:

1. Add the VFS dependency to your project. For example, if you are using Maven, add the following dependency to your pom.xml file:

dependency> groupId>org.apache.commonsgroupId> artifactId>commons-vfs2artifactId> version>2.8.0version> dependency>

FileSystemManager fsManager = VFS.getManager();

FileSystemOptions opts = new FileSystemOptions(); opts.setRootURI("ram:///"); FileSystem ramFileSystem = fsManager.createFileSystem(opts);

FileObject file = fsManager.resolveFile(ramFileSystem.getRoot(), "newFile.txt"); file.createFile();

OutputStream os = file.getContent().getOutputStream(); String content = "This is the content of the file"; os.write(content.getBytes()); os.close();

InputStream is = file.getContent().getInputStream(); byte[] buffer = new byte[1024]; int bytesRead = is.read(buffer); while (bytesRead != -1)  System.out.println(new String(buffer, 0, bytesRead)); bytesRead = is.read(buffer); > is.close();

That's it! You have successfully created a new java.io.File in memory using a virtual file system (VFS).

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Create a Java File object (or equivalent) using a byte array in memory (without a physical file)

I want to create a Java File object in memory (without creating a physical file) and populate its content with a byte array. Can this be done? The idea is to pass it to a Spring InputStreamSource . I'm trying the method below, but it returns saying "the byte array does not contain a file name.".

MimeMessage message = mailSender.createMimeMessage(); MimeMessageHelper helper = new MimeMessageHelper(message); helper.setFrom("no-reply@example.com", "xyz"); helper.setTo(email); helper.setText(body,true); helper.setSubject(subject); helper.addInline("cImage", new InputStreamResource(new ByteArrayInputStream(imageByteArr))); mailSender.send(message); 

5 Answers 5

Can you paste the full stack trace? There is no such thing as an "in memory" file. Using a ByteArrayInputStream should be sufficient.

You need to implement Resource#getFilename(). Try the following:

helper.addInline("cImage", new ByteArrayResource(imageByteArr) < @Override public String getFilename() < return fileName; >>); 

java.lang.IllegalStateException: resource loaded from byte array does not carry a filename at org.springframework.core.io.AbstractResource.getFilename(AbstractResource.java:148) at org.springframework.mail.javamail.MimeMessageHelper.addInline(MimeMessageHelper.java:922)

Nope. I get - java.lang.IllegalStateException: abc.png does not carry a filename at org.springframework.core.io.AbstractResource.getFilename(AbstractResource.java:148) at org.springframework.mail.javamail.MimeMessageHelper.addInline(MimeMessageHelper.java:922)

Maybe its worth trying a different overload of the method:

addInline(String contentId, InputStreamSource inputStreamSource, String contentType) 

addInline("cImage", new InputStreamSource () < final private InputStream src = new ByteArrayInputStream(imageByteArr); public InputStream getInputStream() >, "image/jpeg"); // or whatever image type you use 

It is important to create the MimeMessageHelper object correctly to support attachments and inline resources.

Example: MimeMessageHelper helper = new MimeMessageHelper(message, true, "UTF-8");

In this example since multipart is set to true MULTIPART_MODE_MIXED_RELATED will be used and attachments and inline resouces will be supported.

Have you tried changing the resource you feed to addInline()? If you wanted the resource to be in memory, I would have tried a org.springframework.core.io.ByteArrayResource.

Update: I think you might need to use the DataSource version of the addInline() method and then use a byte array bound data source object to feed the data into the helper class. I would try the following:

MimeMessage message = mailSender.createMimeMessage(); MimeMessageHelper helper = new MimeMessageHelper(message); helper.setFrom("no-reply@example.com", "xyz"); helper.setTo(email); helper.setText(body,true); helper.setSubject(subject); // use javax.mail.util.ByteArrayDataSource ByteArrayDataSource imgDS = new ByteArrayDataSource( imageByteArr, "image/png"); helper.addInline("cImage", imgDS); mailSender.send(message); 

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Is it possible to create an in-memory File object in Java?

One of the optimization projects I'm working on right now makes extensive use of EPANet. We make repeated calls to two simulation methods in EPANet to understand how water flows through a water distribution network. HydraulicSim is one of the classes we make use of. See the overloaded simulate methods:

public void simulate(File hyd) throws ENException < . >public void simulate(OutputStream out) throws ENException, IOException < . >public void simulate(DataOutput out) throws ENException, IOException

public void simulate(File hydFile, File qualFile) throws IOException, ENException < . >void simulate(File hydFile, OutputStream out) throws IOException, ENException

1. Create two File objects, hydFile and qualFile .
2. Call HydraulicSim.simulate on hydFile .
3. Call QualitySim.simulate on hydFile and qualFile .
4. Delete the files.

The problem is that we have to do this a lot of times. For a large problem, we can do this hundreds of thousands or even millions of times. You can imagine the slowdown repeatedly creating/writing/deleting these files causes.

So my question is this: Is it possible for me to create these files so that they reside only in memory and never touch the disk? Each file is very small (I'm talking a few hundred bytes), so throwing them into memory won't be a problem; I just need to figure out how. I searched around and didn't really find much, save for MappedByteBuffer, but I'm not sure how, or if it's even possible, to create a File from that class.

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