How to create a file for serialization?

Question

I need to write a method to take an array of objects and serialize them into a file. My question is how do I check to see if the file is there and if the file is not, it will create a file then serialize the objects to it?

@smacaz take a look at [this](https://meta.stackexchange.com/a/61072) ;) — Cardinal System, Feb 23 '18 at 21:29

score 2 · Answer 1 · edited Feb 23 '18 at 21:38

2

As long as the object is serializable, and you have Apache Commons OI and Apache Commons Lang, you can use SerializationUtils and FileUtils:

File file = new File(<file>)
if(!file.exists()){
    FileUtils.writeByteArrayToFile (file, SerializationUtils.serialize(<obj>));
}

edited Feb 23 '18 at 21:38

Cardinal System

2,749
3
21
42

answered Feb 23 '18 at 21:23

Sashi

1,977
16
15

score 1 · Answer 2 · answered Feb 23 '18 at 21:18

1

Use File#exists and File#createNewFile:

File file = new File("C:/...");

if(file.exists()){
   // ...
} else {
   file.createNewFile();
   // ...
}

answered Feb 23 '18 at 21:18

Cardinal System

2,749
3
21
42

Yes, so if this program is ran on another users console it can read it from the program and not from a path stored on the computer. Does that make sense? Thanks for the feedback! – Quinton Thompson Feb 23 '18 at 21:24
@QuintonThompson you can read a resource file using `Class#getResourceAsStream(String)`, although *creating* a resource file is a whole nother topic. See [Add files in jar during runtime](https://stackoverflow.com/questions/6763449/add-files-in-jar-during-runtime) – Cardinal System Feb 23 '18 at 21:27
1

@QuintonThompson So when you clearly say 'file' in your question you now mean 'not a file'. Please clarify. – user207421 Feb 23 '18 at 21:27
`File.createNewFile()` does not accomplish the stated objective. – user207421 Feb 23 '18 at 21:39

score 1 · Answer 3 · answered Feb 23 '18 at 21:20

Besides new File("path").exists(), that has already been pointed out there is Files.exists(path) that takes the Path object that you will need to construct. Which one you want to use depends on the way the file path arrives in your application (whichever is simpler is the correct solution for you).

score -1 · Answer 4 · answered Feb 23 '18 at 21:25

Using Java 8 File API

final Path path = Paths.get(URI.create("file:/C:/temp/test.txt"));
final ByteArrayOutputStream out = new ByteArrayOutputStream();
final ObjectOutputStream objOut = new ObjectOutputStream(out);
// add your objects here
objOut.writeObject(new Object());
Files.write(path, out.toByteArray(), StandardOpenOption.CREATE, StandardOpenOption.TRUNCATE_EXISTING);

How to create a file for serialization?

4 Answers4