CODEWITHSUNDEEP
java
Ben Dilts
Ben Dilts

Reputation: 10745

Construct a java.io.File that reads from a byte array

I'm using an API that requires a java.io.File as input, but I'd like to just send the API a byte array. Is there a way to create a File object that will read from my byte array rather than from disk?

Upvotes: 4

Views: 3458

Answers (4)

Sam Stern
Sam Stern

Reputation: 25134

You could open a temporary file: open temp file in java

Then if take your byte[] and write it all to the temporary file through some OutputStream/Writer you should have what you need. Then you can pass the file without creating it on disk.

Upvotes: 2

brimborium
brimborium

Reputation: 9512

String strFilePath = "filePath";
byte[] writeBytes = ...;
try {
  FileOutputStream fos = new FileOutputStream(strFilePath);
  fos.write(byteArray);
  fos.close();
} catch(FileNotFoundException ex) {
  // handle FileNotFoundException here
} catch(IOException ioe) {
  // handle IOException here
}

EDIT: Usually I am not a fan of just giving code, but this whole streaming stuff used to confuse me quite a bit, so here is a solution ;).

EDIT2: Ok, I just realized, that this is exactly, what the OP did NOT want to do. So this answer is not really useful...

Upvotes: 0

Jack
Jack

Reputation: 133609

Java works with InputStream (or OutpusStream) approach. Using a java.io.File already enforces too much, then a File class is not anything on which you can read or write: it just represents an entry in a file system (from which you can then, but just then, obtain its associated input and output streams).

So you should have a method which accepts a generic InputStream. If the API just supports a File object then it's clearly poorly designed (or you are trying to use them in a non allowed way).

Upvotes: 0

Attila
Attila

Reputation: 28772

A File does not read anything. It represents a file (or directory) object in the Operating System (or virtual file, etc.). To read from it, you wil have to create Readers or Streams, which the API will do on its own. If the API does not provide methods to pass in a Reader or InputStream, you are out of luck.

Upvotes: 6

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