CODEWITHSUNDEEP
javaandroidjava-io
tzj
tzj

Reputation: 93

Error when converting file to byte array. java.ioFileNotFoundException

I am trying to make user pick an image from gallery, display it in imageView, and at the same time, create a byte array of the same file.

After debugging:

'new File(filepath)' executed no problem, with a valid filesystem path.

but it always skips 'ByteArrayOutputStream baos = new ByteArrayOutputStream();' to the IOException with a java.ioFileNotFoundException

any idea what im doing wrong?

or is there a more efficient way i can just convert a byte of a file to a HEX character, and send each HEX char until the last byte of the file?

try {
                    File myFile = new File(imageUri.getPath());
                    String filepath = myFile.getAbsolutePath();
                    Log.d("onActivityResult", "filepath: " + filepath);
                    FileOutputStream fos = new FileOutputStream ( new File(filepath) );
                    ByteArrayOutputStream baos = new ByteArrayOutputStream();

                    // Put data in your baos

                    baos.writeTo(fos);
                } catch(IOException ioe) {
                    // Handle exception here
                    ioe.printStackTrace();
                    Toast.makeText(this, "Byte buffer error.", Toast.LENGTH_LONG).show();
                }

Upvotes: 0

Views: 542

Answers (1)

CommonsWare
CommonsWare

Reputation: 1006514

any idea what im doing wrong?

You think that a Uri always points to a file.

If the scheme of the Uri is file, then getPath() will be a filesystem path. Depending on how you got the Uri, that filesystem path may be usable.

Most of the time, the Uri will have a different scheme, typically content. For such a Uri, getPath() is meaningless, just as getPath() is meaningless for a Uri like https://stackoverflow.com/questions/46457384/error-when-converting-file-to-byte-array-java-iofilenotfoundexception.

For a Uri with a scheme of content (or file, or android.resource), use a ContentResolver and openInputStream() to get an InputStream on the content identified by the Uri.

Even if you fix that, you will crash most of the time with an OutOfMemoryError, as you will not have a single contiguous block of free heap space to load the entire content that the user chose.

Upvotes: 2

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