Image Uri to bytesarray

Question

I currently have two activities. One for pulling the image from the SD card and one for Bluetooth connection.

I have utilized a Bundle to transfer the Uri of the image from activity 1.

Now what i wish to do is get that Uri in the Bluetooth activity to and convert it into a transmittable state via Byte Arrays i have seen some examples but i can't seem to get them to work for my code!!

Bundle goTobluetooth = getIntent().getExtras();
    test = goTobluetooth.getString("ImageUri");

is what i have to pull it across. What would be the next step?

score 119 · Accepted Answer · answered Apr 24 '12 at 11:37

119

From Uri to get byte[] I do the following things,

InputStream iStream =   getContentResolver().openInputStream(uri);
byte[] inputData = getBytes(iStream);

and the getBytes(InputStream) method is:

public byte[] getBytes(InputStream inputStream) throws IOException {
      ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
      int bufferSize = 1024;
      byte[] buffer = new byte[bufferSize];

      int len = 0;
      while ((len = inputStream.read(buffer)) != -1) {
        byteBuffer.write(buffer, 0, len);
      }
      return byteBuffer.toByteArray();
    }

answered Apr 24 '12 at 11:37

user370305

108,599
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getContentResolver().openInputStream(test); is receiving an error saying that it is not applicable for arguments for string!! In reference to my code above its states the the uri is in string form how do i change this to coincide with the code you have stated above! – user1314243 Apr 24 '12 at 12:05
test is a string variable you have to pass Uri. Make Uri from test and then pass it to method.. – user370305 Apr 24 '12 at 12:06
Still getting an error saying uri can't be resolved to variable!! – user1314243 Apr 24 '12 at 12:18
1

You have to write this line, InputStream iStream = getContentResolver().openInputStream(Uri.parse(test)); – user370305 Apr 24 '12 at 12:20
Actually instead of passing string Uri from activity 1 you have to get byte[] from it in that activity and then you have to just pass the byte[] using intent to activity 2. – user370305 Apr 24 '12 at 12:28
i just need to now get the byte array to work within my transmission phase now. It used to send txt messages need to change it to the array just created: Thread.sleep(1000); OutputStream outStream; outStream = BT_Socket.getOutputStream(); byte[] ByteArray = ; outStream.write(ByteArray); outStream.flush(); Thread.sleep(1500); – user1314243 Apr 24 '12 at 12:35
i just need to get the byte array into the transmission code i stated above. it keeps coming up with errors when i try and include it in byte[] ByteArray = inputdata; etc – user1314243 Apr 24 '12 at 12:44
If the above answer works for you then accept it as a correct answer and make a new question on your current problem. – user370305 Apr 24 '12 at 12:49
this can cause an OutOfMemory error and crash at `while ((len = inputStream.read(buffer)) != -1) ` if the image/video/file is too large. have a different solution? – CQM Feb 24 '15 at 16:22
3

`InputStream iStream =...` line gives me `FileNotFoundException` – mrid Oct 04 '17 at 13:12
@mrid just hit Alt+Enter and surround with try/catch. – DJTano Apr 10 '19 at 18:23

Peter F · Answer 2 · 2023-06-05T08:22:47.360

25

Kotlin is very concise here:

@Throws(IOException::class)
private fun readBytes(context: Context, uri: Uri): ByteArray? = 
    context.contentResolver.openInputStream(uri)?.use { it.buffered().readBytes() }

Kotlin has convenient extension functions for InputStream like buffered,use , and readBytes.

buffered decorates the input stream as BufferedInputStream
use handles closing the stream
readBytes does the main job of reading the stream and writing into a byte array

Error cases:

IOException can occur during the process (like in Java)
openInputStream can return null. If you call the method in Java you can easily oversee this. Think about how you want to handle this case.

edited Jun 05 '23 at 08:22

answered Jan 18 '19 at 09:16

Peter F

3,633
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2

I think it's better to do `context.contentResolver.openInputStream(uri)?.use { it.buffered().readBytes() }` instead, this way you can ensure the stream is closed even if the buffer fails somehow (although improbable). – rtsketo Jul 29 '22 at 06:47
@rtsketo I agree with your suggestion. I updated the answer accordingly. Thanks! – Peter F Jun 05 '23 at 08:24

score 8 · Answer 3 · answered Oct 15 '19 at 13:01

8

Syntax in kotlin

val inputData = contentResolver.openInputStream(uri)?.readBytes()

answered Oct 15 '19 at 13:01

Lúcio Aguiar

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1

This isn't completely correct, you need to always close `openInputStream(..)` after reading. You can use `use { }` for auto-closing it. Check Peter's [answer](https://stackoverflow.com/a/54250792/2115403). – rtsketo Jul 29 '22 at 06:30

score 5 · Answer 4 · answered Sep 18 '15 at 08:47

Java best practice: never forget to close every stream you open! This is my implementation:

/**
 * get bytes array from Uri.
 * 
 * @param context current context.
 * @param uri uri fo the file to read.
 * @return a bytes array.
 * @throws IOException
 */
public static byte[] getBytes(Context context, Uri uri) throws IOException {
    InputStream iStream = context.getContentResolver().openInputStream(uri);
    try {
        return getBytes(iStream);
    } finally {
        // close the stream
        try {
            iStream.close();
        } catch (IOException ignored) { /* do nothing */ }
    }
}



 /**
 * get bytes from input stream.
 *
 * @param inputStream inputStream.
 * @return byte array read from the inputStream.
 * @throws IOException
 */
public static byte[] getBytes(InputStream inputStream) throws IOException {

    byte[] bytesResult = null;
    ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
    int bufferSize = 1024;
    byte[] buffer = new byte[bufferSize];
    try {
        int len;
        while ((len = inputStream.read(buffer)) != -1) {
            byteBuffer.write(buffer, 0, len);
        }
        bytesResult = byteBuffer.toByteArray();
    } finally {
        // close the stream
        try{ byteBuffer.close(); } catch (IOException ignored){ /* do nothing */ }
    }
    return bytesResult;
}

score 1 · Answer 5 · answered Apr 24 '12 at 12:04

use getContentResolver().openInputStream(uri) to get an InputStream from a URI. and then read the data from inputstream convert the data into byte[] from that inputstream

Try with following code

public byte[] readBytes(Uri uri) throws IOException {
          // this dynamically extends to take the bytes you read
        InputStream inputStream = getContentResolver().openInputStream(uri);
          ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();

          // this is storage overwritten on each iteration with bytes
          int bufferSize = 1024;
          byte[] buffer = new byte[bufferSize];

          // we need to know how may bytes were read to write them to the byteBuffer
          int len = 0;
          while ((len = inputStream.read(buffer)) != -1) {
            byteBuffer.write(buffer, 0, len);
          }

          // and then we can return your byte array.
          return byteBuffer.toByteArray();
        }

Refer this LINKs

Whats the difference between your answer and mine? – user370305 Apr 24 '12 at 12:08 — user370305, Apr 24 '12 at 12:08
The name of method :) – Yehor Hromadskyi Mar 24 '16 at 14:18 — Yehor Hromadskyi, Mar 24 '16 at 14:18

score 0 · Answer 6 · answered Apr 24 '12 at 12:09

This code works for me

Uri selectedImage = imageUri;
            getContentResolver().notifyChange(selectedImage, null);
            ImageView imageView = (ImageView) findViewById(R.id.imageView1);
            ContentResolver cr = getContentResolver();
            Bitmap bitmap;
            try {
                 bitmap = android.provider.MediaStore.Images.Media
                 .getBitmap(cr, selectedImage);

                imageView.setImageBitmap(bitmap);
                Toast.makeText(this, selectedImage.toString(),
                        Toast.LENGTH_LONG).show();
                finish();
            } catch (Exception e) {
                Toast.makeText(this, "Failed to load", Toast.LENGTH_SHORT)
                        .show();

                e.printStackTrace();
            }

score 0 · Answer 7 · edited Dec 13 '16 at 10:10

public void uriToByteArray(String uri)
    {

        ByteArrayOutputStream baos = new ByteArrayOutputStream();
        FileInputStream fis = null;
        try {
            fis = new FileInputStream(new File(uri));
        } catch (FileNotFoundException e) {
            e.printStackTrace();
        }

        byte[] buf = new byte[1024];
        int n;
        try {
            while (-1 != (n = fis.read(buf)))
                baos.write(buf, 0, n);
        } catch (IOException e) {
            e.printStackTrace();
        }
        byte[] bytes = baos.toByteArray();
    }

score 0 · Answer 8 · answered Nov 03 '21 at 18:19

Use the following method to create a bytesArray from a URI in Android studio.

public byte[] getBytesArrayFromURI(Uri uri) {
    try {
        InputStream inputStream = getContentResolver().openInputStream(uri);
        ByteArrayOutputStream byteBuffer = new ByteArrayOutputStream();
        int bufferSize = 1024;
        byte[] buffer = new byte[bufferSize];

        int len = 0;
        while ((len = inputStream.read(buffer)) != -1) {
            byteBuffer.write(buffer, 0, len);
        }

        return byteBuffer.toByteArray();

    }catch(Exception e) {
        Log.d("exception", "Oops! Something went wrong.");
    }
    return null;
}

Image Uri to bytesarray

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