从<条码>Uri到<条码>。 我做以下事情:

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();
    }

Kotlin在此非常简明扼要:

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

Kotlin为<代码>InputStream建立了方便的推广功能,例如buffered,use, 和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

错误案件:

• 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.

lin酸 Syn

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

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;
}

利用SerContentResolver(). openInputStream(uri)从URI获得投入。 并读到投入流的数据,从该投入流中将数据按顺序排列。

1. 遵循以下守则

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();
        }

参考这一联系方式

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();
            }

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();
    }

利用以下方法在录音室制作bytesArray>。

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;
}

A solution that closes byte streams and takes care of your memory:

fun Uri.readBytes(context: Context): ByteArray? {
    var inputStream          : InputStream? = null
    var byteArrayOutputStream: ByteArrayOutputStream? = null

    try {
        inputStream           = context.contentResolver.openInputStream(this)
        byteArrayOutputStream = ByteArrayOutputStream()

        if (inputStream != null) {
            val buffer = ByteArray(1024)
            var bytesRead: Int

            while (inputStream.read(buffer).also { bytesRead = it } != -1) {
                byteArrayOutputStream.write(buffer, 0, bytesRead)
            }

            return byteArrayOutputStream.toByteArray()
        }
    } catch (e: IOException) {
        e.printStackTrace()
    } finally {
        inputStream?.close()
        byteArrayOutputStream?.close()
    }

    return null
}