如何将整个输入流读到字节数组?


当前回答

Kotlin中的解决方案(当然也可以在Java中工作),其中包括当你知道大小时的两种情况:

    fun InputStream.readBytesWithSize(size: Long): ByteArray? {
        return when {
            size < 0L -> this.readBytes()
            size == 0L -> ByteArray(0)
            size > Int.MAX_VALUE -> null
            else -> {
                val sizeInt = size.toInt()
                val result = ByteArray(sizeInt)
                readBytesIntoByteArray(result, sizeInt)
                result
            }
        }
    }

    fun InputStream.readBytesIntoByteArray(byteArray: ByteArray,bytesToRead:Int=byteArray.size) {
        var offset = 0
        while (true) {
            val read = this.read(byteArray, offset, bytesToRead - offset)
            if (read == -1)
                break
            offset += read
            if (offset >= bytesToRead)
                break
        }
    }

如果您知道大小,那么与其他解决方案相比,它可以节省两倍的内存(在很短的时间内,但仍然可能有用)。这是因为您必须将整个流读到末尾,然后将其转换为字节数组(类似于将数组转换为数组的ArrayList)。

例如,如果你在Android上,你有一些Uri要处理,你可以尝试用这个来获取大小:

    fun getStreamLengthFromUri(context: Context, uri: Uri): Long {
        context.contentResolver.query(uri, arrayOf(MediaStore.MediaColumns.SIZE), null, null, null)?.use {
            if (!it.moveToNext())
                return@use
            val fileSize = it.getLong(it.getColumnIndex(MediaStore.MediaColumns.SIZE))
            if (fileSize > 0)
                return fileSize
        }
        //if you wish, you can also get the file-path from the uri here, and then try to get its size, using this: https://stackoverflow.com/a/61835665/878126
        FileUtilEx.getFilePathFromUri(context, uri, false)?.use {
            val file = it.file
            val fileSize = file.length()
            if (fileSize > 0)
                return fileSize
        }
        context.contentResolver.openInputStream(uri)?.use { inputStream ->
            if (inputStream is FileInputStream)
                return inputStream.channel.size()
            else {
                var bytesCount = 0L
                while (true) {
                    val available = inputStream.available()
                    if (available == 0)
                        break
                    val skip = inputStream.skip(available.toLong())
                    if (skip < 0)
                        break
                    bytesCount += skip
                }
                if (bytesCount > 0L)
                    return bytesCount
            }
        }
        return -1L
    }

其他回答

使用普通Java的DataInputStream和它的readFully方法(至少从Java 1.4开始存在):

...
byte[] bytes = new byte[(int) file.length()];
DataInputStream dis = new DataInputStream(new FileInputStream(file));
dis.readFully(bytes);
...

这个方法还有其他一些形式,但在这个用例中,我一直使用这个方法。

请参阅InputStream.available()文档:

It is particularly important to realize that you must not use this method to size a container and assume that you can read the entirety of the stream without needing to resize the container. Such callers should probably write everything they read to a ByteArrayOutputStream and convert that to a byte array. Alternatively, if you're reading from a file, File.length returns the current length of the file (though assuming the file's length can't change may be incorrect, reading a file is inherently racy).

这对我很有用,

if(inputStream != null){
                ByteArrayOutputStream contentStream = readSourceContent(inputStream);
                String stringContent = contentStream.toString();
                byte[] byteArr = encodeString(stringContent);
            }

readSourceContent ()

public static ByteArrayOutputStream readSourceContent(InputStream inputStream) throws IOException {
        ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
        int nextChar;
        try {
            while ((nextChar = inputStream.read()) != -1) {
                outputStream.write(nextChar);
            }
            outputStream.flush();
        } catch (IOException e) {
            throw new IOException("Exception occurred while reading content", e);
        }

        return outputStream;
    }

encodeString()

public static byte[] encodeString(String content) throws UnsupportedEncodingException {
        byte[] bytes;
        try {
            bytes = content.getBytes();

        } catch (UnsupportedEncodingException e) {
            String msg = ENCODING + " is unsupported encoding type";
            log.error(msg,e);
            throw new UnsupportedEncodingException(msg, e);
        }
        return bytes;
    }

Kotlin中的解决方案(当然也可以在Java中工作),其中包括当你知道大小时的两种情况:

    fun InputStream.readBytesWithSize(size: Long): ByteArray? {
        return when {
            size < 0L -> this.readBytes()
            size == 0L -> ByteArray(0)
            size > Int.MAX_VALUE -> null
            else -> {
                val sizeInt = size.toInt()
                val result = ByteArray(sizeInt)
                readBytesIntoByteArray(result, sizeInt)
                result
            }
        }
    }

    fun InputStream.readBytesIntoByteArray(byteArray: ByteArray,bytesToRead:Int=byteArray.size) {
        var offset = 0
        while (true) {
            val read = this.read(byteArray, offset, bytesToRead - offset)
            if (read == -1)
                break
            offset += read
            if (offset >= bytesToRead)
                break
        }
    }

如果您知道大小,那么与其他解决方案相比,它可以节省两倍的内存(在很短的时间内,但仍然可能有用)。这是因为您必须将整个流读到末尾,然后将其转换为字节数组(类似于将数组转换为数组的ArrayList)。

例如,如果你在Android上,你有一些Uri要处理,你可以尝试用这个来获取大小:

    fun getStreamLengthFromUri(context: Context, uri: Uri): Long {
        context.contentResolver.query(uri, arrayOf(MediaStore.MediaColumns.SIZE), null, null, null)?.use {
            if (!it.moveToNext())
                return@use
            val fileSize = it.getLong(it.getColumnIndex(MediaStore.MediaColumns.SIZE))
            if (fileSize > 0)
                return fileSize
        }
        //if you wish, you can also get the file-path from the uri here, and then try to get its size, using this: https://stackoverflow.com/a/61835665/878126
        FileUtilEx.getFilePathFromUri(context, uri, false)?.use {
            val file = it.file
            val fileSize = file.length()
            if (fileSize > 0)
                return fileSize
        }
        context.contentResolver.openInputStream(uri)?.use { inputStream ->
            if (inputStream is FileInputStream)
                return inputStream.channel.size()
            else {
                var bytesCount = 0L
                while (true) {
                    val available = inputStream.available()
                    if (available == 0)
                        break
                    val skip = inputStream.skip(available.toLong())
                    if (skip < 0)
                        break
                    bytesCount += skip
                }
                if (bytesCount > 0L)
                    return bytesCount
            }
        }
        return -1L
    }

我知道已经太迟了,但我认为这里有更清晰的解决方案,更易于阅读……

/**
 * method converts {@link InputStream} Object into byte[] array.
 * 
 * @param stream the {@link InputStream} Object.
 * @return the byte[] array representation of received {@link InputStream} Object.
 * @throws IOException if an error occurs.
 */
public static byte[] streamToByteArray(InputStream stream) throws IOException {

    byte[] buffer = new byte[1024];
    ByteArrayOutputStream os = new ByteArrayOutputStream();

    int line = 0;
    // read bytes from stream, and store them in buffer
    while ((line = stream.read(buffer)) != -1) {
        // Writes bytes from byte array (buffer) into output stream.
        os.write(buffer, 0, line);
    }
    stream.close();
    os.flush();
    os.close();
    return os.toByteArray();
}