如何将整个输入流读到字节数组?


当前回答

如果你碰巧使用谷歌Guava,它将像使用ByteStreams一样简单:

byte[] bytes = ByteStreams.toByteArray(inputStream);

其他回答

这是我的复制粘贴版本:

@SuppressWarnings("empty-statement")
public static byte[] inputStreamToByte(InputStream is) throws IOException {
    if (is == null) {
        return null;
    }
    // Define a size if you have an idea of it.
    ByteArrayOutputStream r = new ByteArrayOutputStream(2048);
    byte[] read = new byte[512]; // Your buffer size.
    for (int i; -1 != (i = is.read(read)); r.write(read, 0, i));
    is.close();
    return r.toByteArray();
}

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;
    }

Java 7及以上版本:

import sun.misc.IOUtils;
...
InputStream in = ...;
byte[] buf = IOUtils.readFully(in, -1, false);

如果有人还在寻找一个没有依赖的解决方案,如果你有一个文件。

DataInputStream

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

ByteArrayOutputStream

 InputStream is = new FileInputStream(file);
 ByteArrayOutputStream buffer = new ByteArrayOutputStream();
 int nRead;
 byte[] data = new byte[(int) file.length()];
 while ((nRead = is.read(data, 0, data.length)) != -1) {
     buffer.write(data, 0, nRead);
 }

RandomAccessFile

 RandomAccessFile raf = new RandomAccessFile(file, "r");
 byte[] data = new byte[(int) raf.length()];
 raf.readFully(data);

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
    }