如何将整个输入流读到字节数组?


当前回答

Java 9最终会给你一个很好的方法:

InputStream in = ...;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
in.transferTo( bos );
byte[] bytes = bos.toByteArray();

其他回答

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
    }

下面的代码

public static byte[] serializeObj(Object obj) throws IOException {
  ByteArrayOutputStream baOStream = new ByteArrayOutputStream();
  ObjectOutputStream objOStream = new ObjectOutputStream(baOStream);

  objOStream.writeObject(obj); 
  objOStream.flush();
  objOStream.close();
  return baOStream.toByteArray(); 
} 

OR

BufferedImage img = ...
ByteArrayOutputStream baos = new ByteArrayOutputStream(1000);
ImageIO.write(img, "jpeg", baos);
baos.flush();
byte[] result = baos.toByteArray();
baos.close();

安全解决方案(正确关闭流):

Java 9及更新版本: 最终字节[]字节; try (inputStream) { 字节= inputStream.readAllBytes(); }


Java 8 and older: public static byte[] readAllBytes(InputStream inputStream) throws IOException { final int bufLen = 4 * 0x400; // 4KB byte[] buf = new byte[bufLen]; int readLen; IOException exception = null; try { try (ByteArrayOutputStream outputStream = new ByteArrayOutputStream()) { while ((readLen = inputStream.read(buf, 0, bufLen)) != -1) outputStream.write(buf, 0, readLen); return outputStream.toByteArray(); } } catch (IOException e) { exception = e; throw e; } finally { if (exception == null) inputStream.close(); else try { inputStream.close(); } catch (IOException e) { exception.addSuppressed(e); } } }


Kotlin(当Java 9+不可访问时): @Throws (IOException::类) fun InputStream.readAllBytes(): ByteArray { val bufLen = 4 * 0x400 // 4KB val buf = ByteArray(bufLen) var readLen: Int = 0 ByteArrayOutputStream()。使用{o -> 这一点。使用{I -> i.read(buf, 0, bufLen)。{readLen = it} != -1) o.write(buf, 0, readLen) } 返回o.toByteArray () } } 避免嵌套使用请看这里。


Scala(当Java 9+不可访问时)(By @Joan。Thx): def readAllBytes(inputStream: inputStream): Array[Byte] = Stream.continually (read)。takeWhile(_ != -1).map(_. tobyte).toArray . take (_

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);

我试图编辑@numan的答案,修复了写垃圾数据,但编辑被拒绝。虽然这段简短的代码并不出色,但我看不到其他更好的答案。以下是我认为最有意义的建议:

ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buffer = new byte[1024]; // you can configure the buffer size
int length;

while ((length = in.read(buffer)) != -1) out.write(buffer, 0, length); //copy streams
in.close(); // call this in a finally block

byte[] result = out.toByteArray();

ByteArrayOutputStream不需要关闭。为了可读性,省略了一些结构