/*InputStream class_InputStream = null;
I am reading class from DB 
class_InputStream = rs.getBinaryStream(1);
Your Input stream could be from any source
*/
int thisLine;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
while ((thisLine = class_InputStream.read()) != -1) {
    bos.write(thisLine);
}
bos.flush();
byte [] yourBytes = bos.toByteArray();

/*Don't forget in the finally block to close ByteArrayOutputStream & InputStream
 In my case the IS is from resultset so just closing the rs will do it*/

if (bos != null){
    bos.close();
}

20年后，终于有了一个不需要第三方库的简单解决方案，这要感谢Java 9:

InputStream is;
…
byte[] array = is.readAllBytes();

还要注意方便的方法readNBytes(byte[] b, int off, int len)和transferTo(OutputStream)来解决重复的需求。

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);

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;
    }

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();
}

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 (_