这个问题的解决方案不是最简单的，但由于没有提到NIO流和通道，这里有一个使用NIO通道和ByteBuffer将流转换为字符串的版本。

public static String streamToStringChannel(InputStream in, String encoding, int bufSize) throws IOException {
    ReadableByteChannel channel = Channels.newChannel(in);
    ByteBuffer byteBuffer = ByteBuffer.allocate(bufSize);
    ByteArrayOutputStream bout = new ByteArrayOutputStream();
    WritableByteChannel outChannel = Channels.newChannel(bout);
    while (channel.read(byteBuffer) > 0 || byteBuffer.position() > 0) {
        byteBuffer.flip();  //make buffer ready for write
        outChannel.write(byteBuffer);
        byteBuffer.compact(); //make buffer ready for reading
    }
    channel.close();
    outChannel.close();
    return bout.toString(encoding);
}

下面是如何使用它的示例：

try (InputStream in = new FileInputStream("/tmp/large_file.xml")) {
    String x = streamToStringChannel(in, "UTF-8", 1);
    System.out.println(x);
}

对于大型文件，此方法的性能应该很好。

将inputStream转换为字符串的方法

public static String getStringFromInputStream(InputStream inputStream) {

    BufferedReader bufferedReader = null;
    StringBuilder stringBuilder = new StringBuilder();
    String line;

    try {
        bufferedReader = new BufferedReader(new InputStreamReader(
                inputStream));
        while ((line = bufferedReader.readLine()) != null) {
            stringBuilder.append(line);
        }
    } catch (IOException e) {
        logger.error(e.getMessage());
    } finally {
        if (bufferedReader != null) {
            try {
                bufferedReader.close();
            } catch (IOException e) {
                logger.error(e.getMessage());
            }
        }
    }
    return stringBuilder.toString();
}

此外，您还可以从指定的资源路径获取InputStream：

public static InputStream getResourceAsStream(String path)
{
    InputStream myiInputStream = ClassName.class.getResourceAsStream(path);
    if (null == myiInputStream)
    {
        mylogger.info("Can't find path = ", path);
    }

    return myiInputStream;
}

要从特定路径获取InputStream，请执行以下操作：

public static URL getResource(String path)
{
    URL myURL = ClassName.class.getResource(path);
    if (null == myURL)
    {
        mylogger.info("Can't find resource path = ", path);
    }
    return myURL;
}

如果需要将字符串转换为没有外部库的特定字符集，则：

public String convertStreamToString(InputStream is) throws IOException {
  try (ByteArrayOutputStream baos = new ByteArrayOutputStream();) {
    is.transferTo(baos);
    return baos.toString(StandardCharsets.UTF_8);
  }
}

我建议使用StringWriter类来解决这个问题。

StringWriter wt= new StringWriter();
IOUtils.copy(inputStream, wt, encoding);
String st= wt.toString();

Kotlin用户只需：

println(InputStreamReader(is).readText())

鉴于

readText()

是Kotlin标准库的内置扩展方法。