这是一个改编自org.apache.commons.io.IOUtils源代码的答案，适用于那些想要apache实现但不想要整个库的人。

private static final int BUFFER_SIZE = 4 * 1024;

public static String inputStreamToString(InputStream inputStream, String charsetName)
        throws IOException {
    StringBuilder builder = new StringBuilder();
    InputStreamReader reader = new InputStreamReader(inputStream, charsetName);
    char[] buffer = new char[BUFFER_SIZE];
    int length;
    while ((length = reader.read(buffer)) != -1) {
        builder.append(buffer, 0, length);
    }
    return builder.toString();
}

这是最适合Android和任何其他JVM的纯Java解决方案。

这个解决方案非常好。。。它简单、快速，适用于大小河流！！（见上文第8号基准）

public String readFullyAsString(InputStream inputStream, String encoding)
        throws IOException {
    return readFully(inputStream).toString(encoding);
}

public byte[] readFullyAsBytes(InputStream inputStream)
        throws IOException {
    return readFully(inputStream).toByteArray();
}

private ByteArrayOutputStream readFully(InputStream inputStream)
        throws IOException {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();
    byte[] buffer = new byte[1024];
    int length = 0;
    while ((length = inputStream.read(buffer)) != -1) {
        baos.write(buffer, 0, length);
    }
    return baos;
}

InputStreamReader i = new InputStreamReader(s);
BufferedReader str = new BufferedReader(i);
String msg = str.readLine();
System.out.println(msg);

这是您的InputStream对象，它将被转换为String

Kotlin用户只需：

println(InputStreamReader(is).readText())

鉴于

readText()

是Kotlin标准库的内置扩展方法。

如果你喜欢冒险，你可以把Scala和Java混合起来，最后得到这样的结果：

scala.io.Source.fromInputStream(is).mkString("")

混合Java和Scala代码和库有其好处。

请参阅此处的完整描述：在Scala中将InputStream转换为String的惯用方法

JDK 7/8应答，关闭流并仍然抛出IOException：

StringBuilder build = new StringBuilder();
byte[] buf = new byte[1024];
int length;
try (InputStream is = getInputStream()) {
  while ((length = is.read(buf)) != -1) {
    build.append(new String(buf, 0, length));
  }
}