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;
}

以下是我经过一些实验后提出的最优雅、纯Java（无库）解决方案：

public static String fromStream(InputStream in) throws IOException
{
    BufferedReader reader = new BufferedReader(new InputStreamReader(in));
    StringBuilder out = new StringBuilder();
    String newLine = System.getProperty("line.separator");
    String line;
    while ((line = reader.readLine()) != null) {
        out.append(line);
        out.append(newLine);
    }
    return out.toString();
}

我会使用一些Java8技巧。

public static String streamToString(final InputStream inputStream) throws Exception {
    // buffering optional
    try
    (
        final BufferedReader br
           = new BufferedReader(new InputStreamReader(inputStream))
    ) {
        // parallel optional
        return br.lines().parallel().collect(Collectors.joining("\n"));
    } catch (final IOException e) {
        throw new RuntimeException(e);
        // whatever.
    }
}

基本上与其他一些答案相同，只是更简洁。

下面是如何仅使用JDK使用字节数组缓冲区来实现这一点。这实际上就是commons-ioIOUItils.copy（）方法的工作原理。如果要从Reader而不是InputStream复制，可以用char[]替换byte[]。

import java.io.ByteArrayOutputStream;
import java.io.InputStream;

...

InputStream is = ....
ByteArrayOutputStream baos = new ByteArrayOutputStream(8192);
byte[] buffer = new byte[8192];
int count = 0;
try {
  while ((count = is.read(buffer)) != -1) {
    baos.write(buffer, 0, count);
  }
}
finally {
  try {
    is.close();
  }
  catch (Exception ignore) {
  }
}

String charset = "UTF-8";
String inputStreamAsString = baos.toString(charset);

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();
}

此代码段位于\sdk\samples\android-19\connection\NetworkConnect\NetworkConnectSample\src\main\java.com\example\android\NetworkConnect\MainActivity.java中，该代码段根据Apache许可证2.0版授权，由Google编写。

/** Reads an InputStream and converts it to a String.
 * @param stream InputStream containing HTML from targeted site.
 * @param len Length of string that this method returns.
 * @return String concatenated according to len parameter.
 * @throws java.io.IOException
 * @throws java.io.UnsupportedEncodingException
 */
private String readIt(InputStream stream, int len) throws IOException, UnsupportedEncodingException {
    Reader reader = null;
    reader = new InputStreamReader(stream, "UTF-8");
    char[] buffer = new char[len];
    reader.read(buffer);
    return new String(buffer);
}