如果您有java.io.InputStream对象,应该如何处理该对象并生成String?


假设我有一个包含文本数据的InputStream,我想将其转换为String,例如,我可以将其写入日志文件。

获取InputStream并将其转换为字符串的最简单方法是什么?

public String convertStreamToString(InputStream is) {
// ???
}

当前回答

您可以使用Apache Commons。

在IOUItils中,您可以找到toString方法以及三个有用的实现。

public static String toString(InputStream input) throws IOException {
        return toString(input, Charset.defaultCharset());
}

public static String toString(InputStream input) throws IOException {
        return toString(input, Charset.defaultCharset());
}

public static String toString(InputStream input, String encoding)
            throws IOException {
        return toString(input, Charsets.toCharset(encoding));
}

其他回答

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;
}

如果不能使用Commons IO(FileUtils/IOUtils/CopyUtils),下面是一个使用BufferedReader逐行读取文件的示例:

public class StringFromFile {
    public static void main(String[] args) /*throws UnsupportedEncodingException*/ {
        InputStream is = StringFromFile.class.getResourceAsStream("file.txt");
        BufferedReader br = new BufferedReader(new InputStreamReader(is/*, "UTF-8"*/));
        final int CHARS_PER_PAGE = 5000; //counting spaces
        StringBuilder builder = new StringBuilder(CHARS_PER_PAGE);
        try {
            for(String line=br.readLine(); line!=null; line=br.readLine()) {
                builder.append(line);
                builder.append('\n');
            }
        } 
        catch (IOException ignore) { }

        String text = builder.toString();
        System.out.println(text);
    }
}

或者,如果你想要原始速度,我会根据Paul de Vrieze的建议(避免使用StringWriter(内部使用StringBuffer))提出一个变体:

public class StringFromFileFast {
    public static void main(String[] args) /*throws UnsupportedEncodingException*/ {
        InputStream is = StringFromFileFast.class.getResourceAsStream("file.txt");
        InputStreamReader input = new InputStreamReader(is/*, "UTF-8"*/);
        final int CHARS_PER_PAGE = 5000; //counting spaces
        final char[] buffer = new char[CHARS_PER_PAGE];
        StringBuilder output = new StringBuilder(CHARS_PER_PAGE);
        try {
            for(int read = input.read(buffer, 0, buffer.length);
                    read != -1;
                    read = input.read(buffer, 0, buffer.length)) {
                output.append(buffer, 0, read);
            }
        } catch (IOException ignore) { }

        String text = output.toString();
        System.out.println(text);
    }
}

如果您正在使用Google Collections/Guava,您可以执行以下操作:

InputStream stream = ...
String content = CharStreams.toString(new InputStreamReader(stream, Charsets.UTF_8));
Closeables.closeQuietly(stream);

请注意,InputStreamReader的第二个参数(即Charsets.UTF_8)不是必需的,但如果您知道编码(您应该这样做!)

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));
  }
}

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();
}