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

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。

其他回答

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);

一个很好的方法是使用Apache Commons IOUItils将InputStream复制到StringWriter中。。。类似于

StringWriter writer = new StringWriter();
IOUtils.copy(inputStream, writer, encoding);
String theString = writer.toString();

甚至

// NB: does not close inputStream, you'll have to use try-with-resources for that
String theString = IOUtils.toString(inputStream, encoding);

或者,如果不想混合流和写入器,可以使用ByteArrayOutputStream。

如果使用流读取器,请确保在结束时关闭流

private String readStream(InputStream iStream) throws IOException {
    //build a Stream Reader, it can read char by char
    InputStreamReader iStreamReader = new InputStreamReader(iStream);
    //build a buffered Reader, so that i can read whole line at once
    BufferedReader bReader = new BufferedReader(iStreamReader);
    String line = null;
    StringBuilder builder = new StringBuilder();
    while((line = bReader.readLine()) != null) {  //Read till end
        builder.append(line);
        builder.append("\n"); // append new line to preserve lines
    }
    bReader.close();         //close all opened stuff
    iStreamReader.close();
    //iStream.close(); //EDIT: Let the creator of the stream close it!
                       // some readers may auto close the inner stream
    return builder.toString();
}

编辑:在JDK7+上,可以使用trywithresources构造。

/**
 * Reads the stream into a string
 * @param iStream the input stream
 * @return the string read from the stream
 * @throws IOException when an IO error occurs
 */
private String readStream(InputStream iStream) throws IOException {

    //Buffered reader allows us to read line by line
    try (BufferedReader bReader =
                 new BufferedReader(new InputStreamReader(iStream))){
        StringBuilder builder = new StringBuilder();
        String line;
        while((line = bReader.readLine()) != null) {  //Read till end
            builder.append(line);
            builder.append("\n"); // append new line to preserve lines
        }
        return builder.toString();
    }
}
String inputStreamToString(InputStream inputStream, Charset charset) throws IOException {
    try (
            final StringWriter writer = new StringWriter();
            final InputStreamReader reader = new InputStreamReader(inputStream, charset)
        ) {
        reader.transferTo(writer);
        return writer.toString();
    }
}

纯Java标准库解决方案-无库自Java 10以来-Reader#transferTo(Java.io.Writer)无环溶液无新行字符处理

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;
}