How to check if InputStream is Gzipped?

Question

Is there any way to check if InputStream has been gzipped? Here\'s the code:

public static InputStream decompressStream(InputStream input) {
    try {
               


        

Answer 1

Wrap the original stream in a BufferedInputStream, then wrap that in a GZipInputStream. Next try to extract a ZipEntry. If this works, it's a zip file. Then you can use "mark" and "reset" in the BufferedInputStream to return to the initial position in the stream, after your check.

Answer 2

I believe this is simpliest way to check whether a byte array is gzip formatted or not, it does not depend on any HTTP entity or mime type support

public static boolean isGzipStream(byte[] bytes) {
      int head = ((int) bytes[0] & 0xff) | ((bytes[1] << 8) & 0xff00);
      return (GZIPInputStream.GZIP_MAGIC == head);
}

Answer 3

It's not foolproof but it's probably the easiest and doesn't rely on any external data. Like all decent formats, GZip too begins with a magic number which can be quickly checked without reading the entire stream.

public static InputStream decompressStream(InputStream input) {
     PushbackInputStream pb = new PushbackInputStream( input, 2 ); //we need a pushbackstream to look ahead
     byte [] signature = new byte[2];
     int len = pb.read( signature ); //read the signature
     pb.unread( signature, 0, len ); //push back the signature to the stream
     if( signature[ 0 ] == (byte) 0x1f && signature[ 1 ] == (byte) 0x8b ) //check if matches standard gzip magic number
       return new GZIPInputStream( pb );
     else 
       return pb;
}

(Source for the magic number: GZip file format specification)

Update: I've just dicovered that there is also a constant called GZIP_MAGIC in GZipInputStream which contains this value, so if you really want to, you can use the lower two bytes of it.

Answer 4

This function works perfectly well in Java:

public static boolean isGZipped(File f) {   
    val raf = new RandomAccessFile(file, "r")
    return GZIPInputStream.GZIP_MAGIC == (raf.read() & 0xff | ((raf.read() << 8) & 0xff00))
}

In scala:

def isGZip(file:File): Boolean = {
   int gzip = 0
   RandomAccessFile raf = new RandomAccessFile(f, "r")
   gzip = raf.read() & 0xff | ((raf.read() << 8) & 0xff00)
   raf.close()
   return gzip == GZIPInputStream.GZIP_MAGIC
}

Answer 5

Building on the answer by @biziclop - this version uses the GZIP_MAGIC header and additionally is safe for empty or single byte data streams.

public static InputStream maybeDecompress(InputStream input) {
    final PushbackInputStream pb = new PushbackInputStream(input, 2);

    int header = pb.read();
    if(header == -1) {
        return pb;
    }

    int b = pb.read();
    if(b == -1) {
        pb.unread(header);
        return pb;
    }

    pb.unread(new byte[]{(byte)header, (byte)b});

    header = (b << 8) | header;

    if(header == GZIPInputStream.GZIP_MAGIC) {
        return new GZIPInputStream(pb);
    } else {
        return pb;
    }
}

Answer 6

SimpleMagic is a Java library for resolving content types:

<!-- pom.xml -->
    <dependency>
        <groupId>com.j256.simplemagic</groupId>
        <artifactId>simplemagic</artifactId>
        <version>1.8</version>
    </dependency>

---

import com.j256.simplemagic.ContentInfo;
import com.j256.simplemagic.ContentInfoUtil;
import com.j256.simplemagic.ContentType;
// ...

public class SimpleMagicSmokeTest {

    private final static Logger log = LoggerFactory.getLogger(SimpleMagicSmokeTest.class);

    @Test
    public void smokeTestSimpleMagic() throws IOException {
        ContentInfoUtil util = new ContentInfoUtil();
        InputStream possibleGzipInputStream = getGzipInputStream();
        ContentInfo info = util.findMatch(possibleGzipInputStream);

        log.info( info.toString() );
        assertEquals( ContentType.GZIP, info.getContentType() );
    }