Need to improve performance in file unzipping from DB

Question

I am getting zipped blob from db and using that blob in below way,

Ex:-

byte[] inputBlob = blobfile.getBytes(1, (int) blobfile.length());

After getting the blob, the way i got the zippedStream and passed it into another Class method(unzipper).

Ex:-

ByteArrayOutputStream zippedStream = null;
InputStream byteInputStream = null;
IParser parser = null;
byte[] buffer = null;
try {
        zippedStream = new ByteArrayOutputStream();
        byteInputStream = new ByteArrayInputStream(blob);
        blob = null;
        int bytes_read;
        buffer = new byte[byteInputStream.available()];

        while ((bytes_read = byteInputStream.read(buffer)) > 0) {
            zippedStream.write(buffer, 0, bytes_read);
        }
        buffer = null;
        byteInputStream.close();
        byteInputStream = null;

    }catch(Exception e){
        e.printStackTrace();
    }

unzipper method: Ex:-

byte[] buffer = new byte[1024];
try {
    InputStream decodedInput = new ByteArrayInputStream(zippedStream.toByteArray());
    zippedStream.close();
    zippedStream = null;
    GZIPInputStream unzippedStream = new GZIPInputStream(decodedInput);
    decodedInput.close();
    decodedInput = null;
    int bytes_read;
    unzippedOutputstream = new ByteArrayOutputStream();
    while ((bytes_read = unzippedStream.read(buffer)) > 0) {
        unzippedOutputstream.write(buffer, 0, bytes_read);
    }
    buffer = null;
    unzippedStream.close();
    unzippedStream = null;
} catch (Exception ex) {
    logger.setException(ex);
    logger.error("unzipper", generateMsg("Exception occurred"));
}

Using this way my application got stucked some time, and performance was so bad. Is there any optimize way to get the zippedstream file and unzipping that easily?

Mate, your formatting.. noone should even think about reading this. — huidube, Nov 07 '14 at 07:47

score 0 · Accepted Answer · answered Nov 07 '14 at 08:39

0

Is all this buffering really needed. Can you IParser parse a Stream?

 InputStream zippedStream = ...
 IParser parser = ...
 parser.parse(new GZIPInputStream(zippedStream));

This will read compressed data, uncompressing as it goes which is much more efficient.

answered Nov 07 '14 at 08:39

Peter Lawrey

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Can you give a idea for avoiding the above method performance? – Boo Balan Nov 11 '14 at 12:24
I don't know what you mean. Can the parser take an input stream? – Peter Lawrey Nov 11 '14 at 12:28

Need to improve performance in file unzipping from DB

1 Answers1