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Here is the java code that, that publishes data to Redis

import com.streambase.sb.util.ByteOrderedDataOutput;

byte[] valuebuffer=null;
ByteOrderedDataOutput boutput = new ByteOrderedDataOutput(0,tuple.getByteOrder());      
tuple.serialize(boutput,0); 
valuebuffer = boutput.getBuffer();
byte[] keybuffer = null;
String keyvalue = redisStream+"."+keyFieldStr;
keybuffer = keyvalue.getBytes();
strLuaCommands += "redis.call('set',KEYS["+ (++keyCount) +"],ARGV["+ (++argCount) +"])";
keys.add(keybuffer);
args.add(valuebuffer);

I was able to get the data through python struct, but this is not in correct format.

import redis, struct
redis_client = redis.StrictRedis(host="abc.com", port=6379,  db=0)
temp = redis_client.get('samplekey')
struct.unpack("<" + ("s" * (len(temp))), temp)
Krishna Kumar
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1 Answers1

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Tuple.serialize() uses the com.streambase.sb.util.ByteOrderedDataOutput class, which has never been part of the StreamBase public API. Therefore the Tuple.serialize() methods shouldn't be considered part of the public API, either.

Also, there's no particular reason to believe that the Python struct.unpack() method knows how to understand StreamBase's ByteOrderedDataOutput, whatever that is. So it's not surprising that what you are unpacking is not what you want.

One sort of quick-to-imagine workaround would be to use the StreamBase Python Operator to convert your StreamBase tuple into Python objects, and then use a Python script to write whatever you want to write into redis. Then, since you will have now encoded the stuff and decoded the stuff with the same Python complementary library functions, you'll have a much better chance of not mangling your data.

sbarber2
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