hbase scan timerange return old version

Question

I have one question about hbase scan by using timerange. I create a 'test' table,it has one family 'cf' and one version , after I put 4 rows data in that table, and scan that table by using timerange, however, I get a old version row within the timerange.

for example:

 create 'test',{NAME=>'cf',VERSIONS=>1}
 put 'test','row1','cf:u','value1' 
 put 'test','row2','cf:u','value2'
 put 'test','row3','cf:u','value3'
 put 'test','row3','cf:u','value4'

and then I scan this table,the following is the output:

 hbase(main):008:0> scan 'test'
 ROW                                      COLUMN+CELL                                                                                                          
 row1                                    column=cf:u, timestamp=1340259691771, value=value1                                                                   
 row2                                    column=cf:u, timestamp=1340259696975, value=value2                                                                   
 row3                                    column=cf:u, timestamp=1340259704569, value=value4   

it it right,row3 have the newest version.

however,If I use scan it with timerange I get this:

  hbase(main):010:0> scan 'test',{TIMERANGE=>[1340259691771,1340259704569]}
  ROW                                      COLUMN+CELL                                                                                                          
  row1                                    column=cf:u, timestamp=1340259691771, value=value1                                                                   
  row2                                    column=cf:u, timestamp=1340259696975, value=value2                                                                   
  row3                                    column=cf:u, timestamp=1340259701085, value=value3     

it return row3 old version, but this table I set version equal 1

if I increase maxtimestamp ,I get :

  hbase(main):011:0> scan 'test',{TIMERANGE=>[1340259691771,1340259704570]}
  ROW                                      COLUMN+CELL                                                                                                          
  row1                                    column=cf:u, timestamp=1340259691771, value=value1                                                                   
  row2                                    column=cf:u, timestamp=1340259696975, value=value2                                                                   
  row3                                    column=cf:u, timestamp=1340259704569, value=value4                                                                   

3 row(s) in 0.0330 seconds

It is right,I can understand it.

What I want is scan a table within a timerange,it return only newest version, I know there is a TimestampsFilter, however that filter only support specific timestamp ,not time range.

Is there any way to scan a table within a timerange and only return newest verion?

I try to write my own timerangefilter,the following is my code.

import java.io.DataInput;
import java.io.DataOutput;
import java.io.IOException;
import java.util.ArrayList;

import org.apache.hadoop.hbase.KeyValue;
import org.apache.hadoop.hbase.filter.Filter;
import org.apache.hadoop.hbase.filter.FilterBase;
import org.apache.hadoop.hbase.filter.ParseFilter;

import com.google.common.base.Preconditions;  

public class TimeRangeFilter extends FilterBase {

private long minTimeStamp = Long.MIN_VALUE;
private long maxTimeStamp = Long.MAX_VALUE;

public TimeRangeFilter(long minTimeStamp, long maxTimeStamp) {
    Preconditions.checkArgument(maxTimeStamp >= minTimeStamp, "max timestamp %s must be big than min timestamp %s", maxTimeStamp, minTimeStamp);
    this.maxTimeStamp = maxTimeStamp;
    this.minTimeStamp = minTimeStamp;
}

@Override
public ReturnCode filterKeyValue(KeyValue v) {
    if (v.getTimestamp() >= minTimeStamp && v.getTimestamp() <= maxTimeStamp) {
        return ReturnCode.INCLUDE;
    } else if (v.getTimestamp() < minTimeStamp) {
        // The remaining versions of this column are guaranteed
        // to be lesser than all of the other values.
        return ReturnCode.NEXT_COL;
    }
    return ReturnCode.SKIP;
}

public static Filter createFilterFromArguments(ArrayList<byte[]> filterArguments) {
    long minTime, maxTime;
    if (filterArguments.size() < 2)
        return null;
    minTime = ParseFilter.convertByteArrayToLong(filterArguments.get(0));
    maxTime = ParseFilter.convertByteArrayToLong(filterArguments.get(1));
    return new TimeRangeFilter(minTime, maxTime);
}

@Override
public void write(DataOutput out) throws IOException {
    // TODO Auto-generated method stub
    out.writeLong(minTimeStamp);
    out.writeLong(maxTimeStamp);
}

@Override
public void readFields(DataInput in) throws IOException {
    // TODO Auto-generated method stub
    this.minTimeStamp = in.readLong();
    this.maxTimeStamp = in.readLong();
}

}

I add this jar into hbase HBASE_CLASSPATH in hbase-env.sh, however,I get the following error:

org.apache.hadoop.hbase.client.ScannerCallable@a9255c, java.io.IOException: IPC server unable to read call parameters: Error in readFields

Answer 1

Dape,

When you set the max versions to 1 and have more than one entry for a cell, Hbase tombstones the older cells and gets and scans cannot see them unless ofcourse you specify a particular timestamp range which qualifies only one cell. The tombstoned cells are only deleted after a Major_compact is run on the table, which is when the older cells would stop popping up.

To always get the latest cells from a scan all you need to do is use the method below -

    Result.getColumnLatest(family, qualifier)