I have created a column family with Comparator_type="LexicalUUIDType", Default_validation_class="UTF8Type" and Key_validation_class="UTF8Type".
And set TimeUUID as column_name within the column family above. It's insertion runs very well, but how can I get the columns? I can't set the correct column_name! The following are the code:
ColumnPath path = new ColumnPath();
path.setColumn_family("test");
path.setColumn(("44c32fe1-38a4-11e1-a06a-485d60c81a3e".getBytes()));
ColumnOrSuperColumn or = new ColumnOrSuperColumn();
try {
or = client.get(ByteBuffer.wrap("key").getBytes()), path, ConsistencyLevel.ONE);
} catch (InvalidRequestException e) {
...
data in Cassandra DB:
=> (column=44c32fe0-38a4-11e1-a06a-485d60c81a3e, value=32, timestamp=1325881397726)
=> (column=44c32fe1-38a4-11e1-a06a-485d60c81a3e, value=33, timestamp=1325881397726)
=> (column=44c32fe2-38a4-11e1-a06a-485d60c81a3e, value=34, timestamp=1325881397727)
=> (column=44c37e00-38a4-11e1-a06a-485d60c81a3e, value=35, timestamp=1325881397728)
=> (column=44c37e01-38a4-11e1-a06a-485d60c81a3e, value=36, timestamp=1325881397728)
...
And the exception informations:
InvalidRequestException(why:LexicalUUID should be 16 or 0 bytes (36))
at org.apache.cassandra.thrift.Cassandra$get_result.read(Cassandra.java:6490)
at org.apache.cassandra.thrift.Cassandra$Client.recv_get(Cassandra.java:519)
at org.apache.cassandra.thrift.Cassandra$Client.get(Cassandra.java:492)
at test.cassandra.MainTest.query(MainTest.java:118)
...
That's why? I can't execute single query or slice query now. How can I execute query by key and column name with uuid? Thank in advance!