How do I insert a row with a TimeUUIDType column in Cassandra?

Question

In Cassandra, I have the following Column Family: ``` <ColumnFamily CompareWith="TimeUUIDType" Name="Posts"/> ``` I'm trying to insert a record into it as follows using a C++ generated function generated by Thrift: ``` ColumnPath new_col; new_col.__isset.column = true; /* this is required! */ new_col.column_family.assign("Posts"); new_col.super_column.assign(""); new_col.column.assign("1968ec4a-2a73-11df-9aca-00012e27a270"); client.insert("Keyspace1", "somekey", new_col, "Random Value", 1234, ONE); ``` However, I'm getting the following error: "UUIDs must be exactly 16 bytes" I've even tried the Cassandra CLI with the following command: ``` set Keyspace1.Posts['somekey']['1968ec4a-2a73-11df-9aca-00012e27a270'] = 'Random Value' ``` but I still get the following error: ``` Exception null InvalidRequestException(why:UUIDs must be exactly 16 bytes) at org.apache.cassandra.thrift.Cassandra$insert_result.read(Cassandra.java:11994) at org.apache.cassandra.thrift.Cassandra$Client.recv_insert(…

Accepted Answer

Thrift is a binary protocol; 16 bytes means 16 bytes. "1968ec4a-2a73-11df-9aca-00012e27a270" is 36 bytes. You need to get your library to give you the raw, 16 bytes form.

I don't use C++ myself, but "version 1 uuid" is the magic string you want to google for when looking for a library that can do this. http://www.google.com/search?q=C%2B%2B+version+1+uuid

How do I insert a row with a TimeUUIDType column in Cassandra?

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