FAQ
Hi,

I am running cassandra version 1.2.19 and cqlsh version 3.1.8 in my setup

cqlsh:apprepo> select * from test_proxy_revisions_r21 limit 10 ;
Traceback (most recent call last):
   File "./cqlsh", line 1039, in perform_statement_untraced
     self.cursor.execute(statement, decoder=decoder)
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/cursor.py",
line 81, in execute
     return self.process_execution_results(response, decoder=decoder)
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/thrifteries.py",
line 116, in process_execution_results
     self.get_metadata_info(self.result[0])
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/cursor.py",
line 97, in get_metadata_info
     name, nbytes, vtype, ctype = self.get_column_metadata(colid)
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/cursor.py",
line 104, in get_column_metadata
     return self.decoder.decode_metadata_and_type(column_id)
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/decoders.py",
line 40, in decode_metadata_and_type
     valdtype = cqltypes.lookup_casstype(validator)
   File "./../lib/cql-internal-only-1.4.1.zip/cql-1.4.1/cql/cqltypes.py",
line 145, in lookup_casstype
     raise ValueError("Don't know how to parse type string %r: %s" %
(casstype, e))
ValueError: Don't know how to parse type string
'org.apache.cassandra.db.marshal.DynamicCompositeType(s=>org.apache.cassandra.db.marshal.UTF8Type)':
weird characters '=>org.apache.cassandra.db.marshal.UTF8Type)' at end

Any clue?

Regards,

Kaushal

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postedMay 19, '15 at 7:44a
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