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Php SimpleCassie Cassandra TimeUUID

I created the Users column family using the following command:

create column family Users with comparator=TimeUUIDType and default_validation_class=UTF8Type;

Then I insert the column into the "Users". He shows the following.

RowKey: jsmith

=>(column=66829930-515b-11e0-8443-0f82b246fa40, value=hello, timestamp=1300451382)

I want to access it using SimpleCassie. Team:

$data = $cassie->keyspace('Keyspace1')->cf('Users')->key('jsmith')->column('66829930-515b-11e0-8443-0f82b246fa40')->value();

(I also tried: $data = $cassie->keyspace('Keyspace1')->cf('Users')->key('jsmith')->column($cassie->uuid('66829930-515b-11e0-8443-0f82b246fa40')->__toString())->value();)

However, they do not work. It always returns NULL. How can I get the column value (hello) that I want?

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1 answer

Try:

$cassie->keyspace('Keyspace1')->cf('Users')->key('jsmith')->column($cassie->uuid('66829930-515b-11e0-8443-0f82b246fa40')->uuid)->value();)

Cassandra expects a binary representation of the UUID, not a hexadecimal / string representation.

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