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From Jonathan Ellis <jbel...@gmail.com>
Subject Re: How to insert a row with a TimeUUIDType column in C++
Date Tue, 01 Jun 2010 06:01:19 GMT
http://php.net/manual/en/function.pack.php

2010/5/31 刘大伟 <liudaweisir@gmail.com>:
> How can I get 16 bytes timeUUID ?
>
>
> string(36) "4698cc00-6d2f-11df-8c7f-9f342400a648" TException: UUIDs must be
> exactly 16 bytes Error:
>
>
> On Fri, Apr 23, 2010 at 5:59 PM, Olivier Rosello <orosello@corp.free.fr>
> wrote:
>>
>> Here is my test code :
>>
>> ColumnPath new_col;
>> new_col.__isset.column = true; /* this is required! */
>> new_col.column_family.assign("Incoming");
>> new_col.column.assign("1968ec4a-2a73-11df-9aca-00012e27a270");
>> client.insert("MyKeyspace", "somekey", new_col, "Random Value",
>> time(NULL), ONE);
>>
>> I didn't found in the C++ Cassandra/Thrift API how to specify TimeUUID
>> bytes (16) to the column name. The ColumnPath type get only a string field
>> for the name "column".
>>
>> With a String like this example shows, the TimeUUID is a 36 chars String
>> and this code throws a Exception "UUIDs must be exactly 16 bytes".
>>
>> I didn't found a function like "client.insert_timeuuid_column" which
>> convert the column name to an uint8_t[16]... or anything else which could
>> help me.
>>
>> Cheers,
>>
>> Olivier
>>
>>
>>
>> --
>> Olivier
>>
>>
>
>
>
> --
> 执著而努力着      david.liu
>
>



-- 
Jonathan Ellis
Project Chair, Apache Cassandra
co-founder of Riptano, the source for professional Cassandra support
http://riptano.com

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