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From 刘大伟 <liudawei...@gmail.com>
Subject Re: How to insert a row with a TimeUUIDType column in C++
Date Tue, 01 Jun 2010 03:41:01 GMT
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

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