That is not a valid statement, were you getting an error ? 

This creates the ProductCategory CF .

create column family ProductCategory
    with column_type = 'Super'
    and comparator = UTF8Type;

This creates the ProductCatID CF

create column family ProductCatId
WITH comparator = UTF8Type
AND key_validation_class=UTF8Type
AND column_metadata = [
{column_name: subProdName, validation_class: UTF8Type}
{column_name: lenght, validation_class: UTF8Type}
{column_name: width, validation_class: UTF8Type}
];

See the help in the CLI for more info. 

Hope that helps. 

-----------------
Aaron Morton
Freelance Cassandra Developer
@aaronmorton

On 1/10/2011, at 3:38 AM, Ramesh S wrote:

I am using 0.8.6.

On Thu, Sep 29, 2011 at 11:10 PM, Yi Yang <i@iyyang.com> wrote:
Which version are you using? In my memory 0.8.3 cannot do it correctly but later versions fixed the bug.

從我的 BlackBerry® 無線裝置


From: Ramesh S <investtrco@gmail.com>
Date: Thu, 29 Sep 2011 15:23:29 -0500
Subject: create super column family for

I am trying to create a super column family using Cli command.
But I am not getting it.

The structure is 

<<SCF>>ProductCategory
<<SuperColumnName>>#ProductType
<<RowKey>>#productCatId
+subProdName
+lenght
+width

I tried a lot many ways but I can't find the right way to get this done.
Something like this give me error - mismatched input 'column' expecting Identifier
create column family ProductCategory
    with column_type = 'Super'
    and comparator = UTF8Type
with column family productCatId
WITH comparator = UTF8Type
AND key_validation_class=UTF8Type
AND column_metadata = [
{column_name: subProdName, validation_class: UTF8Type}
{column_name: lenght, validation_class: UTF8Type}
{column_name: width, validation_class: UTF8Type}
];
    
Appreciate any help

regards
Ramesh