Saturday, July 16, 2011

MDX Root() function doesn't support multiselect

One of my other favourite MDX functions is Root(). It's a really easy way to get all the filters off a dimension. I often use it in calculations defined in the cube.

However the Root() function does not support multi-select and you might get this error message.

#Error Query (6, 30) The MDX function ROOT failed because the coordinate for the 'Region' attribute contains a set.
Fortunately, there is a workaround. Unfortunately, it means not using the Root() function and going back to the .[All] member of all the attributes (or all the ones being used in the query) of the dimension.
For example, the following query will hit the error.
WITH
MEMBER [Client Host].Region.[AfricaAsia] AS
AGGREGATE({[Client Host].Region.&[Africa]
,[Client Host].Region.&[Asia]})
MEMBER Measures.[HitsPercent] as
[Measures].[Request Count]/(ROOT([Client Host]), [Measures].[Request Count])
SELECT
{Measures.[HitsPercent]}
ON COLUMNS
,[Date].[Month].[Month]
ON ROWS
FROM [EasternMining]
WHERE
([Client Host].Region.[AfricaAsia])


The fix is simply to change Root([Client Host]) to [Client Host].Region.[All]. ie.
WITH
MEMBER [Client Host].Region.[AfricaAsia] AS
AGGREGATE({[Client Host].Region.&[Africa]
,[Client Host].Region.&[Asia]})
MEMBER Measures.[HitsPercent] as
// [Measures].[Request Count]/(ROOT([Client Host]), [Measures].[Request Count])
[Measures].[Request Count]/([Client Host].Region.[All], [Measures].[Request Count])
,format_string="#,#0.00%"
,non_empty_behavior="measures.[Request Count]"
SELECT
{Measures.[HitsPercent]}
ON COLUMNS
,[Date].[Month].[Month]
ON ROWS
FROM [EasternMining]
WHERE
([Client Host].Region.[AfricaAsia])


Note, I have only put the .[All] member in for the Region attribute, whereas you really need to add in a .[All] member for all the dimension attributes to be equivalent to the Root() function.

By the way, I don't know why the Root() function doesn't support multi-select. One would imagine that it should.

Because of this limitation, I might be using the Root() function less liberally in calculations stored in the cube.

1 comment:

  1. Hi Richard,

    the ROOT is also not so good for the performance (see Mosha's answer here http://social.msdn.microsoft.com/Forums/en/sqlanalysisservices/thread/3b724376-aa6d-4bcc-afcb-a2ac07c94df5).

    Michael.

    ReplyDelete