MDAC 2.5 SDK - OLE DB Programmer's Reference
Chapter 26: Mapping MDX to SQL Statements

HIERARCHIZE Function

The first method discussed is how HIERARCHIZE works for a set whose dimensionality is 1. This method will be generalized to sets with arbitrary dimensionality.

Sets with Dimensionality = 1

Consider HIERARCHIZE(S), where S = {Kansas, USA, Canada, Buffalo, Topeka}. As detailed in the section "Literal Sets," create two tables—S1 and S2—that contain the fully qualified member names in the sets S1 and S2, respectively.

Note   In a full-fledged MDX expression, there might be no need to do this because the input set to HIERARCHIZE might be derived from another expression.

Gather information as follows:

1. Use the following query to verify that all members belong to the same dimension. If the returned count is greater than 1, it is an invalid set.

   SELECT COUNT DISTINCT COMPONENT(S.Name, -1)
       FROM S
   

2. Get the dimension name by using the following query:

   SELECT DISTINCT COMPONENT(S.Name, -1)
       FROM S
   

3. Get the level number of each level represented in the input set:

   SELECT DISTINCT LEVEL(S.Name)
       FROM S
   

The objective is to create a table T that looks like this:

For each member of set S, there should be a RANKX column that contains the rank of each of its ancestors. To get the hierarchized set from T, just use the following:

SELECT Name, Rank
FROM 
SELECT * FROM T
   RANK ROWS AS Rank RANKORDER BY Rank1, Rank2, Rank3 
ORDER BY Rank

Note   The number of RANKX columns in T is m + 1, where m = maximum level in the set S. This is the case even if there are unrepresented levels in S, such as when there are members from the COUNTRY and CITY levels but none from the STATE level.

To create the table, perform the following steps:

1. Define a table that has elements of set S and its levels:

   CREATE LOCAL TEMPORARY VIEW S1(Name)
   AS
   SELECT S.Name
   FROM S JOIN members AS M ON(S.Name = M.Level_Name)
   

2. Now make a UNION of three SELECT operations. (Three is the level of the lowermost member in S.)

   Note   The WHERE clause on each UNION iterates from Level = 0 through the level of the lowermost member in S1. Also, the ANCESTOR function in the FROM iterates from 1 through 1 + the level of the lowermost member in S1.

   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1, 
       NULL AS Rank2, NULL AS Rank3
   FROM S1 JOIN Members AS M1 ON S1.Name = M1.Member_Name
   WHERE S1.Level = 0
   UNION
   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1, 
       M2.Natural_Sort_Rank AS Rank2, NULL AS Rank3
   FROM (S1 JOIN Members AS M2 ON S1.Name = M2.Member_Name) 
   JOIN Members AS M1 ON ANCESTOR(S1.Name, 1)= M1.Member_Name
   WHERE S1.Level = 1 
   UNION
   SELECT S1.*, M1.Natural_Sort_Rank AS Rank1, 
       M2.Natural_Sort_Rank AS Rank2, M3.Natural_Sort_Rank AS M3
   FROM (((S1 JOIN Members AS M3 ON S1.Name = M3.Name)
   JOIN Members AS M2 ON ANCESTOR(S1.Name, 2) = M2.Name)
   JOIN Members AS M1 ON ANCESTOR(S1.Name, 3) = M1.Name)
   WHERE S1.Level = 2
   

Generalizing for Sets with Arbitrary Dimensionality

Consider HIERARCHIZE(S), where:

S = {(Kansas, 1996), (Buffalo, 1995.Q4), (USA, 1995.Mar), (Buffalo, 1995), (USA, 1995), (Kansas, 1996.Q4), (Kansas, 1996.Q1), (USA, 1995.Q1)}

The steps in hierarchizing this set are:

1. Let S(Name1, Name2, Rank) be the table associated with the set S.
   
   
2. Let S1, S2 be tables:

• First step:

  CREATE LOCAL TEMPORARY VIEW S1(Name, Rank)
  AS
  SELECT DISTINCT Name1, Rank FROM S
  

• Second step:

  CREATE LOCAL TEMPORARY VIEW S2(Name, Rank)
  AS
  SELECT DISTINCT Name2, Rank FROM S
  

3. HIERARCHIZE S1 and S2 as explained above in "Sets with Dimensionality = 1." Let the resulting hierarchized sets be D1 and D2.
   
   
4. Now use the following query to yield the hierarchized set:

   SELECT S.Name1, S.Name2, NewRank as Rank
   FROM 
       (SELECT S.Name1, S.Name2 
       FROM (S JOIN D1 ON S.Name1 = D1.Name)
           JOIN D2 ON S.Name2 = D2.Name)
       RANK ROWS AS NewRank RANKORDER BY D1.Rank, D2.Rank
   

Generalizing this to sets with dimensionality greater than two should be straightforward.

Note   Even when none of the ancestors of a member are present in the input set, the sorting is done as if the appropriate ancestor were present. Thus, in the above example, Buffalo and NYC appear after LA because California (the parent of LA) sorts before New York (the parent of Buffalo and NYC).