MDAC 2.5 SDK - OLE DB Programmer's Reference
Chapter 26: Mapping MDX to SQL Statements
This section explains how the INTERSECT function handles duplicates.
Consider INTERSECT(S1, S2) where:
S1 = {Kansas, Buffalo, Buffalo, Topeka, Topeka, Wichita, Canada, BC}
S2 = {Buffalo, Topeka, Topeka, Canada, Quebec}
As shown 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.
The following query does the intersect:
SELECT Name, NewRank AS Rank
FROM
(((SELECT Name, Rank FROM S1
WHERE S1.Name IN (SELECT Name FROM S2))
RANKTOP 1 LEVELS RANKORDER BY Rank RANKGROUP BY Name)
RANK ROWS AS NewRank RANKORDER BY Rank)
ORDER BY Rank
In English, the query does the following:
Note You cannot use the INTERSECT operation of SQL-92 because duplicate elimination is not guaranteed to remove duplicates from the tail.
Consider INTERSECT(S1, S2, [ALL]). As in the preceding section, create the tables S1 and S2. Then use the following query:
SELECT Name, NewRank AS Rank
FROM
((SELECT Name, Rank FROM S1
WHERE S1.Name IN (SELECT Name FROM S2))
RANK ROWS AS NewRank RANKORDER BY Rank)
ORDER BY Rank
This is simpler because there is no need to remove duplicates. After the query that finds the intersect, the RANK operation generates a rank value for each of the rows of the resulting table. The outer SELECT produces this table, ordering it by the NewRank (renamed as Rank) column.
Note You cannot use INTERSECT ALL from SQL-92 because its semantics are as follows: If a row r appears n times in S1 and m times in S2, then r appears exactly p times in the result, where p is the lesser of m and n. Moreover, the order of these p rows is not preserved.
Each tuple has more than one element for these sets. Replace the NAME column in the select list of each SELECT statement above with a list of columns (equivalent to the number of dimensions in the input sets). Add as many IN clauses in the WHERE as there are columns, one for each column.