Improvement of VBA code in Access to do something like a pivot table

Question

I have a table of data in MS Access 2007. There are 6 fields per record, thousands of records. I want to make a sort of pivot table like object. That is, if any two rows happens to be the same in the first 4 fields, then they will end up grouped together into one row. The column headers in this pivot table will be the values from the 5th field, and the value in the pivot table will be the 6th field, a dollar amount. Think of the 5th field as letters A, B, C, D, E, F, G. So, the table I start with might have a row with A in the 5th field and $3.48 in the 6th field. Another row may match in the first 4 fields, have B in the 5th field and $8.59 in the 6th field. Another may match in the first 4 fields, have E in the 5th field and $45.20 in the 6th field. I want all these rows to be turned into one row (in a new table) that starts with the first 4 fields where they match, then lists $3.48, $8.59, $0.00, $0.00, $45.20, $0.00, $0.00, corresponding to column headers A, B, C, D, E, F, G (since no records contained C, D, F, G, their corresponding values are $0.00), and then ends with one more field that totals up the money in that row.

Currently, I have some VBA code that does this, written by someone else a few years ago. It is extremely slow and I am hoping for a better way. I asked a previous question (but not very clearly so I was advised to create a new question), where I was asking if there was a better way to do this in VBA. My question asked about reading and writing large amounts of data all at once in Access through VBA, which I know is a good practice in Excel. That is, I was hoping to take my original table and just assign the entire thing to an array all at once (as in Excel, instead of cell by cell), then work with that array in VBA and create some new array and then write that entire array all at once to a new table (instead of record by record, field by field). From the answers in that question, it seems like that is not really a possibility in Access, but my best bet might be to use some sort of query. I tried the Query Wizard and found the Cross Tab query which is close to what I describe above. But, there appears to be a max of 3 fields used in the Row Heading, whereas here I have 4. And, instead of putting $0.00 when a value is not specified (like C, D, F, G in my example above), it just leaves a blank.

Update (in response to Remou's comment to give sample data): Here is some sample data.

ID   a   b   c   d   e   f
 7   1   2   3   5   A   5
 8   1   2   3   5   B  10
 9   1   2   3   5   C  15
10   1   2   3   5   D  20
11   1   2   3   5   E  25
12   1   2   4   4   A  16
13   1   2   4   4   B  26
14   1   3   3   7   D  11
15   1   3   3   7   B  11

The result should be:

a   b   c   d  an  bn  cn  dn  en   Total

1   2   3   5   5  10  15  20  25      75
1   2   4   4  16  26   0   0   0      42
1   3   3   7   0  11   0  11   0      22

But, when I copy and paste the SQL given by Remou, the only output I get is

a   b   c   d  an  bn  cn  dn  en

1   2   3   5   5  10  15  20  25

Answer 1

This is, I think, what you want, but it would be better to consider database design, because this is a spreadsheet-like solution.

SELECT t0.a,
       t0.b,
       t0.c,
       t0.d,
       Iif(Isnull([a1]), 0, [a1]) AS an,
       Iif(Isnull([b1]), 0, [b1]) AS bn,
       Iif(Isnull([c1]), 0, [c1]) AS cn,
       Iif(Isnull([d1]), 0, [d1]) AS dn,
       Iif(Isnull([e1]), 0, [e1]) AS en
FROM   (((((SELECT DISTINCT t.a,
                           t.b,
                           t.c,
                           t.d
           FROM   table3 t) AS t0
           LEFT JOIN (SELECT t.a,
                              t.b,
                              t.c,
                              t.d,
                              t.f AS a1
                       FROM   table3 t
                       WHERE  t.e = "A") AS a0
             ON ( t0.d = a0.d )
                AND ( t0.c = a0.c )
                AND ( t0.b = a0.b )
                AND ( t0.a = a0.a ))
          LEFT JOIN (SELECT t.a,
                             t.b,
                             t.c,
                             t.d,
                             t.f AS b1
                      FROM   table3 t
                      WHERE  t.e = "B") AS b0
            ON ( t0.d = b0.d )
               AND ( t0.c = b0.c )
               AND ( t0.b = b0.b )
               AND ( t0.a = b0.a ))
         LEFT JOIN (SELECT t.a,
                            t.b,
                            t.c,
                            t.d,
                            t.f AS c1
                     FROM   table3 t
                     WHERE  t.e = "C") AS c0
           ON ( t0.d = c0.d )
              AND ( t0.c = c0.c )
              AND ( t0.b = c0.b )
              AND ( t0.a = c0.a ))
        LEFT JOIN (SELECT t.a,
                           t.b,
                           t.c,
                           t.d,
                           t.f AS d1
                    FROM   table3 t
                    WHERE  t.e = "D") AS d0
          ON ( t0.d = d0.d )
             AND ( t0.c = d0.c )
             AND ( t0.b = d0.b )
             AND ( t0.a = d0.a ))
       LEFT JOIN (SELECT t.a,
                          t.b,
                          t.c,
                          t.d,
                          t.f AS e1
                   FROM   table3 t
                   WHERE  t.e = "E") AS e0
         ON ( t0.d = e0.d )
            AND ( t0.c = e0.c )
            AND ( t0.b = e0.b )
            AND ( t0.a = e0.a ); 

Table3

ID  a   b   c   d   e   f
1   1   2   3   4   a   €10.00
2   1   2   3   4   b   €10.00
3   1   2   3   4   c   €10.00
4   1   2   3   4   d   €10.00
5   1   2   3   4   e   €10.00
6   1   2   3   5   a   €10.00
7   1   2   3   5   b   
8   1   2   3   5   c   €10.00
9   1   2   3   5   d   €10.00
10  1   2   3   5   e   €10.00

Result

There are two rows, because there are only two different sets in the first four columns.

a   b   c   d   an      bn      cn      dn      en
1   2   3   4   €10.00  €10.00  €10.00  €10.00  €10.00
1   2   3   5   €10.00  €0.00   €10.00  €10.00  €10.00

The way the sql above is supposed to work, is that it selects each of the four definition columns and the currency column from the table where the sort column has a particular sort letter and labels the currency column with the sort letter, each of these sub queries are then assembled, however, you can take a sub query and look at the results. The last one is the part between the parentheses:

INNER JOIN (SELECT t.a,
                          t.b,
                          t.c,
                          t.d,
                          t.f AS e1
                   FROM   table3 t
                   WHERE  t.e = "E") AS e0