Pivoting Table Data — Horizontal Display

Question:
[Joe Celko’s Milestones Puzzle]

This puzzle, in a little different form, came from Brian Young. His system tracks a series of dates (milestones) for each particular type of service (tos) that they sell. These dates constitute the schedule for the delivery of the service and vary with the type of service they are delivering. Their management would like to see a schedule for each shop horizontally, which I must admit is a reasonable request. They also want to be able to specify which task code (tos) to display.

Brain ran across a clever solution to this problem by Steve Roti (which is actually in the SQLServer manual p161), but it relies on the SUM function and a multiplication by 1 to yield the correct result. (That Roti guy is very clever!). Unfortunately, this technique doesn’t work with dates. So here is the table structure:

    CREATE TABLE Schedule        (shop CHAR (3) NOT NULL,        order CHAR (10) NOT NULL,        schseq SMALLINT NOT NULL CHECK (schseq IN (1,2,3),        tos CHAR (2) NOT NULL,        schactdate DATE);

Where schseq is encoded as:

    (1 = ‘processed’)    (2 = ‘completed’)    (3 = ‘confirmed’)

The data normally appears like this:

    Schedule    shop  order      schseq    tos    schactdate    ===============================================    002  4155526710  1         01     1994-Jul-16     002  4155526710  2         01     1994-Jul-30     002  4155526710  3         01     1994-Oct-01    002  4155526711  1         01     1994-Jul-16     002  4155526711  2         01     1994-Jul-30     002  4155526711  3         01     NULL

This is the way they would like it to appear, assuming they want to look at (tos = 01):

    order       processed    completed    confirmed    ================================================    4155526710  1994-Jul-16  1994-Jul-16  1994-Oct-01    4155526711  1994-Jul-16  1994-Jul-16  NULL

Answer:
In SQL-92, this is easy and very fast:

    SELECT order, (SELECT schactdate            FROM Schedule AS S1            WHERE S1.schseq = 1                AND S1.order = S0.order) AS processed,        (SELECT schactdate            FROM Schedule AS S2            WHERE S2.schseq = 2                AND S2.order = S0.order) AS completed,        (SELECT schactdate            FROM Schedule AS S3            WHERE S3.schseq = 3                AND S3.order = S0.order) AS confirmed        FROM Schedule AS S0        WHERE tos = :n; — set task code

But since Brian is working with SQL Server, which is not even close to the old SQL-89 standard, we have to fake it with UNIONS, which will run forever:

    INSERT INTO Work (order, processed, completed, confirmed)        SELECT order, NULL, NULL, NULL            FROM Schedule AS S0            WHERE tos = :n — set task code        UNION        SELECT order, schactdate, NULL, NULL            FROM Schedule AS S1            WHERE S1.schseq = 1                AND S1.order = :n                AND tos = :nn — set task code        UNION        SELECT order, NULL, schactdate, NULL            FROM Schedule AS S2            WHERE S2.schseq = 2                AND S2.order = :n            AND tos = :nn — set task code        UNION        SELECT order, NULL, NULL, schactdate            FROM Schedule AS S3            WHERE S3.schseq = 3                AND S3.order = :n                AND tos = :nn — set task code

This simple UNION might have to be broken down into four INSERTs.

The final query is simply:

    SELECT order, MAX(processed), MAX(completed), MAX(confirmed)        FROM Work        GROUP BY order; 

The MAX() function picks the highest non-NULL value in the group, which also happens to be the only non-NULL value in the group.

Puzzle provided courtesy of:
Joe Celko
[email protected]