Question

Possible Duplicate:
SQL “transform” query

I can’t figure out how to compose a SQL statement to display the different values of a table as separate columns…here’s an example:

Table A - Project
= = = = = = = = = = = = = =
ID | Project
= = = = = = = = = = = = = =
1  | Google

Table B - Tasks
= = = = = = = = = = = = = = = = = = = = = = = = = = = =
ID  | Project ID    | Task_name | Date Complete
= = = = = = = = = =  = = = = = = = = = = = = = = = = = =
1   |  1            | Build     | 10/30/12
2   |  1            | Revise    | 11/15/12
3   |  1            | Launch    | 11/30/12

SQL outputs in the following format:

Website   | Build Date  | Revise Date   | Launch Date
= = = = = = = = = =  = = = = = = = = = = = = = = = = = =
Google    | 10/30/12    | 11/15/12      | 11/30/12  

Answer 1

This is known as a PIVOT but MySQL does not have a PIVOT function but you can replicate it using an aggregate function and a CASE statement:

select p.project as website,
  max(case when t.task_name = 'build' then date_format(`Date Complete`, '%m/%d/%y') end) BuildDate,
  max(case when t.task_name = 'Revise' then date_format(`Date Complete`, '%m/%d/%y') end) ReviseDate,
  max(case when t.task_name = 'Launch' then date_format(`Date Complete`, '%m/%d/%y') end) LaunchDate
from project p
inner join tasks t
  on p.id = t.`Project ID`
group by p.project

See SQL Fiddle with Demo

Result:

| WEBSITE | BUILDDATE | REVISEDATE | LAUNCHDATE |
-------------------------------------------------
|  Google |  10/30/12 |   11/15/12 |   11/30/12 |