MySQL Order By in Subquery - mysql

Question

I'm trying to solve this problem on Hacker Rank:
Input (Projects table)
Output:
2015-10-28 2015-10-29
2015-10-30 2015-10-31
2015-10-13 2015-10-15
2015-10-01 2015-10-04
So what the problem is asking for is to treat consecutive end dates as part of one project and return the start and end dates of projects ordered by the date differences in ascending order. As you can see from the above example, tasks 1,2,3 are the in the same project, tasks 4,5 are in the same project and tasks 7 and 8 are their own projects.
This is one of the solutions I found:
set #sdate = null;
set #nextdate = null;
select sd, max(ed) ed2
from (
select if(#nextdate = start_date, #sdate, #sdate := start_date) as sd,
#nextdate := end_date as ed
from Projects
order by start_date
) tmp
group by sd
order by datediff(max(ed), sd)
It is using variables to store the previous end date and compare it to the current row, but I'm confused by the order by clause in the subquery:
If I take out the 'order by start_date' in the subquery, the result it returns will be wrong -- I was under the impression that in MySQL the ordering of subqueries is ignored?
My understanding was that order by is executed after the select so here it would be ordering the results from select in the subquery, but it seems like it's actually ordering the source table (Properties) before the select statement -- am I correct?
Could someone help me understand why this is the case? Thanks

Answer 1

The ORDER BY in the subquery is necessary for the variable assignment to work. This is similar to the idea of assigning row numbers in older MySQL version. When you run:
SELECT *
FROM Projects;
Without the ORDER BY, you'll get almost every Start_date value in ascending order except for the Start_date of 2015-11-04 until 2015-11-07 which gives you result in this order:
...
ID Start_date End_date
16 2015-11-04 2015-11-05
10 2015-11-07 2015-11-08
15 2015-11-06 2015-11-07
11 2015-11-05 2015-11-06
...
It start from 4 but then the next one is 7,6 then 5. This breaks the variable assignment. If you run the query like this:
SELECT *,IF(#nextdate = start_date, #sdate, #sdate := start_date) AS sd,
#nextdate := end_date AS ed
FROM Projects;
You can see the difference in the results where the Start_date is in the correct order vs with the incorrect order:
+-----+------------+------------+------------+------------+
| ID | Start_date | End_date | sd | ed |
+-----+------------+------------+------------+------------+
| 1 | 2015-10-01 | 2015-10-02 | 2015-10-01 | 2015-10-02 |
| 24 | 2015-10-02 | 2015-10-03 | 2015-10-01 | 2015-10-03 |
| 2 | 2015-10-03 | 2015-10-04 | 2015-10-01 | 2015-10-04 |
| 23 | 2015-10-04 | 2015-10-05 | 2015-10-01 | 2015-10-05 |
.......
| 3 | 2015-10-11 | 2015-10-12 | 2015-10-11 | 2015-10-12 |
| 22 | 2015-10-12 | 2015-10-13 | 2015-10-11 | 2015-10-13 |
.......
| 16 | 2015-11-04 | 2015-11-05 | 2015-11-04 | 2015-11-05 |
| 10 | 2015-11-07 | 2015-11-08 | 2015-11-07 | 2015-11-08 |
| 15 | 2015-11-06 | 2015-11-07 | 2015-11-06 | 2015-11-07 |
| 11 | 2015-11-05 | 2015-11-06 | 2015-11-05 | 2015-11-06 |
+-----+------------+------------+------------+------------+
With the subquery returning the result like above, whereby 3 of the rows should actually generate sd=2015-11-04 but instead have different value, the outer query GROUP BY will then gave 14 rows of results. By adding the ORDER BY Start_date in the subquery, you'll get the result for 2015-11-04 until 2015-11-07 like this instead:
+----+------------+------------+------------+------------+
| ID | Start_date | End_date | sd | ed |
+----+------------+------------+------------+------------+
| 16 | 2015-11-04 | 2015-11-05 | 2015-11-04 | 2015-11-05 |
| 11 | 2015-11-05 | 2015-11-06 | 2015-11-04 | 2015-11-06 |
| 15 | 2015-11-06 | 2015-11-07 | 2015-11-04 | 2015-11-07 |
| 10 | 2015-11-07 | 2015-11-08 | 2015-11-04 | 2015-11-08 |
+----+------------+------------+------------+------------+
So, it's not actually the ordering that makes the answer wrong but it's the extra rows in the end result.
Here's a fiddle to play around