ORDER BY clause in SQL query

Question

I have a SQL query ending with:

...
ORDER BY event_date ASC

So that I get these results:

|-------------------------------------|
| location_id | event_date | event_id |
|-------------------------------------|
|     112     | 2014-06-01 |    501   |
|-------------------------------------|
|      19     | 2014-06-04 |    508   |
|-------------------------------------|
|     112     | 2014-06-17 |    667   |
|-------------------------------------|
|      19     | 2014-07-07 |    434   |
|-------------------------------------|

I was wondering if I can use a different ORDER BY clause to get the following results:

|-------------------------------------|
| location_id | event_date | event_id |
|-------------------------------------|
|     112     | 2014-06-01 |    501   |
|-------------------------------------|
|     112     | 2014-06-17 |    667   |
|-------------------------------------|
|      19     | 2014-06-04 |    508   |
|-------------------------------------|
|      19     | 2014-07-07 |    434   |
|-------------------------------------|

---

EDIT:

In other words, I want the earliest event_date (with location_id = X), and then all the other rows with location_id = X, ordered by event_date .
Then the next earliest event_date (with location_id != X = Y), and again all the rows with location_id = Y, ordered by event_date .
And so on...

Answer 1

You can join onto a derived table that calculates the first event_date for each location and use that for ordering.

SELECT E1.* 
FROM   Events E1 
       JOIN (SELECT location_id, 
                    MIN(event_date) AS min_date 
             FROM   Events 
             GROUP  BY location_id) AS E2 
         ON E1.location_id = E2.location_id 
ORDER  BY E2.min_date, 
          E1.location_id, -- In case two locations have same MIN(event_date)
          E1.event_date 

Answer 2

您应该使用此ORDER BY子句获取所需的结果：

ORDER BY location_id DESC, event_date ASC