SQL GROUP BY with "default values"

I'm trying to create SELECT statement with a GROUP BY clause, which should return "default values".

Imagine the following simple MySQL table:

CREATE TABLE `tracker` (
  `id` INTEGER PRIMARY KEY auto_increment,
  `date` DATETIME NOT NULL,
  `customer_id` INTEGER NOT NULL
);

The table contains only one record:

INSERT INTO `tracker` (`date`, `customer_id`) VALUES('2010-05-03', 1);

After wards I'm executing the following SQL query:

SELECT DATE(`date`), COUNT(customer_id) FROM tracker
WHERE DATE(`date`) >= '2010-05-01' AND DATE(`date`) <= '2010-05-05'
GROUP BY DATE(`date`) ORDER BY DATE(`date`);

And get the expected result set:

+----+---------------------+-------------+
| id | date                | customer_id |
+----+---------------------+-------------+
|  1 | 2010-05-10 00:00:00 |           1 |
+----+---------------------+-------------+

H开发者_C百科owever, I would like the result set to look like this:

+--------------+--------------------+
| DATE(`date`) | COUNT(customer_id) |
+--------------+--------------------+
| 2010-05-01   |                  0 |
| 2010-05-02   |                  0 |
| 2010-05-03   |                  1 |
| 2010-05-04   |                  0 |
| 2010-05-05   |                  0 |
+--------------+--------------------+

Is it possible to achieve this behavior?

---

You could build a temporary table of the valid dates in the range and then incorporate that into your query - that's about the only way forward that I can immediately see...

Martin

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As Martin said, the best solution is to create a temp table with dates.

Then there's 2 approaches:

• Do an outer join with that temp table and do a group by on result, OR

• group by on the original table + UNION select date,0 as count from date_table d where not exists (select 1 from customer c where c.date=d.date)