I would like to ask a quick question regarding a mysql query.
I have a table named trans :
+----+---------------------+------+-------+----------+----------+
| ID | Date | User | PCNum | Customer | trans_In |
+----+---------------------+------+-------+----------+----------+
| 8 | 2013-01-23 16:24:10 | test | PC2 | George | 10 |
| 9 | 2013-01-23 16:27:22 | test | PC2 | Nick | 0 |
| 10 | 2013-01-24 16:28:48 | test | PC2 | Ted | 10 |
| 11 | 2013-01-25 16:36:40 | test | PC2 | Danny | 10 |
+----+---------------------+------+-------+----------+----------+
and another named customers :
+----+---------+-----------+
| ID | Name | Surname |
+----+---------+-----------+
| 1 | George | |
| 2 | Nick | |
| 3 | Ted | |
| 4 | Danny | |
| 5 | Alex | |
| 6 | Mike | |
.
.
.
.
+----+---------+-----------+
I want to view the sum of trans_in column for specific customers in a date range BUT ALSO include in the result set, those customers that haven’t got any records in the selected date range. Their sum of trans_in could appear as NULL or 0 it doesn’t matter…
I have the following query :
SELECT
`Date`,
Customer,
SUM(trans_in) AS 'input'
FROM trans
WHERE Customer IN('George','Nick','Ted','Danny')
AND `Date` >= '2013-01-24'
GROUP BY Customer
ORDER BY input DESC;
But this will only return the sum for ‘Ted’ and ‘Danny’ because they only have transactions after the 24th of January…
How can i include all the customers that are inside the WHERE IN (…) function, even those who have no transactions in the selected date range??
I suppose i’ll have to join them somehow with the customers table but i cannot figure out how.
Thanks in advance!!
🙂
In order to include all records from one table without matching records in another, you have to use a LEFT JOIN.
Of course, I would mention that you should be referencing customer by ID, and not by name in your related table. Your current setup leads to information duplication. If the customer changes their name, you now have to update all related records in the trans table instead of just in the customer table.