尝试COALESCE:
DB::raw('COALESCE(COUNT(*), 0) as `orders_per_month`')
此功能允许您定义的默认值时,否则它是NULL。
http://dev.mysql.com/doc/refman/5.7/en/comparison-operators.html#function_coalesce
UPDATE:
我认为,如果你选择哪个具有上个月订单唯一的客户,那么你肯定不会得到这些客户没有订单。
你可以实现你需要的东西像什么:
mysql> select * from cu;
+----+----------+
| id | name |
+----+----------+
| 1 | Google |
| 2 | Yahoo |
| 3 | Mirosoft |
+----+----------+
3 rows in set (0.00 sec)
mysql> select * from so;
+----+-------------+---------------------+-------+
| id | customer_id | created_at | price |
+----+-------------+---------------------+-------+
| 1 | 1 | 2016-08-23 12:12:12 | 2 |
| 2 | 1 | 2016-09-24 12:14:13 | 3 |
| 3 | 2 | 2016-09-25 00:00:00 | 5 |
| 4 | 2 | 2016-09-12 09:00:00 | 3 |
+----+-------------+---------------------+-------+
4 rows in set (0.00 sec)
mysql> select cu.id as customer_id, coalesce(sum(so.price),0) as total, coalesce(count(so.customer_id),0) as orders_per_month from cu left join so on (cu.id = so.customer_id) where so.created_at >= '2016-09-01 00:00:00' or so.created_at is null group by cu.id;
+-------------+-------+------------------+
| customer_id | total | orders_per_month |
+-------------+-------+------------------+
| 1 | 3 | 1 |
| 2 | 8 | 2 |
| 3 | 0 | 0 |
+-------------+-------+------------------+
3 rows in set (0.00 sec)
然而,得到的查询将不会为这种或那种方式非常有效,你必须扫描所有客户,并加入让那些没有命令。获得带有订单的客户列表并单独订购可能会更快,并将其与UNION或通过应用程序代码合并。
mysql> select customer_id, sum(total) as total, sum(orders_per_month) as orders_per_month from (select id as customer_id, 0 as total, 0 as orders_per_month from cu union all select customer_id, sum(so.price) as total, count(so.customer_id) as orders_per_month from so where created_at >= '2016-09-01 00:00:00'group by customer_id) agg group by customer_id;
+-------------+-------+------------------+
| customer_id | total | orders_per_month |
+-------------+-------+------------------+
| 1 | 3 | 1 |
| 2 | 8 | 2 |
| 3 | 0 | 0 |
+-------------+-------+------------------+
3 rows in set (0.00 sec)
与所有可能的值创建某种日历表。做外部连接。 – jarlh
或者从您的客户列表中进行选择,并在您的纸质桌上进行左连接。 – aynber