Question

我基本上有两张桌子：

1）计费费用：这里我们存储餐馆的ID（restid），费用id（chargeid），充电时间（费用发生时的时间），chargeamount（实际费用的金额）.charge id是外键billingchargedetails表。

2）billingchargedetails：这里我们存储所有可能费用的详细信息。 chargeid（主键int），chargename（费用名称），perdaycost（每天的费用）

我的期望：

每家餐厅每次收费的总费用汇总报告。

表格中的当前条目是：

select * from billingcharges;
+--------+----------+---------------+--------------+
| restid | chargeid | chargetime    | chargeamount |
+--------+----------+---------------+--------------+
|      1 |        1 | 1536363636363 |          700 |
|      2 |        1 | 1536363636363 |          500 |
|      1 |        1 | 1568789654123 |          500 |
+--------+----------+---------------+--------------+

select * from billingchargedetails;

+----------+--------------------+------------------+
| chargeid | chargename         | chargecostperday |
+----------+--------------------+------------------+
|        1 | Base Charge        |               50 |
|        2 | Spotlight Listing  |               50 |
|        3 | Gold Notification  |              500 |
|        4 | Discount (FIRST50) |               18 |
+----------+--------------------+------------------+

对chargeid的一个简单联接最终没有按照预期给我数量和总和。所以我需要某种形式的左或右外连接，我知道并尝试了很多

我尝试了左连接，如下所示：

select restid, B.chargeid, chargename, count(B.chargeid) as qty,
  sum(ifnull(chargeamount,0)) as total 
from billingcharges as B 
left join billingchargedetails as C on B.chargeid=C.chargeid 
group by restid,B.chargeid;

+--------+----------+-------------+-----+-------+
| restid | chargeid | chargename  | qty | total |
+--------+----------+-------------+-----+-------+
|      1 |        1 | Base Charge |   2 |  1200 |
|      2 |        1 | Base Charge |   1 |   500 |
+--------+----------+-------------+-----+-------+

这确实起作用并且总结了一些东西，但每家餐馆都有丢失的费用。即使它们不存在于计费表中，即左表，我需要它与数量0和总计0。

我尝试了一个右连接，mysql从左表中的非现有条目中选择了一个随机值，如下所示：

select restid, B.chargeid, chargename, count(B.chargeid) as qty,
  sum(ifnull(chargeamount,0)) as total 
from billingcharges as B 
right join billingchargedetails as C on B.chargeid=C.chargeid 
group by restid,B.chargeid;

+--------+----------+-------------------+-----+-------+
| restid | chargeid | chargename        | qty | total |
+--------+----------+-------------------+-----+-------+
|   NULL |     NULL | Spotlight Listing |   0 |     0 |
|      1 |        1 | Base Charge       |   2 |  1200 |
|      2 |        1 | Base Charge       |   1 |   500 |
+--------+----------+-------------------+-----+-------+

预期输出类似于：

 restid chargeid  chargename   qty totalamount
  1        1      Base Charge   2   1200
  1        2      Spotlight     0   0
  1        3      Gold          0   0
  1        4      Discount      0   0
  2        1      Base Charge   1   500
  2        2      Spotlight     0   0
  2        3      Gold          0   0
  2        4      Discount      0   0
  'same as above expected for each restid in billingcharges'

Answer 1

在进行外部联接之前，您需要生成餐馆的交叉产品以收取类型费用。

以下内容（但我没有测试过）：

SELECT R.restid, D.chargename, COUNT(B.chargeid) AS qty,
  SUM(IFNULL(B.chargeamount, 0)) AS total
FROM (SELECT DISTINCT restid FROM billingcharges) AS R
CROSS JOIN billingchargedetails AS D
LEFT JOIN billingcharges AS B ON R.restid=B.restid AND D.chargeid=B.chargeid
GROUP BY R.restid, D.chargename;

在这个例子中，R和D的交叉积是每个餐馆与每种收费类型交叉。

当然，并非每家餐厅都有这些费用。因此，对于计费费用的外部联接，找到对于餐馆和餐馆的每个相应组合存在的那些行。充电类型。

MYSQL连接查询无法按预期工作

1 个答案: