MySQL Left Регистрация Count Sum с повторяющимися строками

Question

У меня есть три таблицы: 1. магазинов 2. фруктовых 3. Вегета

Я оставил присоединиться stores таблицы в то время как fruity и vegeta как союз.

Количество и сумма, которые мне нужно группировать по order_id и vendor_id, но у них все еще есть повторяющиеся строки.

Вот таблицы:

mysql> SELECT * FROM stores; 
+-----+-----------------+ 
| id | store_name  | 
+-----+-----------------+ 
| 701 | Machette Grill | 
| 702 | Mateau Conserva | 
+-----+-----------------+ 

mysql> SELECT * FROM fruity; 
+------+----------+-----------+----------+-------+ 
| id | order_id | vendor_id | store_id | sales | 
+------+----------+-----------+----------+-------+ 
| 1816 |  86 |   1 |  701 | 1000 | 
| 1817 |  86 |  11 |  701 | 1000 | 
| 1818 |  86 |  12 |  701 | 1000 | 
| 1819 |  86 |   1 |  702 | 1000 | 
| 1820 |  86 |   1 |  702 | 1000 | 
| 1821 |  86 |  11 |  702 | 1000 | 
| 1822 |  86 |  12 |  702 | 1000 | 
| 1823 |  86 |   1 |  702 | 1000 | 
| 1824 |  86 |   1 |  702 | 1000 | 
| 1825 |  86 |   1 |  702 | 1000 | 
| 1826 |  86 |  11 |  702 | 1000 | 
| 1827 |  86 |  12 |  702 | 1000 | 
| 1828 |  86 |   1 |  701 | 1000 | 
+------+----------+-----------+----------+-------+ 

mysql> SELECT * FROM vegeta; 
+----+----------+-----------+----------+-------+ 
| id | order_id | vendor_id | store_id | sales | 
+----+----------+-----------+----------+-------+ 
| 15 |  86 |  11 |  701 | 2000 | 
| 16 |  86 |  12 |  702 | 2000 | 
| 17 |  86 |  11 |  701 | 2000 | 
| 18 |  86 |  12 |  702 | 2000 | 
| 19 |  86 |  11 |  701 | 2000 | 
| 20 |  86 |  12 |  702 | 2000 | 
+----+----------+-----------+----------+-------+ 

код я столкнулся ниже:

SELECT 
s.order_id, 
s.store_id, 
c.store_name, 
s.vendor_id, 
s.fruity_count, 
s.vegeta_count, 
s.fruity_sum, 
s.vegeta_sum 
FROM stores AS c 
LEFT JOIN (
    SELECT 
    order_id AS order_id, 
    store_id AS store_id, 
    vendor_id AS vendor_id, 
    COUNT(sales) AS fruity_count, 
    0 AS vegeta_count, 
    SUM(sales) AS fruity_sum, 
    0 AS vegeta_sum 
    FROM fruity 
    WHERE order_id = 86 
    GROUP BY store_id,vendor_id 
    UNION 
    SELECT 
    order_id AS order_id, 
    store_id AS store_id, 
    vendor_id AS vendor_id, 
    0 AS fruity_count, 
    COUNT(sales) AS vegeta_count, 
    0 AS fruity_sum, 
    SUM(sales) AS vegeta_sum 
    FROM vegeta 
    WHERE order_id = 86 
    GROUP BY store_id,vendor_id) AS s ON s.store_id = c.id 
WHERE s.order_id = 86 
ORDER BY s.store_id ASC; 

Я просто нужен результат, как это показано ниже:

+----------+----------+-----------------+-----------+--------------+--------------+------------+------------+ 
| order_id | store_id | store_name  | vendor_id | fruity_count | vegeta_count | fruity_sum | vegeta_sum | 
+----------+----------+-----------------+-----------+--------------+--------------+------------+------------+ 
|  86 |  701 | Machette Grill |   1 |   2 |   0 |  2000 |   0 | 
|  86 |  701 | Machette Grill |  11 |   1 |   3 |  1000 |  6000 | 
|  86 |  701 | Machette Grill |  12 |   1 |   0 |  1000 |   0 | 
|  86 |  702 | Mateau Conserva |  12 |   2 |   3 |  2000 |  6000 | 
|  86 |  702 | Mateau Conserva |   1 |   5 |   0 |  5000 |   0 | 
|  86 |  702 | Mateau Conserva |  11 |   2 |   0 |  2000 |   0 | 
+----------+----------+-----------------+-----------+--------------+--------------+------------+------------+ 

Все, что я хочу для этого нужно перемещать повторяющиеся строки до заполнения vegeta_count и vegeta_sum. Не создавать новую строку для их подсчета.

Пожалуйста, помогите. Спасибо.

http://sqlfiddle.com/#!9/99387/1/0

Answer 1

Вы должны использовать group by в внешнем запросе, как это:

SELECT 
s.order_id, 
s.store_id, 
c.store_name, 
s.vendor_id, 
SUM(s.fruity_count) AS fruity_count, 
SUM(s.vegeta_count) AS vegeta_count, 
SUM(s.fruity_sum) AS fruity_sum, 
SUM(s.vegeta_sum) AS vegeta_sum 
FROM stores AS c 
LEFT JOIN (
    SELECT 
    order_id AS order_id, 
    store_id AS store_id, 
    vendor_id AS vendor_id, 
    COUNT(sales) AS fruity_count, 
    0 AS vegeta_count, 
    SUM(sales) AS fruity_sum, 
    0 AS vegeta_sum 
    FROM fruity 
    WHERE order_id = 86 
    GROUP BY store_id,vendor_id 
    UNION 
    SELECT 
    order_id AS order_id, 
    store_id AS store_id, 
    vendor_id AS vendor_id, 
    0 AS fruity_count, 
    COUNT(sales) AS vegeta_count, 
    0 AS fruity_sum, 
    SUM(sales) AS vegeta_sum 
    FROM vegeta 
    WHERE order_id = 86 
    GROUP BY store_id,vendor_id) AS s ON s.store_id = c.id 
WHERE s.order_id = 86 
GROUP BY store_id,vendor_id 
ORDER BY s.store_id ASC 

SQLFiddle Demo