daily_average
列始终返回零。默认时间戳值是过去一周。在获取每日平均订单价值方面,我对此做错了什么?
SELECT
SUM(price+shipping_price) AS total_sales,
COUNT(id) AS total_orders,
AVG(price+shipping_price) AS order_total_average,
(SELECT
SUM(quantity)
FROM `order_product`
INNER JOIN `order` ON (
`order`.id = order_product.order_id AND
`order`.created >= '.$startTimestamp.' AND
`order`.created <= '.$endTimestamp.' AND
`order`.type_id = '.$type->getId().' AND
`order`.fraud = 0
)
) as total_units,
SUM(price+shipping_price)/DATEDIFF('.$endTimestamp.', '.$startTimestamp.') as daily_average
FROM `order`
WHERE created >= '.$startTimestamp.' AND
created <= '.$endTimestamp.' AND
fraud = 0 AND
type_id = '.$type->getId().'
答案 0 :(得分:0)
您正在使用聚合函数(SUM,COUNT,AVG)而没有聚合命令(分组依据)。我认为你的SQL比它需要的更复杂(不需要内部选择)。
这是一个应该工作的SQL命令(很难在没有测试数据的情况下进行测试;))
SELECT
COUNT(id) total_orders,
SUM(finalprice) total_sales,
AVG(finalprice) order_average,
SUM(units) total_units,
SUM(finalprice)/DATEDIFF('.$endTimestamp.', '.$startTimestamp.') daily_average
FROM (
SELECT
o.id id,
o.price+o.shipping_price finalprice,
SUM(p.quantity) units
FROM order o INNER JOIN order_product p ON p.order_id=o.id
WHERE o.created>='.$startTimestamp.'
AND o.created<='.$endTimestamp.'
AND o.fraud=0
AND o.type_id='.$type->getId().'
GROUP BY p.order_id
) t;
答案 1 :(得分:0)
在该部门中投射其中一个元素是否适合您?
SELECT
SUM(price+shipping_price) AS total_sales,
COUNT(id) AS total_orders,
AVG(price+shipping_price) AS order_total_average,
(SELECT
SUM(quantity)
FROM `order_product`
INNER JOIN `order` ON (
`order`.id = order_product.order_id AND
`order`.created >= '.$startTimestamp.' AND
`order`.created <= '.$endTimestamp.' AND
`order`.type_id = '.$type->getId().' AND
`order`.fraud = 0
)
) as total_units,
CAST(SUM(price+shipping_price) AS float)/DATEDIFF('.$endTimestamp.', '.$startTimestamp.') as daily_average
FROM `order`
WHERE created >= '.$startTimestamp.' AND
created <= '.$endTimestamp.' AND
fraud = 0 AND
type_id = '.$type->getId().'