Question

以下是我的查询，给出了相应值之和的结果。如何计算每列预期Id的平均值？

Select 
   count(StoreId) as TotalReview, 
   sum(Case OnTimeDelivery when 'Within Time' then 1 else 0 end) as OnTimeDeliveryWithinTime,
   sum(Case OnTimeDelivery when 'Little Delay' then 1 else 0 end) as  OnTimeDeliveryLittleDelay,
   sum(Case OnTimeDelivery when 'Excess Delay' then 1 else 0 end) as OnTimeDeliveryExcessDelay,
from 
   StoreReviews 
where 
   StoreId = 1

Answer 1

您需要使用GROUP BY子句

SELECT storeId, COUNT(ontimedelivery), AVG(<column>)
FROM storereviews
WHERE storeid=1
GROUP BY ontimedelivery

Answer 2

您可以根据自己的查询获得每个商店的SUM，但您必须添加GROUP BY，如：

SELECT Storeid, COUNT(StoreId) as TotalReview, 
SUM(Case OnTimeDelivery WHEN 'Within Time' THEN 1 ELSE 0 END) as OnTimeDeliveryWithinTime,
SUM(Case OnTimeDelivery WHEN 'Little Delay' THEN 1 ELSE 0 END) as  OnTimeDeliveryLittleDelay,
SUM(Case OnTimeDelivery WHEN 'Excess Delay' THEN 1 ELSE 0 END) as OnTimeDeliveryExcessDelay
FROM StoreReviews 
GROUP BY Storeid

现在，如果您希望平均值不是针对每个商店，而是针对所有商店并且基于OnTimeDelivery，Within Time或{ {1}}或Little Delay您可以像这样使用Excess Delay：

AVG

查看我的Fiddle Demo

来自sql表的审阅者的平均得分

2 个答案: