Question

我需要针对“按日期售出的特定产品的平均数量。每天售出的至少一种产品”编写SQL查询。

def print_dirs(directories):
    try:
        dirpath, dirnames, files = next(directories)
        print(dirpath)              # print current path; no need for join here
        for _ in dirnames:          # once for each sub-directory...
            print_dirs(directories) # ... recursively call generator
        for name in files:          # now, print files in current directory
            print(os.path.join(dirpath, name))
    except StopIteration:
        pass

print_dirs(os.walk(topdir))

P.S。 PID表示产品ID。

此查询可以吗？这应该给出正确的答案吗？

Answer 1

考虑使用distinct列的date个计数

SELECT PID, 
       COUNT(PID)/COUNT(distinct date_) as "Avg.Product Sold By Days"
  FROM SOLD
 GROUP BY PID;

Answer 2

您可以在下面尝试此sql查询。基本上，它将为您拥有的每种产品返回平均“销售”数。它将按每个不同的产品ID分组。请向我们提供您的表格等数据结构。

  SELECT product_ID, trans_date
           Sum(sales_of_product) / COUNT(DISTINCT sold_transaction) AS 'avg'
    FROM SOLD
    GROUP BY product_ID

MySQL-当天售出的特定产品的平均数量

2 个答案: