计算SQL百分比

Question

我正在尝试使用另一个表创建表（列），但是在获取所需的列时遇到了麻烦。 请指导。

表名称：交易，客户，Prod_cat_info

SELECT DISTINCT T.cust_id, 
                c.gender, 
                C.city_code, 
                Datediff(mm, c.dob, Getdate())             AS AGE_IN_MONTHS, 
                (SELECT Count(transaction_id) 
                 FROM   transactions 
                 GROUP  BY tran_date)                      AS 
                basket_count_per_day, 
                (SELECT Sum(total_amt) 
                 FROM   transactions)                      AS Tot_sal_amt, 
                (SELECT Sum(qty) 
                 FROM   transactions)                      AS tot_sal_qnty, 
                (SELECT Count(DISTINCT prod_cat) 
                 FROM   dbo.prod_cat_info)                 AS Unq_cat_cnt, 
                (SELECT DISTINCT prod_subcat 
                 FROM   dbo.prod_cat_info)                 AS Unq_Scat_cnt, 
                (SELECT DISTINCT store_type 
                 FROM   dbo.transactions)                  AS Unq_chnl_cnt, 
                (SELECT Max(tran_date) 
                 FROM   dbo.transactions)                  AS 
                Last_Transaction_date, 
                (SELECT Avg(qty) / Count(tran_date) 
                 from transactions)                        AS 
                                                           Avg_basket_qty, 
                (SELECT Avg(total_amt) / Count(tran_date) 
                 from transactions)                                          
                                                         AS Avg_basket_val 
INTO   customer_360 
FROM   dbo.customer AS c 
       INNER JOIN dbo.transactions AS t 
               ON t.cust_id = c.customer_id 
       INNER JOIN dbo.prod_cat_info AS pci 
               ON pci.prod_cat_code = t.prod_cat_code   

错误信息：

消息8120，第16级，状态1，第1行
列'dbo.Transactions.cust_id'在选择列表中无效，因为它既不包含在聚合函数中也不在GROUP BY子句中。

Answer 1

最后两列是聚合查询。 因为您有一个没有FROM的SELECT，它将查询外部查询中指定的表。

            (SELECT Avg(qty) / Count(tran_date))       AS Avg_basket_qty, 
            (SELECT Avg(total_amt) / Count(tran_date)) AS Avg_basket_val 

在这种情况下，您要汇总qty，tran_date和total_amount，因此每隔一列就需要分组或汇总。

添加以下内容应使其成为有效的查询，但是它可能无法满足您的要求！

GROUP BY T.cust_id, c.Gender, C.city_code, DATEDIFF(mm,c.DOB,GETDATE())     

Answer 2

GROUP BY产生的行对于在该子句下指定的列中保存的值的组合是唯一的。 您的整体查询应该看起来像这样，在GROUP BY子句中列出了“非聚合列”

SELECT
      T.cust_id    --<< this is a "non-aggregating" column
    , c.gender     --<< this is a "non-aggregating" column
    , C.city_code  --<< this is a "non-aggregating" column
    , DATEDIFF(mm, c.dob, GETDATE())    AS AGE_IN_MONTHS --<< this is a "non-aggregating" column

    , SUM(total_amt)                    AS Tot_sal_amt
    , SUM(qty)                          AS tot_sal_qnty
    , COUNT(DISTINCT store_type)        AS Unq_chnl_cnt
    , MAX(tran_date)                    AS Last_Transaction_date
    , AVG(qty) / COUNT(tran_date)       AS Avg_basket_qty
    , AVG(total_amt) / COUNT(tran_date) AS Avg_basket_val 

--INTO customer_360
FROM dbo.customer AS c
INNER JOIN dbo.transactions AS t ON t.cust_id = c.customer_id
GROUP BY
      T.cust_id      --<< list each non-aggregating column here
    , c.gender       --<< list each non-aggregating column here
    , C.city_code    --<< list each non-aggregating column here
    , DATEDIFF(mm, c.dob, GETDATE())  --<< list each non-aggregating column here

• 请注意，如果您在同一选择查询中（正确）使用GROUP BY ，则“ 完全选择” 完全是完全多余的 。 同样，“选择不同的对象”所做的事情少于GROUP BY对您的影响（例如SUM / MIN / MAX / AVG值）。

进一步说明：

AVG(qty) / COUNT(tran_date)将产生一个整数结果，您可能想这样做AVG（qty）/（COUNT（tran_date）* 1.0）

此INNER JOIN dbo.prod_cat_info AS pci ON pci.prod_cat_code = t.prod_cat_code已删除INNER JOIN dbo.prod_cat_info AS pci ON pci.prod_cat_code = t.prod_cat_code可能会乘以结果。 将上面的查询与包含此联接的另一个查询进行比较。 如果聚合受到影响，那么您将需要对该表进行子查询并将其联接。 因此，以下几列也丢失了：

            (SELECT Count(DISTINCT prod_cat) 
             FROM   dbo.prod_cat_info)                 AS Unq_cat_cnt, 
            (SELECT DISTINCT prod_subcat 
             FROM   dbo.prod_cat_info)                 AS Unq_Scat_cnt,

这些也已删除，因为我不确定您的意图是什么：

            (SELECT Count(transaction_id) 
             FROM   transactions 
             GROUP  BY tran_date)  AS  basket_count_per_day,

这样的子查询不允许返回多个值； 这将做。 因此，这将导致错误，您需要确定如何将“每日平均值”合并到一个表中，每个客户一行。