在 SQL 查询方面需要一些帮助答案

【问题标题】：Need some assistance with SQL Queries在 SQL 查询方面需要一些帮助
【发布时间】：2019-11-02 00:33:26
【问题描述】：

我目前正在使用 OracleSQL 解决两个不同的查询问题，我需要一些帮助。

第一季度。在这里，两个查询完美地分开工作。但是，我需要将它们放在一个表格/图表中，而不是有两个单独的表格/图表。我错过了什么/我需要做什么来合并两者？主要区别是“WHERE Email_List="

---显示客户数量和平均总支出，按客户细分谁在/不在我们的电子邮件列表中。

SELECT COUNT(DISTINCT Customer_Number) Cust_No_w_Email
      ,AVG(Total_Spending) Avg_Spend_w_Email  
FROM CUST_FILE
WHERE Email_List=1;

SELECT COUNT(DISTINCT Customer_Number) Cust_No_wo_Email
      ,AVG(Total_Spending) Avg_Spend_wo_Email
FROM CUST_FILE
WHERE Email_List=0;

第二季度。与上面的类似，下面的模板是我能得到的最接近的模板......

---显示食品支出的总和（提示：SUM 函数），按工作和按客户是否在我们的电子邮件列表中。仅显示食品支出总和大于 5000 的组。按食品支出总和从高到低对组进行排序。

SELECT Job, Email_List AS W_Email, SUM(Food) AS Food_Exp
FROM CUST_FILE
WHERE Email_List=1
GROUP BY Job, Email_List, Food
HAVING SUM(Food)>5000
ORDER BY Job, Food DESC;

SELECT Job, Email_List AS WO_Email, SUM(Food) AS Food_Exp
FROM CUST_FILE
WHERE Email_List=0
GROUP BY Job, Email_List, Food
HAVING SUM(Food)>5000
ORDER BY Food DESC;

如果你能帮助我，你就是在拯救一个灵魂。提前谢谢！

【问题讨论】：

我已删除（显然）未使用的产品的标签。请仅标记您的问题涉及的产品。
样本数据和期望的结果会很有帮助。您在这里似乎有多个问题，这使得问题过于宽泛。

标签： sql oracle

【解决方案1】：

第一次查询可以使用GROUP BY:

SELECT Email_List,
       COUNT(DISTINCT Customer_Number) as Cust_No_w_Email,
       AVG(Total_Spending) as Avg_Spend_w_Email  
FROM CUST_FILE
WHERE Email_List IN (0, 1);

一个问题就足够了。

【讨论】：

【解决方案2】：

对于第一季度，您可以通过两种方式组合查询：

如果您可以在单行的多列中输出结果，则可以使用条件聚合。
如果您需要将结果保存在不同的行中，您可以在 group by 中包含 email_list 列

这两种解决方案都比单独查询更有效，因为您现在查询表一次而不是两次。

对于第一个选项，这看起来像：

SELECT COUNT(DISTINCT CASE WHEN email_list = 1 THEN customer_number END) cust_no_w_email,
       AVG(CASE WHEN email_list = 1 THEN total_spending END) avg_spend_w_email,
       COUNT(DISTINCT CASE WHEN email_list = 0 THEN customer_number END) cust_no_wo_email,
       AVG(CASE WHEN email_list = 0 THEN total_spending END) avg_spend_wo_email
FROM   cust_file;

对于第二个选项：

SELECT CASE WHEN email_list = 1 THEN 'Y'
            WHEN email_list = 0 THEN 'N'
       END email_present,
     COUNT(DISTINCT customer_number) cust_no_count,
       AVG(total_spending) avg_spend
FROM   cust_file
GROUP BY CASE WHEN email_list = 1 THEN 'Y'
              WHEN email_list = 0 THEN 'N'
         END
ORDER BY CASE WHEN email_list = 1 THEN 'Y'
              WHEN email_list = 0 THEN 'N'
         END;

对于第二季度，您可以在单个查询中完成，如下所示：

SELECT job,
     CASE WHEN email_list = 1 THEN 'Y'
            WHEN email_list = 0 THEN 'N'
       END email_present,
     SUM(food) AS food_exp
FROM   cust_file
GROUP  BY job,
        CASE WHEN email_list = 1 THEN 'Y'
               WHEN email_list = 0 THEN 'N'
          END
HAVING SUM(food) > 5000
ORDER  BY CASE WHEN email_list = 1 THEN 'Y'
               WHEN email_list = 0 THEN 'N'
          END,
          SUM(food) DESC;

【讨论】：

【解决方案3】：

对于第一个问题，您可以使用以下查询 -

SELECT 'customers on list', COUNT(DISTINCT Customer_Number) Cust_No_w_Email
      ,AVG(Total_Spending) Avg_Spend_w_Email  
FROM CUST_FILE
WHERE Email_List = 1
UNION ALL
SELECT 'customers not on list', COUNT(DISTINCT Customer_Number) Cust_No_w_Email
      ,AVG(Total_Spending) Avg_Spend_w_Email  
FROM CUST_FILE
WHERE Email_List = 0

对于第二个问题，您可以尝试以下查询 -

SELECT 'Customer in list', Job, Email_List AS W_Email, SUM(Food) AS Food_Exp
FROM CUST_FILE
WHERE Email_List=1
GROUP BY Job, Email_List
HAVING SUM(Food)>5000
UNION ALL
SELECT 'Customer not in list', Job, Email_List, SUM(Food)
FROM CUST_FILE
WHERE Email_List=0
GROUP BY Job, Email_List
HAVING SUM(Food)>5000
ORDER BY Food_Exp DESC;

【讨论】：