Question

我有一个简单的数据库来模拟电影租赁服务。我有以下3个选择查询，它们可以自行完成：

计算所有租金收入：

SELECT SUM(PRICE) AS RENTAL_REVENUE
    FROM RENTAL

输出：

    +----------------+
    | RENTAL_REVENUE |
    +----------------+
    |      39.92     |
    +----------------+

计算所支付滞纳金的所有收入：

SELECT SUM(NULLIF(LATEFEE, 0)) AS PAID_LATE_FEES
    FROM RENTAL
    WHERE RETURNED = 1

输出：

    +----------------+
    | PAID_LATE_FEES |
    +----------------+
    |       2.99     |
    +----------------+

计算所有尚未支付的滞纳金的总和：

SELECT SUM(NULLIF(LATEFEE, 0)) AS OUTSTANDING_LATE_FEES
    FROM RENTAL
    WHERE RETURNED = 0

输出：

    +-----------------------+
    | OUTSTANDING_LATE_FEES |
    +-----------------------+
    |          5.98         |
    +-----------------------+

我想将这3个查询的结果合并到一个包含3列的表中，如下所示：

+----------------+----------------+-----------------------+
| RENTAL_REVENUE | PAID_LATE_FEES | OUTSTANDING_LATE_FEES |
+----------------+----------------+-----------------------+
|      39.92     |       2.99     |          5.98         |
+----------------+----------------+-----------------------+

我能够通过以下查询实现此目的，但1 = 1无意义向我暗示可能有更好的方法。

SELECT rental_revenue + paid_late_fees + outstanding_late_fees AS TOTAL_REVENUE, 
       rental_revenue, 
       paid_late_fees, 
       outstanding_late_fees 
FROM   (SELECT SUM(price) AS RENTAL_REVENUE 
        FROM   rental) 
       inner join (SELECT SUM(Nullif(latefee, 0)) AS PAID_LATE_FEES 
                   FROM   rental 
                   WHERE  returned = 1) 
               ON 1 = 1 
       inner join (SELECT SUM(Nullif(latefee, 0)) AS OUTSTANDING_LATE_FEES 
                   FROM   rental 
                   WHERE  returned = 0) 
               ON 1 = 1;

有更好的方法吗？

Answer 1

您可以使用SUM或CASE-WHEN选择所有记录并在DECODE内构建过滤器：

SELECT SUM(price) AS rental_revenue,
       SUM(CASE 
             WHEN returned = 1 AND latefee <> 0 THEN latefee
             ELSE 0
           END) AS paid_late_fees,
       SUM(CASE
             WHEN returned = 0 AND latefee <> 0 THEN latefee
             ELSE 0
           END) AS outstanding_late_fees             
FROM rental

Answer 2

您只需使用case表达式或decode()函数

即可

select sum(price) as return_revenue
     , sum(case 
             when returned = 1 
             then latefee
           end)  as paid_late_fees
     , sum(case
            when returned = 0
            then latefee
           end)  as outstanding_late_fees           
  from rental

将3列总和合并为一个表

2 个答案: