ORACLE 合并两个 SELECT 和 UNION 相加答案

【问题标题】：ORACLE merge two SELECTs with UNION adding upORACLE 合并两个 SELECT 和 UNION 相加
【发布时间】：2014-07-28 04:14:23
【问题描述】：

我有这个 SQL：

SELECT 
c.customer_code,                           
SUM(units) AS tot_units,
SUM(total_amount) AS tot_money,
null as units_to_date,
null as amount_to_date,
FROM customers c
join transactions t on t.customer_code = c.customer_code
WHERE customer_active='S'
GROUP BY c.customer_code
UNION
SELECT 
c.customer_code,               
null AS tot_units,
null AS tot_money,
SUM(units) as units_to_date,
SUM(total_amount) as amount_to_date, 
FROM customers c
join transactions t on t.customer_code = c.customer_code
WHERE customer_active='S' and t.transaction_date >= (trunc(current_date) - (60 * INTERVAL '1' DAY))
GROUP BY c.customer_code;

结果是：

CUSTOMER_CODE, TOT_UNITS, TOT_AMOUNT, TO_DATE_UNITS, TO_DATE_AMOUNT
0000001        450        300         null           null          
0000001        null       null        30             15

我需要的结果是：

CUSTOMER_CODE, TOT_UNITS, TOT_AMOUNT, TO_DATE_UNITS, TO_DATE_AMOUNT
0000001        450        300         30             15

我尝试过使用 UNION，但它不起作用。

【问题讨论】：

但是你为什么大喊大叫？

标签： sql oracle select union

【解决方案1】：

另一个简单的解决方案；

WITH CUSTOMER AS (
  SELECT C.CUSTOMER_CODE,
         SUM(UNITS) AS TOT_UNITS,
         SUM(TOTAL_AMOUNT) AS TOT_MONEY,
         NULL AS UNITS_TO_DATE,
         NULL AS AMOUNT_TO_DATE,
    FROM CUSTOMERS C JOIN TRANSACTIONS T ON T.CUSTOMER_CODE = C.CUSTOMER_CODE
   WHERE CUSTOMER_ACTIVE = 'S'
GROUP BY C.CUSTOMER_CODE
UNION
  SELECT C.CUSTOMER_CODE,
         NULL AS TOT_UNITS,
         NULL AS TOT_MONEY,
         SUM(UNITS) AS UNITS_TO_DATE,
         SUM(TOTAL_AMOUNT) AS AMOUNT_TO_DATE,
    FROM CUSTOMERS C JOIN TRANSACTIONS T ON T.CUSTOMER_CODE = C.CUSTOMER_CODE
   WHERE     CUSTOMER_ACTIVE = 'S'
         AND T.TRANSACTION_DATE >= (TRUNC(CURRENT_DATE) - (60 * INTERVAL '1' DAY))
GROUP BY C.CUSTOMER_CODE)
    SELECT CUSTOMER_CODE,
           MAX(TOT_UNITS) AS TOT_UNITS,
           MAX(TOT_MONEY) AS TOT_MONEY,
           MAX(UNITS_TO_DATE) AS UNITS_TO_DATE,
           MAX(AMOUNT_TO_DATE) AS AMOUNT_TO_DATE
      FROM CUSTOMER
  GROUP BY CUSTOMER_CODE

【讨论】：

【解决方案2】：

扩展@VJHil 的答案，可以摆脱union。诀窍是使用case 过滤掉您想要的日期范围之外的所有内容：

SELECT   c.customer_code,
         SUM (units) AS tot_units,
         SUM (total_amount) AS tot_money,
         SUM (
            CASE
               WHEN t.transaction_date >=
                       (TRUNC (CURRENT_DATE) - (60 * INTERVAL '1' DAY)) THEN
                  units
               ELSE
                  NULL
            END)
            AS units_to_date,
         SUM (
            CASE
               WHEN t.transaction_date >=
                       (TRUNC (CURRENT_DATE) - (60 * INTERVAL '1' DAY)) THEN
                  total_amount
               ELSE
                  NULL
            END)
            AS amount_to_date
FROM     customers c JOIN transactions t ON t.customer_code = c.customer_code
WHERE    customer_active = 'S'
GROUP BY c.customer_code

这应该比任何两次访问数据的解决方案都要好。

【讨论】：

我同意您的方法，并且您“忠实”他关于求和 units_to_date 和 ammount_to_date 的原始条件。当然，对 NULLS 求和将导致 NULLS。因此我认为这在技术上是正确的，但我认为最终用户会想要零来代替 NULLS。
@Patrick：在这种情况下，我倾向于使用null 而不是零，因为它适用于sum 和count，所以它更实用一些。如果该列为空时需要非空结果，则可以使用nvl 或coalesce 进行转换。

【解决方案3】：

我认为你没有任何理由使用联合。试试这个，我刚刚删除了空值。

SELECT 
c.customer_code,               
SUM(units) AS tot_units,
SUM(total_amount) AS tot_money,
SUM(units) as units_to_date,
SUM(total_amount) as amount_to_date, 
MIN(transaction_date) AS first_transaction_date
FROM customers c
join transactions t on t.customer_code = c.customer_code
WHERE customer_active='S' and t.transaction_date >= (trunc(current_date) - (60 * INTERVAL '1' DAY))
GROUP BY c.customer_code;

【讨论】：

你好。谢谢，但在 total_amount 中，无论购买日期如何，我都需要该客户购买的 total_amount，而在 amount_to_date 中，我只需要一段时间内购买的金额。

【解决方案4】：

可以拆分为两个视图并加入它们。然后使用最适合的NVL() 或MAX()。

WITH V1 AS (
SELECT 
c.customer_code,                           
SUM(units) AS tot_units,
SUM(total_amount) AS tot_money,
null as units_to_date,
null as amount_to_date,
FROM customers c
join transactions t on t.customer_code = c.customer_code
WHERE customer_active='S'
GROUP BY c.customer_code
),
V2 AS (
SELECT 
c.customer_code,               
null AS tot_units,
null AS tot_money,
SUM(units) as units_to_date,
SUM(total_amount) as amount_to_date, 
FROM customers c
join transactions t on t.customer_code = c.customer_code
WHERE customer_active='S' and t.transaction_date >= (trunc(current_date) - (60 * INTERVAL '1' DAY))
GROUP BY c.customer_code)
SELECT 
 V1.CUSTOMER_CODE, NVL(V1.TOT_UNITS,V2. TOT_UNITS), NVL(V1.TOT_AMOUNT,V2. TOT_AMOUNT), NVL(V1.TO_DATE_UNITS,V2. TO_DATE_UNITS) TO_DATE_AMOUNT
FROM V1, V2
WHERE V1.CUSTOMER_CODE = V2.CUSTOMER_CODE

【讨论】：

进一步测试这个解决方案我发现当 V2 没有记录时（因为在时间间隔内没有购买）我没有得到任何记录。我将尝试使用外部连接。
好的，你也可以考虑艾伦的回答，因为这听起来更好！除非你的桌子不是很大。