SELECT c.customer_name,
sc.customer_id,
SUM(
(nvl(sc.year_1_new, 0) + nvl(sc.year_2_new, 0) + nvl(sc.year_3_new, 0) +
nvl(sc.year_4_new, 0) + nvl(sc.year_5_new, 0)) * sc.suggested_net
) AS ttl_new,
SUM(
(nvl(sc.year_1_nl, 0) + nvl(sc.year_2_nl, 0) + nvl(sc.year_3_nl, 0) +
nvl(sc.year_4_nl, 0) + nvl(sc.year_5_nl, 0)) * sc.suggested_net
) AS ttl_exist,
SUM(ttl_new - ttl_exist) AS ttl_delta
FROM scenario_customers sc,
customers c
WHERE sc.scenario_id = 10
AND sc.customer_id = c.customer_id
GROUP BY sc.customer_id,
c.customer_name
ORDER BY c.customer_name
我希望能够从ttl_exist col中减去ttl_new col,并且当我使用动态名称时出现错误,但是如果只是将两个sum函数的全部内容粘贴到第3个和函数它的工作原理。所以只是想知道这是否可能,它肯定更容易阅读。
这适用于Oracle 8i
答案 0 :(得分:0)
大多数数据库的作用域规则不允许您在同一SELECT语句中使用别名。您可以使用子查询执行此操作:
select c.customer_name, sc.customer_id, ttl_new, ttl_exist, (ttl_new - ttl_exist) as ttl_delta
from (SELECT c.customer_name, sc.customer_id,
SUM((nvl(sc.year_1_new, 0) + nvl(sc.year_2_new, 0) + nvl(sc.year_3_new, 0) + nvl(sc.year_4_new, 0) + nvl(sc.year_5_new, 0)) * sc.suggested_net) AS ttl_new,
SUM((nvl(sc.year_1_nl, 0) + nvl(sc.year_2_nl, 0) + nvl(sc.year_3_nl, 0) + nvl(sc.year_4_nl, 0) + nvl(sc.year_5_nl, 0)) * sc.suggested_net) AS ttl_exist
FROM scenario_customers sc join
customers c
on sc.customer_id = c.customer_id
WHERE sc.scenario_id = 10
GROUP BY sc.customer_id, c.customer_name
) t
ORDER BY c.customer_name
我还修复了你的连接语法。