Question

当我单独运行这一行代码时，我得到了预期的结果1：

sum(case when facilityname like '%AT%' then count(status) else 0 end) as AT_all_status

但是，当我同时运行多个case语句时（如本示例所示），所有case语句都返回NULL：

sum(case when facilityname like '%AT%' then count(status) else 0 end) as AT_all_status,
sum(case when facilityname like '%AT%' and status in ('Current','Approved') then count(status) else 0 end) as AT_approved_current,
sum(case when facilityname like '%CZ%' then count(status) else 0 end) as CZ_all_status,
sum(case when facilityname like '%CZ%' and status in ('Current','Approved') then count(status) else 0 end) as CZ_approved_current,
sum(case when facilityname like '%FGE%' then count(status) else 0 end) as FGE_all_status,
sum(case when facilityname like '%FGE%' and status in ('Current','Approved') then count(status) else 0 end) as FGE_approved_current,
sum(case when facilityname like '%FRA%' then count(status) else 0 end) as FRA_all_status,
sum(case when facilityname like '%FRA%' and status in ('Current','Approved') then count(status) else 0 end) as FRA_approved_current

当他们实际上应该返回自己的数字时，而当他们一个人跑时这样做时，我是否缺少案例陈述行为？

Answer 1

在标准SQL中，这不应返回任何结果（错误除外）：

sum(case when facilityname like '%AT%' then count(status) else 0 end) as AT_all_status

因为聚合函数不能嵌套。 Oracle确实允许嵌套它们，但是我不推荐使用该功能-子查询非常简单。

无论如何，我很确定您的实际意图：

sum(case when facilityname like '%AT%' then 1 else 0 end) as AT_all_status

Answer 2

另一种仅使用一个聚合函数而不嵌套的查询方式如下：

count(case when facilityname like '%AT%' then status end)

与其他case语句一起使用时，CASE表达式返回NULL，在单独时返回预期结果

2 个答案: