Question

我使用以下代码按类型在表中计算记录：

select null link, CASE when SUBSTR(PRJA_TYPE,1,3) ='CHO' then 'CHO'
  else PRJA_TYPE
END as label, 
COUNT(PRJA_TYPE) Contract_Type
from  "SPM"."PROJECT_AGREEMENTS"
where PRJA_ACTIVE = 1 and PRJA_TYPE IS NOT NULL
 group by PRJA_TYPE
order by PRJA_TYPE

它给了我以下结果：

LINK    LABEL   CONTRACT_TYPE

-       Ballpark   14 
-       CHO        7 
-       CHO        7 
-       CHO        6 
-       CHO        30 
-       CHO        11 
-       CHO        9 
-       CHO        1 
-       CHO        2 
-       LOA        9

这部分按预期工作。然后，我需要将Label列中标记为CHO的项目进行汇总。所以我的最终结果将是这样的：

LINK    LABEL   CONTRACT_TYPE

-       Ballpark   14 
-       CHO        73 
-       LOA        9

非常感谢任何帮助。

Answer 1

您需要通过表达式汇总：

select null as link,
       (case when SUBSTR(PRJA_TYPE, 1, 3) = 'CHO' then 'CHO'
             else PRJA_TYPE
        end) as label, 
       COUNT(*) Contract_Type
from  "SPM"."PROJECT_AGREEMENTS"
where PRJA_ACTIVE = 1 and PRJA_TYPE IS NOT NULL
group by (case when SUBSTR(PRJA_TYPE, 1, 3) = 'CHO' then 'CHO'
               else PRJA_TYPE
          end)
order by label;

Answer 2

cte我会猜测

with cte as
(select null link, CASE when SUBSTR(PRJA_TYPE,1,3) ='CHO' then 'CHO'
  else PRJA_TYPE
END as label, 
COUNT(PRJA_TYPE) Contract_Type
from  "SPM"."PROJECT_AGREEMENTS"
where PRJA_ACTIVE = 1 and PRJA_TYPE IS NOT NULL
 group by PRJA_TYPE
order by PRJA_TYPE
)

select link, label, sum(contract_type) contract_type from cte group by link,label

Answer 3

您可以在group-clause中使用case语句：

select 
  null link, 
  CASE when SUBSTR(PRJA_TYPE,1,3) ='CHO' then 'CHO' else PRJA_TYPE END as label, 
COUNT(PRJA_TYPE) Contract_Type
from  "SPM"."PROJECT_AGREEMENTS"
where PRJA_ACTIVE = 1 and PRJA_TYPE IS NOT NULL
group by CASE when SUBSTR(PRJA_TYPE,1,3) ='CHO' then 'CHO' else PRJA_TYPE END
order by PRJA_TYPE

作为替代方案（例如，如果您无法修改查询），您可以使用第二个查询来汇总您的数字：

select link, label, sum(CONTRACT_TYPE)
from (
--your first query
) x
group by link, label

总结CASE语句的结果

3 个答案: