Sql 如何使用分析计数进行条件计数?
使用这个示例,我如何调整sql以报告清单id是否通过了所有测试Sql 如何使用分析计数进行条件计数?,sql,oracle11g,analytics,Sql,Oracle11g,Analytics,使用这个示例,我如何调整sql以报告清单id是否通过了所有测试 with listing_row as ( select 1 as listing_id, 'TEST1' as listing_test, 'Y' as pass_yn from dual union all select 1 as listing_id, 'TEST2' as listing_test, 'Y' as pass_yn from dual union all select 1 as listing_id, '
with listing_row
as
(
select 1 as listing_id, 'TEST1' as listing_test, 'Y' as pass_yn from dual union all
select 1 as listing_id, 'TEST2' as listing_test, 'Y' as pass_yn from dual union all
select 1 as listing_id, 'TEST3' as listing_test, 'Y' as pass_yn from dual union all
select 2 as listing_id, 'TEST1' as listing_test, 'N' as pass_yn from dual union all
select 2 as listing_id, 'TEST2' as listing_test, 'Y' as pass_yn from dual union all
select 2 as listing_id, 'TEST3' as listing_test, 'N' as pass_yn from dual union all
select 3 as listing_id, 'TEST1' as listing_test, 'N' as pass_yn from dual union all
select 3 as listing_id, 'TEST2' as listing_test, 'N' as pass_yn from dual union all
select 3 as listing_id, 'TEST3' as listing_test, 'N' as pass_yn from dual)
select listing_id,
listing_test,pass_yn,
count(*) over (partition by listing_id, pass_yn) as all_y,
count(*) over (partition by listing_id, pass_yn) as all_n
from listing_row
期望的结果
LISTING_ID ALL_Y ALL_N
1 Y N
2 N N
3 N Y
我认为最简单的方法是使用
min()
和max()
:
这使用了一个基于“Y”>N这一事实的技巧。因此,如果取列的
min()
,并且它有任何“N”值,则结果将为“N”。这是使用sum()的另一种解决方案:
select listing_id,
listing_test, pass_yn,
min(pass_yn) over (partition by listing_id) as all_y,
min(case when pass_yn = 'Y' then 'N' else 'Y' end) over (partition by listing_id) as all_n
from listing_row;
SELECT listing_id,
CASE WHEN max(all_test_cnt)-MAX(num_of_test) = 0 THEN 'Y' ELSE 'N' END all_y,
CASE WHEN MAX(num_of_test) = 0 THEN 'Y' ELSE 'N' END all_n
FROM
(SELECT listing_id,
COUNT(DISTINCT listing_test) OVER (PARTITION BY NULL) all_test_cnt,
SUM(CASE WHEN pass_yn = 'Y' THEN 1 ELSE 0 END) OVER (PARTITION BY listing_id)num_of_test
FROM listing_row)
GROUP BY listing_id