仅在满足另一列的条件时计数

Question

只有满足其他列条件时，我才需要计算ID。 ID不是唯一的，可能包括几个步骤。

表看起来像：

rownum | ID | key   | result

1      |100 | step1 | accepted
2      |100 | step2 | accepted
3      |100 | step3 | transfer
4      |101 | step0 | accepted
5      |101 | step1 | accepted
6      |101 | step2 | rejected
7      |102 | step0 | accepted
8      |102 | step1 | accepted
9      |103 | step1 | rejected
10     |104 | step1 | rejected
11     |104 | step1 | rejected
12     |104 | step1 | rejected

在示例中，我有5个ID（但在实际表中有数千个ID），我必须仅COUNT条符合条件的ID。 条件很简单：键<>'step0'，因此我的COUNT脚本应返回值3。

如果我试试

COUNT ID
FROM myTable
WHERE key <> 'step0'

它返回错误的值，因为WHERE子句在COUNT之前应用

任何想法都赞赏。

Answer 1

这是一个不需要嵌套聚合函数且不需要子查询的方法：

select (count(distinct id) -
        count(distinct case when key = 'step0' then id end)
       )
from mytable;

Answer 2

尝试使用不存在的相关子查询

select count(distinct ID)
from tablename a 
   where not exists (select 1 from tablename b where a.id=b.id and key = 'step0')

Answer 3

使用不同

select COUNT (distinct ID)
FROM myTable
WHERE ID not in ( select id from myTable where key = 'step0' and id is not null)

Answer 4

使用带有having子句的分组

select sum(count(distinct ID)) as "Count"
  from myTable
 group by ID
 having sum(case when key = 'step0' then 1 else 0 end)=0;
-- or "having sum( decode(key,'step0',1,0) ) = 0" is also possible specific to Oracle

Count
-----
  3

例如使用反向逻辑，来自key = 'step0'

演示

Answer 5

这应该工作：

SELECT COUNT(COUNT(DISTINCT id)) num_ids
FROM   your_table
GROUP BY id
HAVING MIN(CASE WHEN key = 'step0' THEN key END) IS NULL;

外部COUNT适用于整个查询，包括COUNT(DISTINCT id)的聚合 - 删除它，您将看到三行值为1。