僅在滿足另一列的條件時計數

Question

只有滿足其他列條件時，我才需要計算ID。 ID不是唯一的，可能包括幾個步驟。

表看起來像：

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個ID（但在實際表中有數千個ID），我必須僅COUNT條符合條件的ID。 條件很簡單：鍵<>'step0'，因此我的COUNT腳本應返回值3。

如果我試試

COUNT ID
FROM myTable
WHERE key <> 'step0'

它返回錯誤的值，因為WHERE子句在COUNT之前應用

任何想法都贊賞。

Answer 1

這是一個不需要嵌套聚合函數且不需要子查詢的方法：

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

Answer 2

嘗試使用不存在的相關子查詢

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

使用帶有having子句的分組

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

例如使用反向邏輯，來自key = 'step0'

演示

Answer 5

這應該工作：

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適用於整個查詢，包括COUNT(DISTINCT id)的聚合 - 刪除它，您將看到三行值為1。