SQL查詢計數與Group By的區別

Question

在序數過程中，有許多位置具有成員處於不同步驟。 成員可以處於多個進程中，在一個或多個位置具有不同的進度。

位置和成員之間的連接表可能看起來像這樣稱為步驟。

| id | member_id | step_no | location_id | process_id |
-------------------------------------------------------
| 1  | 1         | 2       | 10          | 57
| 2  | 1         | 5       | 10          | 58
| 3  | 2         | 5       | 11          | 37
| 4  | 2         | 1       | 10          | 57

我無法確定如何根據位置獲得成員在流程中最遠步驟的唯一成員數。

location_id | 1 | 2 | 3 | 4 | 5 |
---------------------------------
 10         | 1 | 0 | 0 | 0 | 1 |
 11         | 0 | 0 | 0 | 0 | 1 |

到目前為止，我有這個查詢：

SELECT count(DISTINCT m.id), l.id, l.name
      FROM members m
      INNER JOIN steps s ON m.id = s.member_id
      INNER JOIN locations l ON s.location_id = l.id
      WHERE step_no = 5
      GROUP BY l.id
      ORDER BY l.name

但是這當然只返回step_no = 5，如果我寫了五個這樣的查詢，那么成員可以在不同的步驟中被計算兩次。

Answer 1

select 
location_id
,sum(case when max_step = 1 then 1 else 0 end) as step_one_cnt
,sum(case when max_step = 2 then 1 else 0 end) as step_two_cnt
,sum(case when max_step = 3 then 1 else 0 end) as step_three_cnt
,sum(case when max_step = 4 then 1 else 0 end) as step_four_cnt
,sum(case when max_step = 5 then 1 else 0 end) as step_five_cnt
FROM
(select
s.location_id, 
s.member_id, 
max(s.step_no) as max_step
FROM steps S
group by 1,2
) as base
group by 1

打破它，基本查詢會給你以下結果：

member_id |  location_id | max_step_no
-------------------------------------------------------

| 1         | 10       | 5 
| 2         | 10       | 1
| 2         | 11       | 5

此子查詢（基礎）上的聚合查詢基本上將結果轉換為您希望看到的形式。 此方法的唯一限制是步驟數在高級中靜態定義。

Answer 2

1-首先在臨時表中選擇以下內容：

select   location_id , step_no , count(member_id) as count 
into     stepsPrime
from     (select member_id, location_id, max(step_no) as step_no
          from steps 
          group by member_id, location_id ) definedTab 
group by location_id, step_no 
order by location_id;

2-使用以下查詢來旋轉上述結果：

select distinct stp.location_id,
       stp1.count as step1, 
       stp2.count as step2, 
       stp3.count as step3, 
       stp4.count as step4, 
       stp5.count as step5 
from stepsPrime stp 
left join stepsPrime stp1 on stp.location_id = stp1.location_id and stp1.step_no = 1 
left join stepsPrime stp2 on stp.location_id = stp2.location_id and stp2.step_no = 2 
left join stepsPrime stp3 on stp.location_id = stp3.location_id and stp3.step_no = 3 
left join stepsPrime stp4 on stp.location_id = stp4.location_id and stp4.step_no = 4 
left join stepsPrime stp5 on stp.location_id = stp5.location_id and stp5.step_no = 5;