依賴於前面的行PostgreSQL的列值

Question

我有五列的表格：id，user_code，created_at，A，和B和start。 id僅僅是一個鍵，user_code識別與該行相關聯的用戶，created_at是一個時間戳，A，B，並start可以獨立取值0或1。

我什麼是運行查詢，其將所述列As和Bs，根據created_at，然後給定行的As值哪些訂單原始表以升序等於先前行的具有相同user_code是有A = 1自從上次start = 1（數量，同樣的事情Bs和B）。

樣品輸入：

id | user_code | created_at | A | B | start 
--------------------------------------------- 
00 |  1  |  t0  | 1 | 0 | 1 
01 |  1  |  t1  | 0 | 0 | 1 
02 |  2  |  t1  | 0 | 0 | 1 
03 |  1  |  t2  | 1 | 0 | 0 
04 |  2  |  t2  | 0 | 1 | 0 
05 |  2  |  t3  | 0 | 1 | 0 
06 |  1  |  t4  | 1 | 1 | 0 
07 |  1  |  t5  | 0 | 1 | 1 

輸出示例：

id | user_code | created_at | A | B | start | As | Bs 
------------------------------------------------------- 
00 |  1  |  t0  | 1 | 0 | 1 | 01 | 00 
01 |  1  |  t1  | 0 | 0 | 1 | 00 | 00 
02 |  2  |  t1  | 0 | 0 | 1 | 00 | 00 
03 |  1  |  t2  | 1 | 0 | 0 | 01 | 00 
04 |  2  |  t2  | 0 | 1 | 0 | 00 | 01 
05 |  2  |  t3  | 0 | 1 | 0 | 00 | 02 
06 |  1  |  t4  | 1 | 1 | 0 | 02 | 01 
07 |  1  |  t5  | 0 | 1 | 1 | 00 | 01 
08 |  2  |  t6  | 1 | 0 | 1 | 01 | 00 

Answer 1

您可以使用COUNT與OVER條款：

SELECT "id", "user_code", "created_at", "A", "B", 
     COUNT(CASE WHEN "A" <> 0 THEN 1 END) 
      OVER (PARTITION BY "user_code" ORDER BY "id") AS As, 
     COUNT(CASE WHEN "B" <> 0 THEN 1 END) 
      OVER (PARTITION BY "user_code" ORDER BY "id") AS Bs 
FROM mytable 

Demo here