的Oracle SQL查詢過程

Question

我有這個表中的Oracle：

Process| State | Date ; 
A| 0 | 19/09/14 ; 
A| 1 | 30/09/14 ; 
A| 2 | 31/09/14 

有了你知道該過程已更改爲相應的狀態

我需要獲得一個表像的日期列這

Process | Time between state 0 and 1 | time between state 1 and 2 

Answer 1

這會給你的PROCESS，CHANGE IN STATE和NUMBER OF DAYS：

SQL> WITH DATA AS(
    2 SELECT 'A' PROCESS, 0 STATE, TO_DATE('19/09/14','dd/mm/rr') DT FROM DUAL UNION ALL 
    3 SELECT 'A',   1,  TO_DATE('30/09/14','dd/mm/rr') FROM DUAL UNION ALL 
    4 SELECT 'A',   2,  TO_DATE('01/10/14','dd/mm/rr') FROM DUAL) 
    5 SELECT PROCESS, 
    6 state 
    7 ||' - ' 
    8 ||new_state change_in_state, 
    9 NEW_DT-dt NUMBER_OF_DAYS 
10 FROM 
11 (SELECT A.*, 
12  LEAD(STATE) OVER(ORDER BY STATE, DT) NEW_STATe, 
13  LEAD(DT) OVER(ORDER BY STATE, DT) NEW_DT 
14 FROM DATA A 
15 ) 
16 WHERE NEW_DT-dt IS NOT NULL 
17/

PROCESS CHANGE_IN_ NUMBER_OF_DAYS 
---------- ---------- -------------- 
A   0 - 1    11 
A   1 - 2    1 

Answer 2

假設進程必須通過狀態0進行，將有1作爲下一個狀態，2作爲下一個狀態;這個查詢使用LEAD（）向前看1和2的狀態值（如果有的話）

查詢

SELECT 
     PROCESS 
    , l1STATEDATE - STATEDATE AS TIME_0_1 
    , l2STATEDATE - L1STATEDATE AS TIME_1_2 
FROM (

     SELECT 
       PROCESS 
      , STATE 
      , STATEDATE 
      --, lead(STATE,1)  over(partition by PROCESS ORDER BY STATEDATE) as l1STATE 
      , lead(STATEDATE,1) over(partition by PROCESS ORDER BY STATEDATE) as l1STATEDATE 
      --, lead(STATE,2)  over(partition by PROCESS ORDER BY STATEDATE) as l2STATE 
      , lead(STATEDATE,2) over(partition by PROCESS ORDER BY STATEDATE) as l2STATEDATE 
     FROM STATEINFO 
    ) 
WHERE STATE = 0 

結果：

| PROCESS | TIME_0_1 | TIME_1_2 | 
|---------|----------|----------| 
|  A |  11 |  1 | 

See the SQLFiddle demo

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注意我使用字段名稱「stateDate」，因爲我不推薦使用SQL關鍵字作爲名稱