如何查找一個月內使用的天數總數？

Question

我將達到一個月內服務使用的總天數。 （開始日期和結束日期是 - 包括兩端）

示例數據1：

User Start_Date  End_Date 
A  01-Jun-2017 30-Jun-2017 
B  06-Jun-2017 30-Jun-2017 

答案：服務使用天= 30天。

示例數據2：

User Start_Date  End_Date 
C  06-Jun-2017 10-Jun-2017 
D  02-Jun-2017 02-Jun-2017 

答案：服務使用天= 6天。

我該如何編寫代碼來查找相同的SQL語言，更適合PLSQL。

Answer 1

測試數據：

CREATE TABLE your_table (usr, start_date, end_date) AS (
    SELECT 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL 
    SELECT 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous 
    SELECT 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL 
    SELECT 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous 
    SELECT 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL 
    SELECT 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous 
    SELECT 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL 
    SELECT 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next 
    SELECT 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL 
    SELECT 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL;   -- Within H and I   

查詢 ：

SELECT SUM(days) AS total_days 
FROM (
    SELECT dt - LAG(dt) OVER (ORDER BY dt) + 1 AS days, 
     start_end 
    FROM (
    SELECT dt, 
      CASE SUM(value) OVER (ORDER BY dt ASC, value DESC, ROWNUM) * value 
      WHEN 1 THEN 'start' 
      WHEN 0 THEN 'end' 
      END AS start_end 
    FROM your_table 
    UNPIVOT (dt FOR value IN (start_date AS 1, end_date AS -1)) 
) 
    WHERE start_end IS NOT NULL 
) 
WHERE start_end = 'end'; 

輸出：

TOTAL_DAYS 
---------- 
     25 

說明：

SELECT dt, value 
FROM your_table 
UNPIVOT (dt FOR value IN (start_date AS 1, end_date AS -1)) 

這將UNPIVOT表，以便開始和結束日期是在同一列（dt）和給出的相應值開始時爲+1，結束日期爲-1。

SELECT dt, 
     SUM(value) OVER (ORDER BY dt ASC, value DESC, ROWNUM) AS total, 
     value 
FROM your_table 
UNPIVOT (dt FOR value IN (start_date AS 1, end_date AS -1)) 

將給出這些生成值的開始和結束日期以及累計總和。範圍的起始位置始終爲value=1和total=1，範圍的末尾始終爲total=0。如果日期在一個範圍的中間，那麼它將有total>1或value=-1和total=1。使用此項，如果乘以value和total，則範圍的開始爲value*total=1，範圍的結尾爲value*total=0，而其他任何值指示範圍中途的日期。

這是什麼這給出：

SELECT dt, 
     CASE SUM(value) OVER (ORDER BY dt ASC, value DESC, ROWNUM) * value 
     WHEN 1 THEN 'start' 
     WHEN 0 THEN 'end' 
     END AS start_end 
FROM your_table 
UNPIVOT (dt FOR value IN (start_date AS 1, end_date AS -1)) 

然後，您可以過濾掉的日期時start_end是NULL這會讓你有一張桌子交替start和end行，您可以使用LAG計算天差數：

SELECT dt - LAG(dt) OVER (ORDER BY dt) + 1 AS days, 
     start_end 
FROM (
    SELECT dt, 
     CASE SUM(value) OVER (ORDER BY dt ASC, value DESC, ROWNUM) * value 
      WHEN 1 THEN 'start' 
      WHEN 0 THEN 'end' 
     END AS start_end 
    FROM your_table 
    UNPIVOT (dt FOR value IN (start_date AS 1, end_date AS -1)) 
) 
WHERE start_end IS NOT NULL 

所有你需要做的就是SUM全部爲end - start的差異;它給出了上面的查詢。

Answer 2

由於@Pravin Satav解決，您的要求，這是不是很清楚，這樣的事情是我從你的描述可以理解：

SELECT sum(CASE WHEN end_date=start_date THEN 1 ELSE (end_date-start_date)+1 END) as total_days 
FROM my_table 
WHERE <conditions that determine your "sample data">;