按组 SQL 查询从上一行中减去日期

Question

数据

id      date        
2380    10/30/12 09:00:00 
2380    10/30/12 09:05:00   
2380    10/30/12 09:10:00   
2380    10/30/12 09:15:00    
2381    10/30/12 10:00:00   
2381    10/30/12 10:05:00  
2381    10/30/12 10:10:00   
2381    10/30/12 10:15:00   
2382    10/30/12 11:00:00
2382    10/30/12 11:05:00
2382    10/30/12 10:10:00
2382    10/30/12 10:15:00

我想要以下解决方案

id      date                 duration        
2380    10/30/12 09:00:00    00:00:00 
2380    10/30/12 09:05:00    00:05:00   
2380    10/30/12 09:10:00    00:10:00
2380    10/30/12 09:15:00    00:15:00
2381    10/30/12 10:00:00    00:00:00
2381    10/30/12 10:05:00    00:05:00
2381    10/30/12 10:10:00    00:10:00
2381    10/30/12 10:15:00    00:15:00
2382    10/30/12 11:00:00    00:00:00
2382    10/30/12 11:05:00    00:05:00
2382    10/30/12 10:10:00    00:10:00
2382    10/30/12 10:15:00    00:10:00

我试图理解以下线程背后的逻辑，但很难理解。

Substract date from previous row by group (using R)

select id, date, date - (select min(date) from date group by id) as duration 
from date

我得到的最接近的是一个 id。

Answer 1

试试下面这个例子，希望这是你想要的输出，

declare @t1 table
(
    id int,
    dtdate datetime
)

insert into @t1 values(2380,'10/30/12 09:00:00') 
insert into @t1 values(2380,'10/30/12 09:05:00')   
insert into @t1 values(2380,'10/30/12 09:10:00')   
insert into @t1 values(2380,'10/30/12 09:15:00')    
insert into @t1 values(2381,'10/30/12 10:00:00')   
insert into @t1 values(2381,'10/30/12 10:05:00')  
insert into @t1 values(2381,'10/30/12 10:10:00')   
insert into @t1 values(2381,'10/30/12 10:15:00')   
insert into @t1 values(2382,'10/30/12 11:00:00')
insert into @t1 values(2382,'10/30/12 11:05:00')
insert into @t1 values(2382,'10/30/12 10:10:00')
insert into @t1 values(2382,'10/30/12 10:15:00')

;WITH CTE AS (
SELECT
rownum = ROW_NUMBER() OVER (partition by id ORDER BY id,dtDate),
id,dtDate
FROM @t1 p
)
SELECT
a.id,
a.dtDate,
CASE WHEN prev.dtdate is NULL THEN '00:00:00' ELSE convert(nvarchar(8),a.dtdate- prev.dtdate,108) END as duration 
FROM CTE a
LEFT JOIN CTE prev ON a.id = prev.id AND prev.rownum = a.rownum - 1

Answer 2

我的方法的关键是找到每个 id 的最小日期值，我称之为 ReferenceDate。然后我加入主表并使用 DATEDIFF() 函数进行日期数学运算，并使用样式 108 使用 CONVERT() 函数将结果转换为 hh:mi:ss强>。这是dbfiddle。

IF OBJECT_ID('tempdb.dbo.#MyTable', 'U') IS NOT NULL
    DROP TABLE #MyTable;

CREATE TABLE #MyTable
(
    id INTEGER NOT NULL
  , date DATETIME NOT NULL
);
INSERT INTO #MyTable (id, date) VALUES (2380, '10/30/12 09:00:00');
INSERT INTO #MyTable (id, date) VALUES (2380, '10/30/12 09:05:00');
INSERT INTO #MyTable (id, date) VALUES (2380, '10/30/12 09:10:00');
INSERT INTO #MyTable (id, date) VALUES (2380, '10/30/12 09:15:00');
INSERT INTO #MyTable (id, date) VALUES (2381, '10/30/12 10:00:00');
INSERT INTO #MyTable (id, date) VALUES (2381, '10/30/12 10:05:00');
INSERT INTO #MyTable (id, date) VALUES (2381, '10/30/12 10:10:00');
INSERT INTO #MyTable (id, date) VALUES (2381, '10/30/12 10:15:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 11:00:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 11:05:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 10:10:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 10:15:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 12:15:00');
INSERT INTO #MyTable (id, date) VALUES (2382, '10/30/12 10:15:30');

SELECT     a.*
         , CONVERT(NVARCHAR(8), a.date - b.ReferenceDate, 108) AS duration
FROM       #MyTable AS a
INNER JOIN (
    SELECT id, MIN(date) AS ReferenceDate 
    FROM #MyTable GROUP BY id) AS b ON a.id = b.id;