sql最大/最小查询和数据转换

Question

更新：更改一次以显示每批货物的时间可能不总是按顺序排列。

这里是我的输入

create table test 
(
    shipment_id int, 
    stop_seq tinyint, 
    time datetime 
) 

insert into test values (1,1,'2009-8-10 8:00:00') 
insert into test values (1,2,'2009-8-10 9:00:00') 
insert into test values (1,3,'2009-8-10 10:00:00') 
insert into test values (2,1,'2009-8-10 13:00:00') 
insert into test values (2,2,'2009-8-10 14:00:00') 
insert into test values (2,3,'2009-8-10 20:00:00') 
insert into test values (2,4,'2009-8-10 18:00:00') 

，我想低于

shipment_id start end 
----------- ----- --- 
    1  8:00 10:00 
    2  13:00 18:00 

我需要从max(stop)排走的时候从min(stop)行每次装运货物的时间和输出分别放置在开始/结束处。我知道这可以很容易地完成多个查询，但我期待看看如果一个选择查询可以做到这一点。

谢谢！

Answer 1

我认为你能做到的唯一方法就是使用子查询。

SELECT shipment_id 
    , (SELECT TOP 1 time 
     FROM test AS [b] 
     WHERE b.shipment_id = a.shipment_id 
     AND b.stop_seq = MIN(a.stop_seq)) AS [start] 
    , (SELECT TOP 1 time 
     FROM test AS [b] 
     WHERE b.shipment_id = a.shipment_id 
     AND b.stop_seq = MAX(a.stop_seq)) AS [end] 
FROM test AS [a] 
GROUP BY shipment_id 

您将需要使用DATEPART函数来削减时间列以获得确切的输出。

Answer 2

难道我的思维纠正你想要的第一时间，而不是在“分钟”时间，最后的时间序列，而不是“最大”时间？

Answer 3

SELECT C.shipment_id, C.start, B2.time AS stop FROM 
( 
    SELECT A.shipment_id, B1.time AS start, A.max_stop_seq FROM 
    (
     SELECT shipment_id, MIN(stop_seq) as min_stop_seq, MAX(stop_seq) as max_stop_seq 
     FROM test 
     GROUP BY shipment_id 
    ) AS A 

    INNER JOIN 

    (
     SELECT shipment_id, stop_seq, time FROM test 
    ) AS B1 

    ON A.shipment_id = B1.shipment_id AND A.min_stop_seq = B1.stop_seq 
) AS C 

INNER JOIN 

(
    SELECT shipment_id, stop_seq, time FROM test 
) AS B2 

ON C.shipment_id = B2.shipment_id AND C.max_stop_seq = B2.stop_seq 

Answer 4

使用公用表表达式（CTE） - 这个工程（至少在我的SQL Server 2008的测试系统）：

WITH SeqMinMax(SeqID, MinID, MaxID) AS 
(
    SELECT Shipment_ID, MIN(stop_seq), MAX(stop_seq) 
    FROM test 
    GROUP BY Shipment_ID 
) 
SELECT 
    SeqID 'Shipment_ID', 
    (SELECT TIME FROM test 
     WHERE shipment_id = smm.seqid AND stop_seq = smm.minid) 'Start', 
    (SELECT TIME FROM test 
     WHERE shipment_id = smm.seqid AND stop_seq = smm.maxid) 'End' 
FROM seqminmax smm 

的SeqMinMax CTE选择最小值和最大值 “stop_seq” 值每个“shipment_id”，然后查询的其余部分建立在这些值上以从表“检索”中检索关联的时间。

CTE在SQL Server 2005上受支持（并且是SQL：2003标准功能 - 实际上没有Microsoft“发明”）。

马克

Answer 5

select t1.shipment_id, t1.time start, t2.time [end] 
from (
    select shipment_id, min(stop_seq) min, max(stop_seq) max 
    from test 
    group by shipment_id 
) a 
inner join test t1 on a.shipment_id = t1.shipment_id and a.min = t1.stop_seq 
inner join test t2 on a.shipment_id = t2.shipment_id and a.max = t2.stop_seq 

Answer 6

我建议你把ROW_NUMBER和枢纽的优势。这可能看起来很乱，但我认为它会表现良好，并且更适应各种假设。例如，它不假定最新的日期时间值对应给定货件的最大stop_seq值。

with test_ranked(shipment_id,stop_seq,time,rankup,rankdown) as (
    select 
    shipment_id, stop_seq, time, 
    row_number() over (
     partition by shipment_id 
     order by stop_seq 
    ), 
    row_number() over (
     partition by shipment_id 
     order by stop_seq desc 
    ) 
    from test 
), test_extreme_times(shipment_id,tag,time) as (
    select 
    shipment_id, 'start', time 
    from test_ranked where rankup = 1 
    union all 
    select 
    shipment_id, 'end', time 
    from test_ranked where rankdown = 1 
) 
    select 
    shipment_id, [start], [end] 
    from test_extreme_times 
    pivot (max(time) for tag in ([start],[end])) P 
    order by shipment_id; 
    go 

PIVOT并不是真的需要，但它很方便。但是，请注意PIVOT表达式中的MAX没有任何用处。每个标签只有一个[时间]值，因此MIN也会起作用。该语法在这个位置需要一个聚合函数。

附录：这里的，如果你有一个出货量桌子，其可以比使用MIN和MAX更有效CptSkippy的解决方案的适应性：

SELECT shipment_id 
    , (SELECT TOP 1 time 
     FROM test AS [b] 
     WHERE b.shipment_id = a.shipment_id 
     ORDER BY stop_seq ASC) AS [start] 
    , (SELECT TOP 1 time 
     FROM test AS [b] 
     WHERE b.shipment_id = a.shipment_id 
     ORDER BY stop_seq DESC) AS [end] 
FROM shipments_table AS [a];