Question

我需要获取TABLE_A中的所有记录，其中至少 2 最后Status是空的（相对于Inspection_Date）和{{1} } Room_ID中不存在。

这是我用作示例的简化表：

TABLE_A ：

TABLE_B

表-B ：

  Room_Id   Status    Inspection_Date
  -------------------------------------
    1        vacant      5/15/2015
    2        occupied    5/21/2015
    2        vacant      1/19/2016
    1        occupied   12/16/2015
    4        vacant      3/25/2016
    3        vacant      8/27/2015
    1        vacant      4/17/2016
    3        vacant     12/12/2015
    3        vacant      3/22/2016
    4        occupied    2/2/2015
    4        vacant      3/24/2015

我的结果应如下所示：

  Room_Id   Status    Inspection_Date
  ------------------------------------
    1        vacant       5/15/2015
    2        occupied     5/21/2015
    2        vacant       1/19/2016
    1        vacant      12/16/2015
    1        vacant       4/17/2016

我试过这种方式，它适用于示例，但不能处理我的数据..逻辑不完整：

   Room_Id  Status  Inspection_Date
   ---------------------------------
    3       vacant      8/27/2015
    3       vacant     12/12/2015
    3       vacant      3/22/2016
    4       occupied    2/2/2015
    4       vacant      3/24/2015
    4       vacant      3/25/2016

这是架构：

 With cteA As
(
Select *, Row_Number() Over (Partition By Room_ID, Status Order By     Inspection_Date Desc) RowNum From Table_A 
)
Select * From Table_A Where Room_Id In
(
Select Room_Id 
    From cteA
    Where Room_Id Not In (Select Room_Id From Table_B) 
        And Status = 'vacant' And RowNum > 1 
)
    Order By Room_Id, Inspection_Date

Answer 1

<强> PLAIN

对于TABLE_A中的每个房间，选择最后一个日期（作为lastDate）
TABLE_A中的每个房间选择上一个日期（作为prevLastDate）
从lastDate获取room_ids，状态为'vacant'（作为lastDateVacant）
从prevLastDate获取room_ids，其状态为'空置'（as prevLastDateVacant）
过滤TABLE_A只包含lastDateVacant和prevLastDateVacant（内部）中的ID
过滤TABLE_A，使其只有不在TABLE_B中的ID（左外+ IS NULL）

结果你有：

WITH lastDate AS (
    SELECT room_id AS room,MAX(inspection_date) AS date
    FROM "TABLE_A"
    GROUP BY room_id
), prevLastDate AS (
    SELECT room_id AS room,MAX(inspection_date) AS date
    FROM "TABLE_A" a
    INNER JOIN lastDate ON a.room_id = lastDate.room and a.inspection_date < lastDate.date
    GROUP BY room_id
), lastDateVacant AS (
    SELECT room_id AS room FROM "TABLE_A"
    WHERE (room_id,inspection_date) IN (
        SELECT room, date FROM lastDate
    ) AND status = 'vacant' 
), prevLastDateVacant AS (
    SELECT room_id AS room FROM "TABLE_A"
    WHERE (room_id,inspection_date) IN (
        SELECT room, date FROM prevLastDate
    ) AND status = 'vacant' 
)

SELECT a.* FROM "TABLE_A" a 
INNER JOIN lastDateVacant 
    ON a.room_id = lastDateVacant.room
INNER JOIN prevLastDateVacant 
    ON a.room_id = prevLastDateVacant.room
LEFT OUTER JOIN "TABLE_B" AS b 
    ON a.room_id = b.room_id    
WHERE b.room_id IS NULL 
ORDER BY a.room_id ASC, a.inspection_date DESC

窗口功能

不确定TSQL的语法是否相同，但这是较短的变体：

按房间划分和按日期排序
检查排名为1和2的ID为“空置”状态的ID，按ID分组并多次出现

带房间AS（选择房间（选择room_id作为房间，状态，inspection_date作为日期， RANK（）OVER（由room_id ORDER BY inspection_date DESC划分）作为排名来自“TABLE_A” ） where（排名在（1,2）和status ='vacant'）按房间分组有计数（）＆gt; 1 ）选择a。 FROM“TABLE_A”a INNER JOIN会议室在a.room_id = room.room LEFT OUTER JOIN“TABLE_B”AS b 在a.room_id = b.room_id
在哪里b.room_id是空的 ORDER BY a.room_id ASC，a.inspection_date DESC

Answer 2

您的条件几乎直接转换为查询。您可以使用窗口函数进行空置计数，将not exists用于与table_b的关系：

select a.*
from (select a.*,
             sum(case when status = 'vacant' then 1 else 0 end) over (partition by room_id) as num_vacant
      from table_a a
      where not exists (select 1
                        from table_b b
                        where b.room_id = a.room_id
                       )
    ) a
where num_vacant >= 2;

编辑：

如果你想让最后两个空缺，你可以找到最后一个非空的记录，然后计算那些大于那个的记录：

select a.*
from (select a.*,
             sum(case when a2.max_nonvacant > a.inspection_date then 0 else 1) over (partition by room_id) as num_vacant_last
      from table_a a outer apply
           (select max(inspection_date) as max_nonvacant
            from table_a a2
            where a2.room_id = a.room_id and a2.status <> 'vacant'
           ) a2
      where not exists (select 1
                        from table_b b
                        where b.room_id = a.room_id
                       )
    ) a
where num_vacant_last >= 2;

Answer 3

我做了这个测试：提取考虑到与inspection_date（降序）相关的最后两个状态（相等状态）的所有room_id：

//example function which receives the sent data as well
function exampleLoggingCallback(sentData, receivedData, textStatus, jqXhr) {
    console.log('Sent data: ' + sentData);
    console.log('Received data: ' + receivedData);
}

//example call to your ajax function:
yourCustomAjaxFunction({foo: 'bar'}, exampleLoggingCallback);

Answer 4

这对我有用，我一次又一次地检查过。

with Rooms as (
select
    Room_Id, Status,
    row_number() over (partition by Room_Id order by Inspection_Date desc) as rn
from TABLE_A
), Candidates as (
select Room_Id from Rooms group by Room_Id
having sum(case when rn in (1, 2) and Status = 'vacant' then 1 else null end) = 2
)
select * from TABLE_A
where Room_Id in (select Room_Id from Candidates except select Room_Id from TABLE_B)
order by Room_Id, Inspection_Date desc

TSQL：给定分区的最高记录（有条件）

4 个答案: