如果coloumn具有相同的值

时间：2016-09-22 07:30:01

标签： sql sql-server sql-server-2008

美好的一天，我正在尝试选择数据。请先查看我的查询和表格。

我有这个查询

select * from 
    ( 
SELECT *    from 
(
 select distinct TOP 201233 row, b.OutletCode as 'Kode Otlet', 
 i.Description as 'Area', 
 c.Nilai,a.Nip, b.Fullname 'Nama Lengkap', f.positioncode as 'Posisi Sebelumnya', 
 case when a.[status]=0 
    then j.ApprovedDate else p.ApprovedDate 
    end as 'Tanggal Upgrade/Demosi Sebelumnya', 
d.positioncode as 'Posisi Baru', a.tanggal as 'Tanggal Upgrade/Demosi' 
from penilaian_header a 
left join Employee b on a.Nip = b.Nip 
left join Position f on b.PositionCode = f.PositionCode 
left join Position d on a.PositionCode = d.PositionCode 
left join arealeader g on g.OutletCode = b.OutletCode 
left join outlet h on g.OutletCode = h.OutletCode 
left join area i on i.areacode = h.areacode 
left join cutoff k on a.periode = k.cutoffcode 
left join 
    (select ROW_NUMBER() OVER(PARTITION BY KodePenilaianH ORDER BY idPenilaiand ASC) AS Row, 
    Nilai,KodePenilaianH from penilaian_Detail ) c on a.KodePenilaian = c.KodePenilaianH 
left join 
    ( SELECT ROW_NUMBER() OVER (PARTITION BY nip ORDER BY ApprovedDate desc) AS rownumber, 
    ApprovedDate, Nip FROM historyposition ) AS p on a.nip=p.nip and p.rownumber = 2 
    left join ( SELECT ROW_NUMBER() OVER (PARTITION BY nip ORDER BY ApprovedDate desc) AS rownumber, ApprovedDate, Nip 
    FROM historyposition ) AS j on a.nip=j.nip and j.rownumber = 1 
    where a.flag = 1 and h.AreaCode like '%%'
     and Periode like '%CO-2016-9-16-15%' and a.nip = '1004863'
     --and tanggal <= k.[to] and tanggal >= k.[from] 
     order by i.Description asc) nilai pivot ( sum(nilai) for row in ([1],[2],[3],[4],[5]) ) piv)A order by Area;

以上查询我得到了这个结果

enter image description here

我有这张表HistoryPosition enter image description here 

KodeMutasiP OldPosition NewPosition ApprovedBy  ApprovedDate    Nip KodePenilaian
HP0000514       P007    P007         0802678    2016-09-15     1004863  PE0000787
HP0000513       P007    P007         0802678    2016-04-04      1004863 PE0000130
NULL            NULL    P007         NULL       2016-04-04      1004863 NULL
NULL            NULL    P041         NULL       2016-01-20      1004863 NULL
NULL            NULL    P007         NULL       2015-02-12      1004863 NULL

确定。所以我要做的是在第一张图片中设置Posisi Sebelumnya和Tanggal Upgrade/Demosi Sebelumnya，并在第二张图片中设置newposition和Approveddate。

但是有了这个条件 如果newposition和nip col在第二张图片中与第二行相同则跳转到第二行，如果第二张与第三张相同则跳转到第四行，如果第四张与第三张相同则选择第三。

我已经编写了查询选择它，这是我的查询

select * INTO #tMP from historyposition where nip = '1004863' order by approveddate desc

declare @NewPosition varchar(50), @NewPositionLast varchar(50), @ApproveDate datetime, @ApproveDateLast datetime
select top 1 @NewPositionLast = NewPosition, @ApproveDateLast=ApprovedDate from #tmp
WHILE EXISTS (SELECT * FROM #tMP)
BEGIN
    select top 1 @NewPosition = NewPosition, @ApproveDate=ApprovedDate from #tmp
    if (@NewPosition = @NewPositionLast)
    begin 
        set @NewPositionLast = @NewPosition 
        set @ApproveDateLast = @ApproveDate
    end 
    else
    begin
        break
    end
    delete top(1) #tMP
END
select  @NewPositionLast , @ApproveDateLast
drop table #tMP

我得到了这个结果

|  P007 |  2016-04-04 00:00:00.000 |

结果就像我期望的那样。

所以我的问题是如何设置Posisi Sebelumnya和Tanggal Upgrade/Demosi Sebelumnya与我上次查询的结果。

抱歉我的英语不好。

这是我的预期结果。

enter image description here

我知道第一张照片也是如此。但它不对。因为在我的第一个查询中我选择了第二行。

select * from 
( 
SELECT * from 
    ( 
select distinct TOP 201233 row, b.OutletCode as 'Kode Otlet', i.Description as 'Area', 
c.Nilai,a.Nip, b.Fullname 'Nama Lengkap', 
case when 
    l.ShortDesc  IS NULL then f.ShortDesc else  l.ShortDesc 
end
    as 'PosisiSebelumnya',
-----
z.approveddate,
------
d.ShortDesc as 'Posisi Baru', a.tanggal as 'Tanggal Upgrade/Demosi' 
from penilaian_header a 
    left join Employee b on a.Nip = b.Nip 
    left join Position f on b.PositionCode = f.PositionCode 
    left join Position d on a.PositionCode = d.PositionCode 
    left join position l on a.posisisaatini = l.positioncode
    left join arealeader g on g.OutletCode = b.OutletCode 
    left join outlet h on g.OutletCode = h.OutletCode 
    left join area i on i.areacode = h.areacode 
    left join cutoff k on a.periode = k.cutoffcode 
    left join 
    (select ROW_NUMBER() OVER(PARTITION BY KodePenilaianH ORDER BY idPenilaiand ASC) AS Row, 
    Nilai,KodePenilaianH from penilaian_Detail ) c on a.KodePenilaian = c.KodePenilaianH 
    left join (
    select top(1) Nip,NewPosition, ApprovedDate
        from (
            select Nip,NewPosition,ApprovedDate
            , grp = row_number() over(order by ApprovedDate desc) - row_number() over(partition by NewPosition order by ApprovedDate desc)
            from HistoryPosition
        ) t
        where grp = 0
        order by ApprovedDate
    )z on z.Nip = a.Nip
    where a.flag = 1 
    and h.AreaCode like '%%' 
    and Periode like '%CO-2016-9-16-15%' 
    and tanggal <= k.[to] and tanggal >= k.[from] order by i.Description asc) nilai 
    pivot ( sum(nilai) for row in ([1],[2],[3],[4],[5]) ) piv)A order by Area;

我明白了。

enter image description here

2 个答案:

答案 0 :(得分：1)

您的上一个脚本会检索具有最小ApprovedDate的行。它应该与

实际相同 
SELECT ApprovedDate, NewPosition
FROM (
   SELECT ROW_NUMBER() OVER (PARTITION BY NewPosition ORDER BY ApprovedDate asc) AS rownumber, ApprovedDate, NewPosition
   FROM historyposition 
   WHERE nip = '1004863') t
WHERE rownumber = 1

尝试在更大的查询中使用此代码段。

修改
当NewPosition被historyposition点亮时，会检测第一组ApprovedDate desc，然后在第一行中找到min ApprovedDate组。

with HistoryPosition as(
    -- sample data
    select sKodeMutasiP,OldPosition,NewPosition,ApprovedBy
       ,cast(ApprovedDate as DATE) ApprovedDate, Nip,KodePenilaian
    from (
        values
         ('HP0000514','P007','P007',0802678,'2016-09-15',1004863,'PE0000787')
        ,('HP0000513','P007','P007',0802678,'2016-04-04',1004863,'PE0000130')
        ,(NULL,NULL,'P007',NULL,'2016-04-04',1004863,NULL)
        ,(NULL,NULL,'P041',NULL,'2016-01-20',1004863,NULL)
        ,(NULL,NULL,'P007',NULL,'2015-02-12',1004863,NULL)
    ) t (sKodeMutasiP,OldPosition,NewPosition,ApprovedBy,ApprovedDate,Nip,KodePenilaian)
)
select top(1) NewPosition, ApprovedDate
from (
    select NewPosition,ApprovedDate
    , grp = row_number() over(order by ApprovedDate desc) - row_number() over(partition by NewPosition order by ApprovedDate desc)
    from HistoryPosition
    where nip = 1004863
) t
where grp = 0
order by ApprovedDate;

<强> EDIT2 在OUTER APPLY中使用此代码段而不是LEFT JOIN，因为它取决于其他表数据：

...
outer apply(
    select top(1) NewPosition, ApprovedDate
    from (
        select NewPosition,ApprovedDate
        , grp = row_number() over(order by ApprovedDate desc) - row_number() over(partition by NewPosition order by ApprovedDate desc)
        from HistoryPosition hp
        where hp.nip = a.nip -- other table 
    ) t
    where grp = 0
    order by ApprovedDate
    ) npa

...

答案 1 :(得分：0)

如果您的问题是，我如何组合两个结果集以使它们出现在一行中，可以使用以下选择：

-- first select
select 1 one, 2 two, 3 three, 0 joinCondition

-- second select
select 'a' a, 'b' b, 'c' c, 0 joinCondition

-- combined
select one, two, three, a, b, c
from (select 1 one, 2 two, 3 three, 0 joinCondition) first
   inner join (select 'a' a, 'b' b, 'c' c, 0 joinCondition) second
      on first.joinCondition = second.joinCondition

我们的想法是，我们为具有相同值的两个选择添加一个虚拟列，以便可以使用它来加入它们

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