根据分组中特定列的最大值从一行中选择一条信息？答案

【问题标题】：Select an information from a row based on the maximum value of specific column in a group by?根据分组中特定列的最大值从一行中选择一条信息？
【发布时间】：2017-12-06 22:17:45
【问题描述】：

我有这个数据


----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Id| Date_Opera | Emitter    |   EmitterIBAN                         |  Receiver    |   ReceiverIBAN                           |         Adresss                          |     Value 
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1, | 2017-07-07 | Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point         0002,        121.72
2, | 2017-09-27 | Keene,         SK81 1004 7484 7505 6308 9259,        Torrance,       RO23 ZWTR OJKK VAU9 T5P4 2GDY,                  35197 Green Ridge Way,                   82.52
3, | 2017-10-17 | Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point         0048,        51.81
4, | 2017-05-01 | Korie,         ME43 9833 9830 7367 4239 60,Roy,      IL69            9686 1536 8102 2219 165,                        5 Swallow Alley,                         88.01
5, | 2017-11-17 | Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point         0001,        133.99
6, | 2017-10-10 | Charmine,      BG92 TOXX 8380 785I JKRQ JS,          Sarette,        MU67 RYRU 9293 5875 6859 7111 075X HR,          8 Sage Place,                            36.30
7, | 2017-07-18 | Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point         0004,        186.99

我想在下面得到这样的结果

计算一对 EmitterIBAN 和 ReceiverIBAN 进行的操作数
计算每对 EmitterIBAN 和 ReceiverIBAN 的总和值
并通过取最大地址值按可以不同的地址分组


-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
sum| Date_Opera | Emitter    |   EmitterIBAN                         |  Receiver    |   ReceiverIBAN                           |         Adresss                          |     SumValue 
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
4, | 2017-11-17 |  Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point           0048,      494,51
1, | 2017-09-27 |  Keene,         SK81 1004 7484 7505 6308 9259,        Torrance,       RO23 ZWTR OJKK VAU9 T5P4 2GDY,                  35197 Green Ridge Way,                   82.52
1, | 2017-05-01 |  Korie,         ME43 9833 9830 7367 4239 60,Roy,      IL69            9686 1536 8102 2219 165,                        5 Swallow Alley,                         88.01
1, | 2017-10-10 |  Charmine,      BG92 TOXX 8380 785I JKRQ JS,          Sarette,        MU67 RYRU 9293 5875 6859 7111 075X HR,          8 Sage Place,                            36.30

所以为了得到这个结果，我使用了这个请求

Select  count(1) as NumberOperation, 
        MAX(Emitter) as EmitterName, 
        EmitterIban, 
        MAX(Receiver) as ReceiverName, 
        ReceiverIban,
        MAX(ReceiverAddress) as ReceiverAddress,
        SUM([Value]) as SumValues
FROM TableEsperadoceTransaction
Group By EmitterIban,
         ReceiverIban

但是现在，我想要的是，而不是像之前的示例那样获取最大地址，我想从具有最大数据时间操作的记录中获取地址。这是我的数据结果的示例


-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
sum| Date_Opera | Emitter    |   EmitterIBAN                         |  Receiver    |   ReceiverIBAN                           |         Adresss                          |     SumValue 
----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
4, | 2017-11-17 |  Ernst,         HR53 8827 2118 4692 8207 5,           Kimbra,         CH20 1042 6T0N MDTG JT47 U,                     3256 Arrowood Point           0002,      494,51
1, | 2017-09-27 |  Keene,         SK81 1004 7484 7505 6308 9259,        Torrance,       RO23 ZWTR OJKK VAU9 T5P4 2GDY,                  35197 Green Ridge Way,                   82.52
1, | 2017-05-01 |  Korie,         ME43 9833 9830 7367 4239 60,Roy,      IL69            9686 1536 8102 2219 165,                        5 Swallow Alley,                         88.01
1, | 2017-10-10 |  Charmine,      BG92 TOXX 8380 785I JKRQ JS,          Sarette,        MU67 RYRU 9293 5875 6859 7111 075X HR,          8 Sage Place,                            36.30

所以我的问题是我该如何提出这样的要求？

PS：我有 2.4 亿条记录

编辑：我有 3 个索引

日期操作
EmitterIban
ReceiverIban

【问题讨论】：

这张表有索引吗？
是的@JohnHC 我在 EmitterIBAN 和 ReceiverIBAN 中有两个索引

标签： sql sql-server tsql sql-server-2005

【解决方案1】：

你可以试试这样的：

Select  count(1) as NumberOperation, 
        MAX(t.Emitter) as EmitterName, 
        t.EmitterIban, 
        MAX(t.Receiver) as ReceiverName, 
        t.ReceiverIban,
        (SELECT TOP 1 x.RecieverAddress 
         FROM TableEsperadoceTransaction AS x 
         WHERE x.EmitterIban=t.EmitterIban AND x.RecieverIban=t.RecieverIban
         ORDER BY Data_Opera DESC) as ReceiverAddress,
        SUM(t.[Value]) as SumValues
FROM TableEsperadoceTransaction AS t
Group By t.EmitterIban,
         t.ReceiverIban;

我将您的MAX(Address) 替换为获取最高地址的子选择，按Data_Opera 排序，条件相同...

顺便说一句：在日期列上放置索引会有所帮助...

更新：这可能会更快...

Select  count(1) as NumberOperation, 
        MAX(t.Emitter) as EmitterName, 
        t.EmitterIban, 
        MAX(t.Receiver) as ReceiverName, 
        t.ReceiverIban,
        (SELECT TOP 1 x.RecieverAddress 
         FROM TableEsperadoceTransaction AS x 
         WHERE x.EmitterIban=t.EmitterIban 
           AND x.RecieverIban=t.RecieverIban
           AND x.Data_Opera=MAX(t.Data_Opera)) as ReceiverAddress,
        SUM(t.[Value]) as SumValues
FROM TableEsperadoceTransaction AS t
Group By t.EmitterIban,
         t.ReceiverIban;

GROUP BY 将允许您直接获取MAX(t.Data_Opera)。使用 三列索引，您应该可以非常快速地获得地址值。

【讨论】：

有了子查询，这不是要命中每一行的索引吗？
我认为引擎将足够聪明，可以内联...您使用ROW_NUMBER() 的方法可能会产生一些开销，因为您必须选择每一行，对它们进行编号并抛出大部分他们又走了。拥有一个包含所有三列和Data_Opera DESC 的索引可能会有所帮助。因此引擎可以在任何情况下选择第一个拟合值......
@Shnugo 我有 3 个索引 Date_Operation、EmitterIban、ReceiverIban 我应该用 3 列创建索引吗？
检查一下，引擎非常智能，结构如此简单（很多，但很简单）。按照上面的建议添加 多列索引 可能会有所帮助...
@Shnugo 我今天早上开始执行请求，我正在等待结果。在我的用例中，我必须从具有相同结构的 4 个表中选择和插入，所以从执行开始到现在已经过去了 6 个小时，这就是为什么它花了很多时间。我想我明天会尝试新版本。我没有什么问题，我对 4 列进行分组，为这 4 列创建索引会更有趣吗？

【解决方案2】：

我认为你应该使用窗口函数（SQL 2012+）：

Select  count(1) as NumberOperation, 
        MAX(t.Emitter) as EmitterName, 
        t.EmitterIban, 
        MAX(t.Receiver) as ReceiverName, 
        t.ReceiverIban,
        FIRST_VALUE(x.RecieverAddress) OVER (PARTITION BY t.EmitterIban, t.ReceiverIban ORDER BY Data_Opera DESC),
        SUM(t.[Value]) as SumValues
FROM TableEsperadoceTransaction AS t
Group By t.EmitterIban,
         t.ReceiverIban;

【讨论】：

很好的解决方案 (+1)，但值得一提的是，这个 FIRST_VALUE() OVER(...) 在 v2012 之前不可用。

【解决方案3】：

我在 CTE 中使用了 row_number()，自加入聚合：

with CTE as
(
select t1.*, row_number() over(partition by EmitterIban, ReceiverIban order by Date_Opera desc)  as rn
from TableEsperadoceTransaction t1
)

select a1.EmitterIban,a1.emitter as EName, 
       a1.ReceiverIban, a1.receiver as RName,
       a1.ReceiverAddress
       max(a2.rn) as NumberOperation,
       sum(a2.value) as SumValues
from CTE a1
inner join CTE a2
on a1.EmitterIban = a2.EmitterIban
and a1.ReceiverIban = a2.ReceiverIban
where a1.rn = 1
group by a1.EmitterIban,a1.emitter, 
         a1.ReceiverIban, a1.receiver,
         a1.ReceiverAddress

【讨论】：